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Space, shape and measurement: Use the Cartesian co-ordinate system to derive and apply equations

# Unit 1: Determine the equation of a circle with any centre

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Find the equation of a circle centred at the origin.
• Find the equation of a circle with centre $\scriptsize (a,b)$.
• Write the equation of the circle in standard form.

## What you should know

Before you start this unit, make sure you can:

• State the theorem of Pythagoras for right-angled triangles. Refer to level 2 subject outcome 3.3 unit 2 if you need help with this.
• Calculate the distance between two points on the Cartesian plane. Refer to level 2 subject outcome 3.3 unit 2 if you need help with this.
• Calculate the midpoint of a line segment between two points on the Cartesian plane. Refer to level 2 subject outcome 3.3 unit 2 if you need help with this.
• Use the Cartesian coordinate system to derive the equation of a straight line through two given points using the gradient-point form or the gradient-intercept form. Refer to level 3 subject outcome 3.2 unit 1 if you need help with this.
• Complete the square of a quadratic expression. Refer to level 3 subject outcome 2.2 unit 3 if you need help with this.
• Find the coordinates of a point $\scriptsize (x,y)$ after it has been reflected about the axes and lines $\scriptsize y=x$ and $\scriptsize y=-x$. Refer to level 2 subject outcome 3.4 unit 1 if you need help with this.

## Introduction

The circle is one of the most common and important geometric shapes in mathematics. It is, after all, what defines one of the most famous numbers in the world, $\scriptsize \pi$ (pi). $\scriptsize \pi$, remember, is the ratio of a circle’s circumference to its diameter (see figure 1) and it pops up in all sorts of interesting and useful places.

The circle, and in particular the unit circle (circle with a radius of $\scriptsize 1$ centred on the origin), is also foundational to trigonometry which is a key link between algebra and geometry.

So, I think you will agree, the circle is important.

However, we did not deal with it when examining functions, because it is not a function. The circle fails the basic test of a function – the vertical line test. Draw any circle on the Cartesian plane and it is easy to see why. Except for two points on each side of the circle, a vertical line cuts the circle twice (see figure 2). Therefore, almost all of the input values correspond to more than one output value.

But just because the circle is not a function does not mean that there is not a relation that relates all the input and output pairs that together make up the circle. In this unit we will discover what this equation is.

## The equation of a circle with centre the origin

By definition, a circle is a flat shape consisting of all the points in a plane that are an equal distance away from its centre. In other words, every point on the circumference of a circle is the same distance (equidistant) from its centre.

You might already know what the equation is for a circle centred on the origin. If not, complete activity 1.1 to find out.

### Activity 1.1: Find the equation of a circle

Time required: 15 minutes

What you need:

• a pen or pencil
• a blank piece of paper or graph paper

What to do:

For this activity, you will need to draw a system of axes. Use a scale of $\scriptsize 1\ \text{cm}=1\ \text{unit}$ and make sure your axes extend about $\scriptsize 5$ units in all directions.

1. Plot the following points on your Cartesian plane:
$\scriptsize O(0,0)$, $\scriptsize A(5,0)$, $\scriptsize B(4,3)$, $\scriptsize C(3,4)$, $\scriptsize D(0,5)$, $\scriptsize E(-3,4)$, $\scriptsize F(-4,3)$, $\scriptsize G(-5,0)$, $\scriptsize H(-4,-3)$, $\scriptsize \displaystyle I(-3,-4)$ $\scriptsize J(0,-5)$, $\scriptsize K(3,-4)$, $\scriptsize L(4,-3)$
2. Determine the following distances using the distance formula $\scriptsize d=\sqrt{{{{{({{x}_{2}}-{{x}_{1}})}}^{2}}+{{{({{y}_{2}}-{{y}_{1}})}}^{2}}}}$:
$\scriptsize OA$, $\scriptsize OB$, $\scriptsize OC$, $\scriptsize OD$, $\scriptsize OE$, $\scriptsize OF$, $\scriptsize OG$, $\scriptsize OH$, $\scriptsize OI$, $\scriptsize OJ$, $\scriptsize OK$, $\scriptsize OL$
3. What do you notice about the length of each line segment? What general term could we give to these line segments?
4. What object do these points form if you join them up with a smooth curve?
5. If the point $\scriptsize P(x,y)$ lies on this shape, use the distance formula to determine an expression for the length of $\scriptsize OP$.
6. Can you deduce an equation for a circle centre the origin?

What did you find?

1. Plot the following points on your Cartesian plane:
2. $\scriptsize OA=\sqrt{{{{{(5-0)}}^{2}}+{{{(0-0)}}^{2}}}}=\sqrt{{25}}=5$
$\scriptsize OB=\sqrt{{{{{(4-0)}}^{2}}+{{{(3-0)}}^{2}}}}=\sqrt{{16+9}}=\sqrt{{25}}=5$
$\scriptsize OC=\sqrt{{{{{(3-0)}}^{2}}+{{{(4-0)}}^{2}}}}=\sqrt{{9+16}}=\sqrt{{25}}=5$
$\scriptsize OD=\sqrt{{{{{(0-0)}}^{2}}+{{{(5-0)}}^{2}}}}\sqrt{{25}}=5$
$\scriptsize OE=\sqrt{{{{{(-3-0)}}^{2}}+{{{(4-0)}}^{2}}}}=\sqrt{{9+16}}=\sqrt{{25}}=5$
$\scriptsize OF=\sqrt{{{{{(-4-0)}}^{2}}+{{{(3-0)}}^{2}}}}=\sqrt{{16+9}}=\sqrt{{25}}=5$
$\scriptsize OG=\sqrt{{{{{(-5-0)}}^{2}}+{{{(0-0)}}^{2}}}}=\sqrt{{25}}=5$
$\scriptsize OH=\sqrt{{{{{(-4-0)}}^{2}}+{{{(-3-0)}}^{2}}}}=\sqrt{{16+9}}=\sqrt{{25}}=5$
$\scriptsize OI=\sqrt{{{{{(-3-0)}}^{2}}+{{{(-4-0)}}^{2}}}}=\sqrt{{9+16}}=\sqrt{{25}}=5$
$\scriptsize OJ=\sqrt{{{{{(0-0)}}^{2}}+{{{(-5-0)}}^{2}}}}=\sqrt{{25}}=5$
$\scriptsize OK=\sqrt{{{{{(3-0)}}^{2}}+{{{(-4-0)}}^{2}}}}=\sqrt{{9+16}}=\sqrt{{25}}=5$
$\scriptsize OL=\sqrt{{{{{(4-0)}}^{2}}+{{{(-3-0)}}^{2}}}}=\sqrt{{16+9}}=\sqrt{{25}}=5$
3. Each of the line segments is the same length ($\scriptsize 5\ \text{units}$). We could call these line segments the radius of a circle.
4. They all form a circle.
5. $\scriptsize OP=\sqrt{{{{{(x-0)}}^{2}}+{{{(y-0)}}^{2}}}}=\sqrt{{{{x}^{2}}+{{y}^{2}}}}$.
6. Since $\scriptsize OP$ is the radius ($\scriptsize r$) of the circle:
$\scriptsize r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}$ or $\scriptsize {{r}^{2}}={{x}^{2}}+{{y}^{2}}$
We have the equation of a circle centre the origin.

In activity 1.1 we discovered what the equation of a circle centre the origin is. This equation is based on the theorem of Pythagoras (see figure 3).
In $\scriptsize \Delta OPQ$:
\scriptsize \begin{align*}O{{P}^{2}}&=O{{Q}^{2}}+P{{Q}^{2}}\quad \text{(Pythagoras)}\\\text{But }OP&=r\text{ and }OQ=x-0\text{ and }PQ=y-0\\\therefore {{r}^{2}}&={{(x-0)}^{2}}+{{(y-0)}^{2}}\\\therefore {{r}^{2}}&={{x}^{2}}+{{y}^{2}}\end{align*}

Equation of a circle centre the origin:
$\scriptsize {{x}^{2}}+{{y}^{2}}={{r}^{2}}$

### Example 1.1

A circle with centre the origin has a radius of $\scriptsize 3\text{ units}$.

1. Determine the equation of the circle.
2. Sketch the circle on the Cartesian plane.
3. Show that the point $\scriptsize (-\sqrt{4},\sqrt{5})$ lies on the circle.

Solutions

1. The circle has its centre the origin. Therefore, its equation is $\scriptsize {{x}^{2}}+{{y}^{2}}={{r}^{2}}$. We are told that the radius is $\scriptsize 3\text{ units}$. Therefore, the equation is $\scriptsize {{x}^{2}}+{{y}^{2}}={{(3)}^{2}}$ or $\scriptsize {{x}^{2}}+{{y}^{2}}=9$.
2. We know that the centre is the origin and the radius is $\scriptsize 3\text{ units}$. Therefore, the circle will pass through the points $\scriptsize (3,0)$, $\scriptsize (0,3)$, $\scriptsize (-3,0)$ and $\scriptsize (0,-3)$.
3. The equation of the circle is $\scriptsize {{x}^{2}}+{{y}^{2}}=9$. Does the point $\scriptsize (-\sqrt{4},\sqrt{5})$ satisfy the equation?
\scriptsize \begin{align*}\text{LHS}&={{x}^{2}}+{{y}^{2}}\\&={{\left( {-\sqrt{4}} \right)}^{2}}+{{\left( {\sqrt{5}} \right)}^{2}}\\&=4+5\\&=9=\text{RHS}\end{align*}
Therefore, the point does lie on the circle.

### Example 1.2

A circle centre the origin passes through the points $\scriptsize A(-6,6)$ and $\scriptsize B(6,-6)$.

1. Plot the points and draw a rough sketch of the circle.
2. Determine the equation of the circle.
3. Calculate the length of $\scriptsize AB$.
4. Explain why $\scriptsize AB$ is a diameter of the circle.

Solutions

1. Here is a sketch of the circle:
2. $\scriptsize {{x}^{2}}+{{y}^{2}}={{r}^{2}}$. We have two possible points we can substitute into the equation. We will substitute $\scriptsize A(-6,6)$.
\scriptsize \begin{align*}{{x}^{2}}+{{y}^{2}} & ={{r}^{2}}\\\therefore {{(-6)}^{2}}+{{6}^{2}} & ={{r}^{2}}\\\therefore 36+36 & ={{r}^{2}}\\\therefore {{r}^{2}} & =72\end{align*}
The equation of the circle is $\scriptsize {{x}^{2}}+{{y}^{2}}=72$.
3. .
\scriptsize \begin{align*}{{d}_{{AB}}}&=\sqrt{{{{{({{x}_{2}}-{{x}_{1}})}}^{2}}+{{{({{y}_{2}}-{{y}_{1}})}}^{2}}}}\\&=\sqrt{{{{{(-6-6)}}^{2}}+{{{(6-(-6))}}^{2}}}}\\&=\sqrt{{{{{(-12)}}^{2}}+{{{12}}^{2}}}}\\&=\sqrt{{144+144}}\\&=\sqrt{{288}}\\&=2\sqrt{{72}}\end{align*}
4. The radius of the circle is $\scriptsize \sqrt{{72}}$. Therefore, $\scriptsize AB$ is twice the radius of the circle. Also, a circle centre the origin is symmetrical about the axes as well as about the lines $\scriptsize y=x$ and $\scriptsize y=-x$. The points $\scriptsize A$ and $\scriptsize B$ are symmetrical about the line $\scriptsize y=-x$ meaning that $\scriptsize AB$ passes through the centre. Therefore, $\scriptsize AB$ is a straight line passing through the centre and twice the radius i.e. $\scriptsize AB$ is the diameter.
.
Note: You could also find the equation of $\scriptsize AB$ and then show that the point $\scriptsize O(0,0)$ lies on this point to demonstrate that $\scriptsize AB$ passes through the centre. In both cases, showing that a line is a diameter requires you to show that it is:
1. a straight line passing through the centre of the circle
2. twice the length of the radius.

An important characteristic of the circle centre the origin is its symmetry. The shape is very symmetrical. It is symmetrical about both axes as well as about the lines $\scriptsize y=x$ and $\scriptsize y=-x$.

### Take note!

The circle centre the origin is symmetrical about both axes as well as about the lines $\scriptsize y=x$ and $\scriptsize y=-x$.

### Take note!

To prove that a line is a diameter you must prove that it is:

1. a straight line passing through the centre of the circle
2. twice the length of the radius.

### Exercise 1.1

1. Given $\scriptsize {{x}^{2}}+{{y}^{2}}=100$:
1. Determine the radius.
2. Make a sketch of the circle.
2. Given $\scriptsize {{x}^{2}}+{{y}^{2}}=\displaystyle \frac{{16}}{9}$:
1. Determine the radius.
2. Make a sketch of the circle.
3. Given $\scriptsize 4{{x}^{2}}+4{{y}^{2}}=80$:
1. Determine the radius.
2. Make a sketch of the circle.
4. Determine the equation of the circle in each case:
1. With centre the origin and a radius of $\scriptsize 6$.
2. With centre $\scriptsize (0,0)$ and $\scriptsize r=\sqrt{{13}}$.
3. With centre the origin passing through $\scriptsize (-3,-7)$.
4. Passing through $\scriptsize (2s,3s)$ and with centre $\scriptsize (0,0)$.
5. .
5. Is $\scriptsize \sqrt{{90}}+{{x}^{2}}-{{y}^{2}}=0$ the equation of a circle with centre the origin?
6. Given a circle with centre the origin, with a radius of $\scriptsize r=\sqrt{{50}}$, determine the coordinates of the points on the circle whose x-values are three times their y-value.

The full solutions are at the end of the unit.

## The equation of a circle centre $\scriptsize (a,b)$

There are two ways to think about the equation of a circle with centre not the origin but any point $\scriptsize (a,b)$ on the Cartesian plane.

The first is to think about translating or shifting the circle centre the origin. Consider the parabola $\scriptsize y=2{{x}^{2}}$. We know that the graph of $\scriptsize y=2{{x}^{2}}+1$ is the same as $\scriptsize y=2{{x}^{2}}$ except that the graph has been shifted one unit up (see figure 4).

We can rewrite the equation $\scriptsize y=2{{x}^{2}}+1$ as $\scriptsize y-1=2{{x}^{2}}$ or $\scriptsize (y-1)=2{{x}^{2}}$.

Now, how do you think the circle defined by $\scriptsize {{x}^{2}}+{{(y-1)}^{2}}=25$ is different from that defined by $\scriptsize {{x}^{2}}+{{y}^{2}}=25$? $\scriptsize {{x}^{2}}+{{(y-1)}^{2}}=25$ is the same except that the circle has been shifted one unit up and the centre is now the point $\scriptsize (0,1)$ (see figure 5).

Consider the parabola $\scriptsize y=2{{x}^{2}}$ again. We know that the graph of $\scriptsize y=2{{(x-1)}^{2}}$ is the same as $\scriptsize y=2{{x}^{2}}$ except that it has been shifted one unit to the right (see figure 6).

In the same way, the circle $\scriptsize {{(x-1)}^{2}}+{{y}^{2}}=25$ is the same as $\scriptsize {{x}^{2}}+{{y}^{2}}=25$ just shifted one unit to the right. The centre is now the point $\scriptsize (1,0)$ (see figure 7).

Therefore, we can generalise and say that the equation $\scriptsize {{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$ is the equation of a circle radius $\scriptsize r$ and with centre $\scriptsize (a,b)$.

The other way to consider this is geometrically. Figure 8 shows a circle centre $\scriptsize (a,b)$ and passing through $\scriptsize p(x,y)$ with a radius $\scriptsize r$.

We can determine the distance $\scriptsize PC$ using the distance formula.
\scriptsize \begin{align*}{{d}_{{PC}}}&=\sqrt{{{{{({{x}_{2}}-{{x}_{1}})}}^{2}}+{{{({{y}_{2}}-{{y}_{1}})}}^{2}}}}\\&=\sqrt{{{{{(x-a)}}^{2}}+{{{(y-b)}}^{2}}}}\end{align*}

But we know that the length of $\scriptsize PC$ is the radius. Therefore $\scriptsize {{d}_{{PC}}}=r$. Therefore:
\scriptsize \begin{align*}r & =\sqrt{{{{{(x-a)}}^{2}}+{{{(y-b)}}^{2}}}}\\\therefore {{r}^{2}} & ={{(x-a)}^{2}}+{{(y-b)}^{2}}\\\therefore {{(x-a)}^{2}}+{{(y-b)}^{2}} & ={{r}^{2}}\end{align*}

With this equation, we can think of the circle with centre the origin as a special case.
\scriptsize \begin{align*}{{(x-a)}^{2}}+{{(y-b)}^{2}} & ={{r}^{2}}\text{ but }(a,b)=(0,0)\\\therefore {{(x-0)}^{2}}+{{(y-0)}^{2}} & ={{r}^{2}}\\\therefore {{x}^{2}}+{{y}^{2}} & ={{r}^{2}}\end{align*}

Equation of a circle with centre at $\scriptsize (a,b)$:
$\scriptsize {{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$

### Example 1.3

$\scriptsize A(4,6)$ is a point on a circle centre $\scriptsize (3,3)$:

1. Determine the equation of the circle.
2. Draw a sketch of the circle.
3. Does the point $\scriptsize B(\displaystyle \frac{3}{2},2)$ lie on the circle?
4. Does the circle cut the y-axis?

Solutions

1. Because the circle has centre $\scriptsize (3,3)$, we know that the equation will be of the form$\scriptsize {{(x-3)}^{2}}+{{(y-3)}^{2}}={{r}^{2}}$. We need to find $\scriptsize r$ which we can do by finding the distance between the point $\scriptsize A(4,6)$ and the centre $\scriptsize (3,3)$.
\scriptsize \begin{align*}r&=\sqrt{{{{{({{x}_{2}}-{{x}_{1}})}}^{2}}+{{{({{y}_{2}}-{{y}_{1}})}}^{2}}}}\\&=\sqrt{{{{{(4-3)}}^{2}}+{{{(6-3)}}^{2}}}}\\&=\sqrt{{1+9}}\\&=\sqrt{{10}}\end{align*}
Therefore, $\scriptsize {{r}^{2}}=10$ and the equation of the circle is $\scriptsize {{(x-3)}^{2}}+{{(y-3)}^{2}}=10$.
2. The easiest way to draw a sketch of a circle centre $\scriptsize (a,b)$ and passing through a point $\scriptsize (x,y)$ is to use a pair of compasses. Set the point on the centre and your pencil on the point to get the radius, and draw the resulting circle.
3. If the point $\scriptsize B(\displaystyle \frac{3}{2},2)$ lies on the circle, it will satisfy the equation of the circle.
\scriptsize \displaystyle \begin{align*}\text{LHS}&={{(x-3)}^{2}}+{{(y-3)}^{2}}\\&={{\left( {\displaystyle \frac{3}{2}-3} \right)}^{2}}+{{(2-3)}^{2}}\\&={{\left( {-\displaystyle \frac{3}{2}} \right)}^{2}}+{{(-1)}^{2}}\\&=\displaystyle \frac{9}{4}+1\\&=\displaystyle \frac{{9+4}}{4}\\&=\displaystyle \frac{{13}}{4}\ne \text{RHS}\end{align*}
Therefore, the point $\scriptsize B(\displaystyle \frac{3}{2},2)$ does not lie on the circle.
4. For the circle to cut the y-axis, there must be a point on the circle with an x-coordinate of zero. We need to substitute $\scriptsize x=0$ into the equation to see if there is at least one real solution for $\scriptsize y$.
\scriptsize \begin{align*}{{(0-3)}^{2}}+{{(y-3)}^{2}} & =10\\\therefore {{(-3)}^{2}}+{{(y-3)}^{2}} & =10\\\therefore 9+{{y}^{2}}-6y+9 & =10\\\therefore {{y}^{2}}-6y+8 & =0\\\therefore (y-4)(y-2) & =0\\\therefore y=4\text{ } & \text{or }y=2\end{align*}
There are two points $\scriptsize (0,2)$ and $\scriptsize (0,4)$ where the circle cuts the y-axis.
Note: You can also find any x- or y-intercepts to help you make an accurate sketch of a circle centre $\scriptsize (a,b)$.

### Example 1.4

Determine the coordinates of the centre of the circle and its radius if the equation of the circle is $\scriptsize 3{{x}^{2}}+12x+3{{y}^{2}}-24y+21=0$.

Solution

This question is basically asking us to rewrite the equation of the circle centre $\scriptsize (a,b)$ in the standard form $\scriptsize {{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$ rather than the current expanded form. Because the standard form of the equation of a circle contains perfect squares, we know that we will need to complete the squares for $\scriptsize x$ and for $\scriptsize y$.

Step 1: Get the coefficients of the $\scriptsize {{x}^{2}}$ and $\scriptsize {{y}^{2}}$ terms equal to $\scriptsize 1$
\scriptsize \begin{align*}3{{x}^{2}}+12x+3{{y}^{2}}-24y+21 & =0\\\therefore {{x}^{2}}+4x+{{y}^{2}}-8y+7 & =0\end{align*}

Step 2: Complete the squares
Remember, to complete the squares, we need to add half the coefficients of the $\scriptsize x$ and $\scriptsize y$ terms squared. Don’t forget to keep the equation balanced by subtracting these values as well.
\scriptsize \begin{align*}{{x}^{2}}+4x+{{y}^{2}}-8y+7 & =0\\\therefore {{x}^{2}}+4x+4+{{y}^{2}}-8y+16+7-4-16 & =0\\\therefore {{(x+2)}^{2}}+{{(y-4)}^{2}} & =13\end{align*}

Step 3: Answer the question
The centre of the circle is the point $\scriptsize (-2,4)$ and the radius is $\scriptsize r=\sqrt{{13}}$.

### Example 1.5

Given a circle centre the origin with $\scriptsize r=9$:

1. Write down the equation of the circle if it is shifted $\scriptsize 3\text{ units}$ to the left and $\scriptsize 5\text{ units}$ down.
2. Draw a sketch of the circle.
3. Write down the equation of the circle that is symmetrical to the circle in question 1 about the line $\scriptsize y=-x$.
4. Sketch the circle in question 3 on the same set of axes as the graph in question 2.

Solutions

1. The equation of the circle we were given is $\scriptsize {{x}^{2}}+{{y}^{2}}=81$. The equation of the shifted circle will be $\scriptsize {{(x+3)}^{2}}+{{(y+5)}^{2}}=81$.
2. We know the centre of the circle but we need at least one point that the circle passes through. An easy way to find this point is to move either horizontally or vertically from the centre a distance of $\scriptsize 9\text{ units}$ (the radius) as indicated below.
3. To find the coordinates of a point symmetrical about the line $\scriptsize y=-x$, each $\scriptsize x$ value must be replaced with $\scriptsize -y$ and each $\scriptsize y$ value replaced with $\scriptsize -x$. Therefore, the equation of the symmetrical circle will be:
\scriptsize \begin{align*}{{\left( {(-y)+3} \right)}^{2}}+{{\left( {(-x)+5} \right)}^{2}} & =81\\\therefore {{(5-x)}^{2}}+{{(3-y)}^{2}} & =81\end{align*}
We can expand the equation and then simplify into standard form.
\scriptsize \begin{align*}{{(5-x)}^{2}}+{{(3-y)}^{2}} & =81\\\therefore 25-10x+{{x}^{2}}+9-6y+{{y}^{2}} & =81\\\therefore {{x}^{2}}-10x+25+{{y}^{2}}-6y+9 & =81\\\therefore {{(x-5)}^{2}}+{{(y-3)}^{2}} & =81\end{align*}
4. Below are the two circles sketched on the same set of axes. Once again, to find another point on the symmetrical circle, we moved $\scriptsize r=9\text{ units}$ horizontally to the right from the centre. The line of symmetry $\scriptsize y=-x$ has also been sketched to better illustrate the symmetry.

### Exercise 1.2

1. Write down the equation of the following circles:
1. Centre $\scriptsize (-3,0)$ and radius $\scriptsize 4\text{ units}$.
2. Centre $\scriptsize C(-5,-1)$ and passing through the point $\scriptsize P(6,-2)$.
3. $\scriptsize r=\sqrt{{11}}$ and centre $\scriptsize (2,-3)$.
2. Determine the centre and radius of the following circles:
1. $\scriptsize \displaystyle 5-{{x}^{2}}-6x-8y-{{y}^{2}}=0$
2. $\scriptsize {{x}^{2}}=21+4y-{{y}^{2}}$
3. $\scriptsize 12x-12y-2{{x}^{2}}-2{{y}^{2}}=24$
3. A circle cuts the x-axis at $\scriptsize A(-3,0)$ and $\scriptsize B(3,0)$. If $\scriptsize r=\sqrt{{21}}$, determine the possible equation(s) of the circle.
4. $\scriptsize P(5,6)$ and $\scriptsize Q(-1,-2)$ are points on a circle. If $\scriptsize PQ$ is a diameter of the circle, determine the equation of the circle.

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• The equation of a circle centre $\scriptsize (a,b)$ is given by $\scriptsize {{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$.
• The equation of a circle centre the origin is a special case and is given by $\scriptsize {{x}^{2}}+{{y}^{2}}={{r}^{2}}$.
• The circle centre the origin is symmetrical about both axes and the lines $\scriptsize y=x$ and $\scriptsize y=-x$.

# Unit 1: Assessment

#### Suggested time to complete: 60 minutes

1. Determine the equation of the circle:
1. With centre the origin and a radius of $\scriptsize \displaystyle \frac{7}{3}$.
2. With centre $\scriptsize (2,-5)$ and $\scriptsize r=2\sqrt{{11}}$.
3. With centre $\scriptsize C(-3,0)$ passing through $\scriptsize R(8,-4)$.
4. Passing through $\scriptsize P(5q,6q)$ and centre $\scriptsize C(0,1)$.
2. Determine the value(s) of $\scriptsize h$ if $\scriptsize (\sqrt{2},h)$ is a point on the circle $\scriptsize {{x}^{2}}+{{y}^{2}}=17$.
3. $\scriptsize P(r,s)$ is a point on the circle with centre at the origin and a diameter of $\scriptsize \text{70 cm}$. Determine the possible coordinates of $\scriptsize P$ if the value of $\scriptsize r$ is four times the value of $\scriptsize s$.

Question 4 adapted from Everything Maths Grade 12 Exercise 7-3 question 6

1. $\scriptsize P(-2,3)$ lies on a circle with centre at $\scriptsize (0,0)$.
1. Determine the equation of the circle.
2. Sketch the circle and label point $\scriptsize P$.
3. If $\scriptsize PQ$ is a diameter of the circle, determine the coordinates of $\scriptsize Q$.
4. Calculate the length of $\scriptsize PQ$.
5. Determine the equation of the line $\scriptsize PQ$.
6. Determine the equation of the line perpendicular to $\scriptsize PQ$ and passing through the point $\scriptsize P$.

Question 5 adapted from Everything Maths Grade 12 Exercise 7-4 question 6

1. A circle with centre $\scriptsize N(4,4)$ passes through the points $\scriptsize K(1,6)$ and $\scriptsize L(6,7)$.
1. Draw a sketch of the circle.
2. Determine the equation of the circle.
3. Determine the coordinates of $\scriptsize M$, the midpoint of $\scriptsize KL$.
4. Show that $\scriptsize MN\bot KL$.
5. Determine the equation of the line $\scriptsize LN$.
6. If $\scriptsize P$ is a point on the circle such that $\scriptsize LP$ is a diameter, determine the coordinates of $\scriptsize P$.
2. A circle passes through the points $\scriptsize A(1,2)$ and $\scriptsize B(-3,2)$. If its centre lies on the line $\scriptsize 6x+3y=-3$, determine the equation of the circle.

Question 7 adapted from Everything Maths Grade 12 Exercise 7-4 question 8

1. A circle with centre $\scriptsize (0,0)$ passes through the point $\scriptsize T(4,3)$.
1. Determine the equation of the circle.
2. If the circle is shifted $\scriptsize 2\text{ units}$ to the right and $\scriptsize 3\text{ units}$ down, determine the new equation of the circle.
3. Draw a sketch of the original circle and the shifted circle on the same system of axes.
4. If the circle in 7 b. is reflected about the x-axis, what will the centre of this new circle be.

Question 8 adapted from NC(V) Mathematics Level 4 Paper 2 November 2016 question 1.5

1. In the figure below two circles with a common centre $\scriptsize M(-1,-2)$ are shown. Radius $\scriptsize MA$ of the larger circle cuts the smaller circle at $\scriptsize B$. Point $\scriptsize A$ has coordinates $\scriptsize (-5,3)$.

1. Determine the equation of the larger circle.
2. If the equation of the smaller circle is $\scriptsize {{x}^{2}}+{{y}^{2}}+2x+4y-4=0$, determine the length $\scriptsize AB$.

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. .
1. .
\scriptsize \begin{align*}{{r}^{2}} & =100\\\therefore r & =10\end{align*}
2. .
2. .
1. .
\scriptsize \begin{align*} r^2&=\displaystyle \frac{16}{9}\\ \therefore r&=\displaystyle \frac{4}{3}\end{align*}
2. .
3. .
1. .
\scriptsize \begin{align*}4{{x}^{2}}+4{{y}^{2}} & =80\\\therefore {{x}^{2}}+{{y}^{2}} & =20\end{align*}
\scriptsize \begin{align*}{{r}^{2}} & =20\\\therefore r & =\sqrt{{20}}=2\sqrt{5}\end{align*}
2. We can find a point, with coordinates that are integers, that lies on the circle to help us sketch the circle more accurately.
\scriptsize \begin{align*}20&=16+4\\&={{4}^{2}}+{{2}^{2}}\end{align*}
4. .
1. $\scriptsize {{x}^{2}}+{{y}^{2}}=36$
2. $\scriptsize {{x}^{2}}+{{y}^{2}}=13$
3. .
\scriptsize \begin{align*}{{(-3)}^{2}}+{{(-7)}^{2}}&=9+49\\&=58\\\therefore {{x}^{2}}+{{y}^{2}} &=58\end{align*}
4. .
\scriptsize \begin{align*}{{(2s)}^{2}}+{{(3s)}^{2}} & =4{{s}^{2}}+9{{s}^{2}}\\&=13{{s}^{2}}\\\therefore {{x}^{2}}+{{y}^{2}} & =13{{s}^{2}}\end{align*}
5. .
\scriptsize \begin{align*}{{(-5t)}^{2}}+{{(-3t)}^{2}} & =25{{t}^{2}}+9{{t}^{2}}\\&=34{{t}^{2}}\\\therefore {{x}^{2}}+{{y}^{2}} & =34{{t}^{2}}\end{align*}
5. .
\scriptsize \begin{align*}\sqrt{90}+x^2-y^2&=0\\ \therefore x^2-y^2&=-\sqrt{90}\\ \therefore y^2-x^2&=\sqrt{90}\end{align*}
This is not the equation of a circle with centre the origin.
6. $\scriptsize {{x}^{2}}+{{y}^{2}}=50$ and $\scriptsize x=3y$. Therefore:
\scriptsize \begin{align*}{{(3y)}^{2}}+{{y}^{2}} & =50\\\therefore 9{{y}^{2}}+{{y}^{2}} & =50\\\therefore 10{{y}^{2}} & =50\\\therefore {{y}^{2}} & =5\\\therefore y & =\pm \sqrt{5}\end{align*}
Substitute $\scriptsize y=+\sqrt{5}$:
\scriptsize \begin{align*}{{x}^{2}}+{{\left( {\sqrt{5}} \right)}^{2}} & =50\\\therefore {{x}^{2}}+5 & =50\\\therefore {{x}^{2}} & =45\\\therefore x & =\pm \sqrt{{45}}\\ & =\pm 3\sqrt{5}\end{align*}
Because the circle is symmetrical about the x-axis, we will get the same result substituting $\scriptsize y=-\sqrt{5}$. Therefore, there are four possible points.
$\scriptsize (3\sqrt{5},\sqrt{5})$, $\scriptsize (-3\sqrt{5},-\sqrt{5})$, $\scriptsize (-3\sqrt{5},\sqrt{5})$ and $\scriptsize (3\sqrt{5},-\sqrt{5})$

Back to Exercise 1.1

### Exercise 1.2

1. .
1. $\scriptsize {{(x+3)}^{2}}+{{y}^{2}}=16$
2. Equation is of the form $\scriptsize {{(x+5)}^{2}}+{{(y+1)}^{2}}={{r}^{2}}$. Length of $\scriptsize CP$ is the radius.
\scriptsize \begin{align*}{{d}_{{CP}}}&=\sqrt{{{{{({{x}_{2}}-{{x}_{1}})}}^{2}}+{{{({{y}_{2}}-{{y}_{1}})}}^{2}}}}\\&=\sqrt{{{{{(-5-6)}}^{2}}+{{{(-1-(-2))}}^{2}}}}\\&=\sqrt{{{{{(-11)}}^{2}}+{{1}^{2}}}}\\&=\sqrt{{121+1}}\\&=\sqrt{{122}}\end{align*}
The equation is $\scriptsize {{(x+5)}^{2}}+{{(y+1)}^{2}}=122$.
3. $\scriptsize {{(x-2)}^{2}}+{{(y+3)}^{2}}=11$
2. Determine the centre and radius of the following circles:
1. .
\scriptsize \begin{align*}5-x^2-6x-8y-y^2&=0\\ x^2+6x+y^2+8y&=5&&\text{Complete the squares}\\ \therefore x^2+6x+9+y^2+8y+16&=5+9+16\\ \therefore (x+3)^2+(y+4)^2&=30\end{align*}
The centre is $\scriptsize (-3,-4)$, the radius is $\scriptsize \sqrt{{30}}$.
2. .
\scriptsize \begin{align*}x^2&=21+4y-y^2\\ x^2+y^2-4y&=21&&\text{Complete the square}\\ \therefore x^2+y^2-4y-+4&=21+4\\ \therefore x^2+(y-2)^2&=25\end{align*}
The centre is $\scriptsize (0,2)$, the radius is $\scriptsize 5$.
3. .
\scriptsize \begin{align*}12x-12y-2{{x}^{2}}-2{{y}^{2}} & =24\\\therefore 2{{x}^{2}}-12x+2{{y}^{2}}+12y & =24\\\therefore {{x}^{2}}-6x+{{y}^{2}}+6y & =12\quad \text{Complete the squares}\\\therefore {{x}^{2}}-6x+9+{{y}^{2}}+6y+9 & =12+9+9\\\therefore {{(x-3)}^{2}}+{{(y+3)}^{2}} & =30\end{align*}
The centre is $\scriptsize (3,-3)$, the radius is $\scriptsize \sqrt{{30}}$.
3. The equation of the circle is $\scriptsize {{(x-a)}^{2}}+{{(y-b)}^{2}}=21$. $\scriptsize A(-3,0)$ and $\scriptsize B(3,0)$ lie on the circle.
Substitute $\scriptsize A(-3,0)$:
\scriptsize \begin{align*}{{(-3-a)}^{2}}+{{(0-b)}^{2}} & =21\\\therefore 9+6a+{{a}^{2}}+{{b}^{2}} & =21\\\therefore {{a}^{2}}+6a+9 & =21-{{b}^{2}}\quad \quad (1)\end{align*}
Substitute $\scriptsize B(3,0)$:
\scriptsize \begin{align*}{{(3-a)}^{2}}+{{(0-b)}^{2}} & =21\\\therefore 9-6a+{{a}^{2}}+{{b}^{2}} & =21\\\therefore {{a}^{2}}-6a+9 & =21-{{b}^{2}}\quad \quad (2)\end{align*}
Set $\scriptsize (1)=(2)$:
\scriptsize \begin{align*}{{a}^{2}}+6a+9 & ={{a}^{2}}-6a+9\\\therefore 12a & =0\\\therefore a & =0\end{align*}
Substitute $\scriptsize a=0$ into $\scriptsize (1)$:
\scriptsize \begin{align*}9 & =21-{{b}^{2}}\\\therefore {{b}^{2}} & =12\\\therefore b & =\pm \sqrt{{12}}\\\therefore b & =\pm 2\sqrt{3}\end{align*}
The possible equations of the circle are $\scriptsize {{x}^{2}}+{{\left( {y-2\sqrt{3}} \right)}^{2}}=21$ or $\scriptsize {{x}^{2}}+{{\left( {y+2\sqrt{3}} \right)}^{2}}=21$.
4. If $\scriptsize PQ$ is a diameter of the circle, then the midpoint of $\scriptsize PQ$ is the centre.
\scriptsize \begin{align*}\text{midpoin}{{\text{t}}_{{PQ}}}&=\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\&=\left( {\displaystyle \frac{{5-1}}{2},\displaystyle \frac{{6-2}}{2}} \right)\\&=(2,2)\end{align*}
The radius of the circle is the distance from the centre, $\scriptsize C$, to either $\scriptsize P$ or $\scriptsize Q$.
\scriptsize \begin{align*}{{d}_{{CP}}}&=\sqrt{{{{{({{x}_{2}}-{{x}_{1}})}}^{2}}+{{{({{y}_{2}}-{{y}_{1}})}}^{2}}}}\\&=\sqrt{{{{{(5-2)}}^{2}}+{{{(6-2)}}^{2}}}}\\&=\sqrt{{{{3}^{2}}+{{4}^{2}}}}\\&=\sqrt{{9+16}}\\&=\sqrt{{25}}\\&=5\end{align*}
Therefore, the equation of the circle is $\scriptsize {{(x-2)}^{2}}+{{(y-2)}^{2}}=25$.

Back to Exercise 1.2

### Unit 1: Assessment

1. .
1. $\scriptsize {{x}^{2}}+{{y}^{2}}=\displaystyle \frac{{49}}{9}$
2. $\scriptsize {{(x-2)}^{2}}+{{(y+5)}^{2}}=44$
3. .
\scriptsize \begin{align*}r&={{d}_{{CR}}}=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-3-8)}}^{2}}+{{{(0-(-4))}}^{2}}}}\\&=\sqrt{{{{{(-11)}}^{2}}+{{4}^{2}}}}\\&=\sqrt{{121+16}}\\&=\sqrt{{137}}\end{align*}
The equation is $\scriptsize {{(x+3)}^{2}}+{{y}^{2}}=137$.
4. .
\scriptsize \begin{align*}r&={{d}_{{CP}}}=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(5q-0)}}^{2}}+{{{(6q-1)}}^{2}}}}\\&=\sqrt{{25{{q}^{2}}+36{{q}^{2}}-12q+1}}\\&=\sqrt{{61{{q}^{2}}-12q+1}}\end{align*}
The equation is $\scriptsize {{x}^{2}}+{{(y-1)}^{2}}=61{{q}^{2}}-12q+1$.
2. Substitute in $\scriptsize (\sqrt{2},h)$:
\scriptsize \begin{align*}{{\left( {\sqrt{2}} \right)}^{2}}+{{h}^{2}} & =17\\\therefore 2+{{h}^{2}} & =17\\\therefore {{h}^{2}} & =15\\\therefore h & =\pm \sqrt{{15}}\end{align*}
3. Diameter is $\scriptsize \text{70 cm}$. Therefore, radius is $\scriptsize \text{35 cm}$.
$\scriptsize r=4s$. Substitute $\scriptsize x=r=4s$ and $\scriptsize y=s$ into $\scriptsize {{x}^{2}}+{{y}^{2}}=1\ 225$.
\scriptsize \begin{align*}{{(4s)}^{2}}+{{(s)}^{2}} & =1\ 225\\\therefore 16{{s}^{2}}+{{s}^{2}} & =1\ 225\\\therefore 17{{s}^{2}} & =1\ 225\\\therefore {{s}^{2}} & =\displaystyle \frac{{1\ 225}}{{17}}\\\therefore s & =\pm \displaystyle \frac{{35}}{{\sqrt{{17}}}}\end{align*}
Therefore:
\scriptsize \begin{align*}r & =4\times \left( {\pm \displaystyle \frac{{35}}{{\sqrt{{17}}}}} \right)\\ & =\pm \displaystyle \frac{{140}}{{\sqrt{{17}}}}\end{align*}
The possible coordinates of $\scriptsize P(r,s)$ are $\scriptsize \left( {\pm \displaystyle \frac{{140}}{{\sqrt{{17}}}},\pm \displaystyle \frac{{35}}{{\sqrt{{17}}}}} \right)$.
4. .
1. .
\scriptsize \begin{align*}{{r}^{2}}&={{(-2)}^{2}}+{{3}^{2}}\\&=4+9\\&=13\end{align*}
The equation of the circle is $\scriptsize {{x}^{2}}+{{y}^{2}}=13$.
2. .
3. If $\scriptsize PQ$ is a diameter then $\scriptsize P$ and $\scriptsize Q$ are symmetrical about the line $\scriptsize y=x$. Therefore $\scriptsize Q(3,-2)$.
4. .
\scriptsize \begin{align*}{{d}_{{PQ}}}&=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-2-3)}}^{2}}+{{{(3-(-2))}}^{2}}}}\\&=\sqrt{{{{{(-5)}}^{2}}+{{5}^{2}}}}\\&=\sqrt{{50}}\\&=5\sqrt{2}\end{align*}
5. .
\scriptsize \begin{align*}{{m}_{{PQ}}}&=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\&=\displaystyle \frac{{3-(-2)}}{{-2-3}}\\&=-\displaystyle \frac{5}{5}\\&=-1\end{align*}
\scriptsize \begin{align*}y-{{y}_{1}} & =m(x-{{x}_{1}})\\\therefore y-3 & =-1\left( {x-(-2)} \right)\\\therefore y-3 & =-(x+2)\\\therefore y-3 & =-x-2\\\therefore y & =-x+1\end{align*}
6. The gradient of the line perpendicular to $\scriptsize PQ$ is $\scriptsize m=1$.
\scriptsize \begin{align*}y-{{y}_{1}} & =m(x-{{x}_{1}})\\\therefore y-3 & =1\left( {x-(-2)} \right)\\\therefore y-3 & =(x+2)\\\therefore y-3 & =x+2\\\therefore y & =x+5\end{align*}
5. .
1. .
2. .
\scriptsize \begin{align*}r={{d}_{{NK}}}&=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(4-1)}}^{2}}+{{{(4-6)}}^{2}}}}\\&=\sqrt{{{{{(-3)}}^{2}}+{{{(-2)}}^{2}}}}\\&=\sqrt{{9+4}}\\&=\sqrt{{13}}\end{align*}
The equation is $\scriptsize {{(x-4)}^{2}}+{{(y-4)}^{2}}=13$.
3. .
\scriptsize \begin{align*}\text{M=midpoin}{{\text{t}}_{{KL}}}&=\left( {\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}}}{2},\displaystyle \frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)\\&=\left( {\displaystyle \frac{{1+6}}{2},\displaystyle \frac{{6+7}}{2}} \right)\\&=(\displaystyle \frac{7}{2},\displaystyle \frac{{13}}{2})\end{align*}
4. .
\scriptsize \begin{align*}{{m}_{{MN}}}&=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\&=\displaystyle \frac{{\displaystyle \frac{{13}}{2}-4}}{{\displaystyle \frac{7}{2}-4}}\\&=\displaystyle \frac{{\displaystyle \frac{5}{2}}}{{-\displaystyle \frac{1}{2}}}\\&=\displaystyle \frac{5}{2}\times \left( {-\displaystyle \frac{2}{1}} \right)=-5\end{align*}
\scriptsize \begin{align*}{{m}_{{KL}}}&=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\&=\displaystyle \frac{{7-6}}{{6-1}}\\&=\displaystyle \frac{1}{5}\end{align*}
$\scriptsize {{m}_{{MN}}}\times {{m}_{{KL}}}=-1$. Therefore, $\scriptsize MN\bot KL$.
5. Equation of LN:
\scriptsize \begin{align*}{{m}_{{LN}}}&=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}\\&=\displaystyle \frac{{7-4}}{{6-4}}\\&=\displaystyle \frac{3}{2}\end{align*}
\scriptsize \begin{align*}y-{{y}_{1}} & =m(x-{{x}_{1}})\\\therefore y-4 & =\displaystyle \frac{3}{2}(x-4)\\\therefore y-4 & =\displaystyle \frac{3}{2}x-6\\\therefore y & =\displaystyle \frac{3}{2}x-2\end{align*}
6. Since $\scriptsize N$ is the centre of the circle passing through point $\scriptsize L$, $\scriptsize LN$ is a radius. Given that $\scriptsize LP$ is a diameter, $\scriptsize L$, $\scriptsize N$ and $\scriptsize P$ lie on the same straight line. Equation of $\scriptsize LP$is thus:
$\scriptsize y =\displaystyle \frac{3}{2}x-2\quad \quad (1)$
$\scriptsize (x-4)^2+(y-4)^2=13\quad \quad (2)$
Substitute $\scriptsize (1)$ into $\scriptsize (2)$:
\scriptsize \begin{align*}{{(x-4)}^{2}}+{{\left( {\left( {\displaystyle \frac{3}{2}x-2} \right)-4} \right)}^{2}} & =13\\\therefore {{x}^{2}}-8x+16+{{\left( {\displaystyle \frac{3}{2}x-6} \right)}^{2}} & =13\\\therefore {{x}^{2}}-8x+16+\displaystyle \frac{9}{4}{{x}^{2}}-18x+36 & =13\\\therefore \displaystyle \frac{{13}}{4}{{x}^{2}}-26x+39 & =0\\\therefore 13{{x}^{2}}-104x+156 & =0\\\therefore {{x}^{2}}-8x+12 & =0\\\therefore (x-2)(x-6) & =0\\\therefore x=2\text{ } & \text{or }x=6\end{align*}
$\scriptsize x=6$ is the x-coordinate of $\scriptsize L$. Therefore $\scriptsize x=2$ is the x-coordinate of $\scriptsize P$.
Substitute $\scriptsize x=2$ into $\scriptsize (1)$:
\scriptsize \begin{align*}y & =\displaystyle \frac{3}{2}(2)-2\\ & =1\end{align*}
$\scriptsize P(2,1)$
6. .
\scriptsize \begin{align*}6x+3y&=-3\\\therefore 3y & =-6x-3\\\therefore y & =-2x-1\end{align*}
Every point on this line has coordinates $\scriptsize \left( {x,-2x-1} \right)$. If the centre is $\scriptsize C$, $\scriptsize {{d}_{{AC}}}={{d}_{{BC}}}$.
\scriptsize \begin{align*}{{d}_{{AC}}}&=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(1-x)}}^{2}}+{{{\left( {2-\left( {-2x-1} \right)} \right)}}^{2}}}}\\&=\sqrt{{1-2x+{{x}^{2}}+{{{\left( {2x+3} \right)}}^{2}}}}\\&=\sqrt{{1-2x+{{x}^{2}}+4{{x}^{2}}+12x+9}}\\&=\sqrt{{5{{x}^{2}}+10x+10}}\end{align*}
\scriptsize \begin{align*}{{d}_{{BC}}}&=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{(-3-x)}}^{2}}+{{{\left( {2-\left( {-2x-1} \right)} \right)}}^{2}}}}\\&=\sqrt{{9+6x+{{x}^{2}}+{{{\left( {2x+3} \right)}}^{2}}}}\\&=\sqrt{{9+6x+{{x}^{2}}+4{{x}^{2}}+12x+9}}\\&=\sqrt{{5{{x}^{2}}+18x+18}}\end{align*}
But $\scriptsize {{d}_{{AC}}}={{d}_{{BC}}}$. Therefore:
\scriptsize \begin{align*}\sqrt{{5{{x}^{2}}+10x+10}} & =\sqrt{{5{{x}^{2}}+18x+18}}\\\therefore 5{{x}^{2}}+10x+10 & =5{{x}^{2}}+18x+18\\\therefore 10x+10 & =18x+18\\\therefore 8x & =-8\\\therefore x & =-1\end{align*}
\scriptsize \begin{align*}y & =-2x-1\\&=-2(-1)-1\\&=1\end{align*}
The centre of the circle is the point $\scriptsize (-1,1)$.
\scriptsize \begin{align*}r&={{d}_{{AC}}}=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{\left( {1-(-1)} \right)}}^{2}}+{{{\left( {2-1} \right)}}^{2}}}}\\&=\sqrt{{4+1}}\\&=\sqrt{5}\end{align*}
The equation of the circle is $\scriptsize {{(x+1)}^{2}}+{{(y-1)}^{2}}=5$
7. .
1. .
\scriptsize \begin{align*}{{r}^{2}}&={{x}^{2}}+{{y}^{2}}\\&={{4}^{2}}+{{3}^{2}}\\&=16+9\\&=25\end{align*}
The equation of the circle is $\scriptsize {{x}^{2}}+{{y}^{2}}=25$.
2. $\scriptsize {{(x-2)}^{2}}+{{(y+3)}^{2}}=25$
3. .
4. The centre of new circle will be $\scriptsize (2,3)$.
8. .
1. The equation has the form $\scriptsize {{(x+1)}^{2}}+{{(y+2)}^{2}}={{r}^{2}}$.
\scriptsize \begin{align*}r&={{d}_{{MA}}}=\sqrt{{{{{({{x}_{1}}-{{x}_{2}})}}^{2}}+{{{({{y}_{1}}-{{y}_{2}})}}^{2}}}}\\&=\sqrt{{{{{\left( {-1-(-5)} \right)}}^{2}}+{{{\left( {-2-3} \right)}}^{2}}}}\\&=\sqrt{{{{4}^{2}}+{{{(-5)}}^{2}}}}\\&=\sqrt{{16+25}}\\&=\sqrt{{41}}\end{align*}
The equation is $\scriptsize {{(x+1)}^{2}}+{{(y+2)}^{2}}=41$.
2. .
\scriptsize \displaystyle \begin{align*}{{x}^{2}}+{{y}^{2}}+2x+4y-4 & =0\\\therefore {{x}^{2}}+2x+{{y}^{2}}+4y & =4\\\therefore {{x}^{2}}+2x+1+{{y}^{2}}+4y+4 & =4+1+4\\\therefore {{(x+1)}^{2}}+{{(y+2)}^{2}} & =9\end{align*}
Therefore, radius of the smaller circle is $\scriptsize 3$.
\scriptsize \begin{align*}AB&=AM-BM\\&=\sqrt{{41}}-3\end{align*}

Back to Unit 1: Assessment


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text:
# Algorithms for finding the self-intersection of a plane curve?

Does anyone here happen to know about algorithms for finding the self-intersection points of a plane curve? For example, find the self-intersection (X, Y) = (0, 0) in the below plane curve.

Edit: Add an actual example. The scattered points are computed numerically. It is not straightforward to control the spacing between each scattered point. The distribution density of points around the intersection (marked by the red circle) can be improved though. Note also that by eye we can tell that the intersection is not at Fg = 0.

1 Like

detect bifurcation? You monitor the spectrum of the jacobian. BifurcationKit or HomotopyContinuation.jl should be able to do it

3 Likes

I’ve gone through BifurcationKit.jl documentation quickly and found it quite interesting. However, I have difficulties understanding the underlying math. The tutorials are all about ODEs and PDEs. Do you have any tutorials for simple functions (such as y^2 = x^3 + x^2). The self-intersection of the above example happens to be the zero point of this function. But the self-intersection can locate anywhere on the plane. In addition, my actual function is more complicated and contains not only polynomials but natural log functions.

This library may be a good option: IntervalRootFinding

1 Like

The implicit equation provided can be parametrized to: X(t) = t^2 - 1; Y(t) = t^3 - t

Then we can solve the system (X(t), Y(t)) = (X(u), Y(u)), whose non-trivial solution (t != u) is u = 1, which gives the crossing point (0, 0).

Disclaimer: not a mathematician, but the above solution is a physical one, we send two particles at different times on the same trajectory and check their collision.

1 Like

True for this simple example. But my actual function is more complicated and has no explicit analytical expression. So trying to find a numerical approach

1 Like

Regarding the recently added scatterplot above, would it be possible to extract two sets of points close to the intersection, fit them independently with linear (or other) functions, and then compute the intersection on that basis?

I’ve gone through BifurcationKit.jl documentation quickly and found it quite interesting. However, I have difficulties understanding the underlying math.

It is to find solutions (x,p) with salar p to an equation F(x, p) = 0

The tutorials are all about ODEs and PDEs.

The first part of the ode problems is to find the zeros of a nonlinear function (not just polynomials) as function of a parameter. For example you could search for y as function of x and detect bifurcations.

Do you have any tutorials for simple functions (such as y^2 = x^3 + x^2). The self-intersection of the above example happens to be the zero point of this function. But the self-intersection can locate anywhere on the plane. In addition, my actual function is more complicated and contains not only polynomials but natural log functions.

1 Like

@rveltz is right that at a self intersection there’s a singularity. So if your curve is the solutions of f(x,y) = 0 then the gradient must be orthogonal to the two linearly independent tanget lines and hence

f_x = f_y = 0

at any self intersection. So you could try to solve the nonlinear system

F = \left ( \begin{array}{c} f_x^2 + f_y^2 \\ f \end{array} \right) = \left ( \begin{array}{ll} 0 \\ 0 \end{array} \right)

Any implementation of Newton’s method, like nsol.jl from SIAMFANLEquations.jl would do that. Even if you only have a function to evaluate f and no analytic expression, that’ll be ok with a numerical Jacobian for Newton as long as f is sufficiently smooth and you pay attention to the difference increments for the first (h) and second (\sqrt{h}) derivatives.

You could think of f(x,y) = 0 as an equation where x is the variable and y a parameter. Then if you have a a point on the curve, you could us BIfurcationKit.jl to do continuation and any singularities will be flagged by the software. Even then, you are not guaranteed to find all solutions.

The tough part in your question is to find all solutions. If you had a polynomial system, then you have a computational algebraic geometry problem and Bertini (google it) would work, but there is no Julia interface AFAIK.

2 Likes

This (the numerical solution) should be an easy task for GMT’s module gmtspatial but there is currently an issue that needs to be fixed (gmtspatial -Ii failes and output strange things · Issue #5889 · GenericMappingTools/gmt · GitHub)

In general, it can be framed as a kind of root-finding problem, as several other posters have suggested, and for which there are many numerical algorithms. Formulating the problem more precisely depends on how your curve is parameterized, which you haven’t specified clearly.

2 Likes
using BifurcationKit, LinearAlgebra, Setfield, Plots
const BK = BifurcationKit

F(x, p) = @. -p.y^2 + x^2 + x^3
J(x, p) = ForwardDiff.jacobian(  z->F(z,p), x)
par = (y = 0., )

opts = BK.ContinuationPar(dsmax = 0.051, dsmin = 1e-3, ds=0.001, maxSteps = 140, pMin = -1., saveSolEveryStep = 1, detectBifurcation = 3)

br, = continuation(F,J, [-1.], par, (@lens _.y),
ContinuationPar(opts, ds = -0.001, maxSteps = 800, newtonOptions = NewtonPar(verbose = false, maxIter = 6), plotEveryStep = 10),
recordFromSolution = (x,p) -> (x = x[1],),
bothside = true,
verbosity = 3, plot = true,
)

plot(br)

The problem is that is gives the fold points as well.

3 Likes

Thanks! This looks great! Will see whether I can formulate my problem to use BifurcationKit.

Here, why do you define another ConinuationPar? Why the first argument is opts? What does it do?

Another good news, I have successfully used this continuation method to draw a binodal line in a ternary phase diagram, which is really awesome! Previously I have to collect all those binodal points by hand which is very cumbersome. It is a more useful application than finding the self-intersection of a plane curve which is only an intermediate step in finding the binodal points. Now, I can simply use BK to find the binodal points for me.

1 Like

Because I typed super fast…sorry. Only one is required

Now, I can simply use BK to find the binodal points for me.

Coool


text:
## Intermediate Algebra for College Students (7th Edition)

$\frac{1}{2}$
The power rule for logarithms says that $\log_b{M^p}=p\log_bM$ i.e. the logarithm of a number with an exponent is the exponent times the logarithm of the number. Also $\log_a a^n=n$ if $a\ne1,a\gt0$. Hence here: $\log_{16}{4}=\log_{16}{16^{\frac{1}{2}}}=\frac{1}{2}$


text:
# Median

## Input

An ordered sequence $S$ of length $n$ such that a comparison $c$ is defined on the component type $C$ of $S$.

## Output

1. If $n$ is odd, the median is the unique value $x \in \{ S[1],...,S[n]\}$ such that
1. $S[i]\lt x$ for at most $\lfloor n/2 \rfloor$ positions $i$ of $S$ and
2. $S[i]\gt x$ for at most $\lfloor n/2 \rfloor$ positions $i$ of $S$.
2. On the other hand, if $n$ is even, the median is $(x+y)/2$, where $x,y \in \{ S[1],...,S[n]\}$ (let $x \le y$) are the unique values such that
1. $S[i]\lt x$ for at most $\lfloor n/2 \rfloor$ positions $i$ of $S$, and
2. $S[i]\gt y$ for at most $\lfloor n/2 \rfloor$ positions $i$ of $S$.

N/A


text:
# When does a singularity start to exist during a black hole formation?

Excuse me if the question is naïve. I am not a physicist.

Let's imagine formation of a black hole from a homogeneous collapsing star. At certain moment when enough of matter is inside of a small enough spherical area the event horizon starts to exist. When does the singularity start to exist and why?

The matter below the event horizon still continues to collapse towards the centre (e.g. in the reference frame of the falling matter). What are the necessary conditions for the singularity to appear?

I have found two similar questions but the answers did not really explain this problem to me:

• The singularity is an unphysical artifact of the theory. One shouldn't take it too seriously. The singularity already disappears in the physical case of a charged rotating black hole, which is a good indicator that it can't be expected to be a stable physical phenomenon. What is "inside" a black hole (and if there is even an "inside") is an open questions, which probably won't find an answer until we can make telescopes that can image the event horizon of a black hole in detail. – CuriousOne Sep 28 '14 at 5:31
• OK, but within the GR theoretical construct how long does it take? I guess it depends where the observer is too. A good question. – Rob Jeffries Sep 28 '14 at 8:53
• @CuriousOne I understand that general relativity fails to describe the interior of a black hole but I think it is still interesting to see which "weirdnesses" show from the failing theory. --- Are there any promising results clarifying more of the problem from the theories trying to unify GR and QFT like the string theory? – pabouk Sep 28 '14 at 8:58
• @pabouk: I can't tell you much about theory, I am an experimentalist. As an experimentalist I can tell you that so far 0 experiments have been performed on black holes. All we have is basically unverified models and a few observations of matter falling into black holes. The signatures of that fall are, of course, generated mostly far away from the event horizon. I think there is one sign that what we are looking at are really black holes, and that's the spectral properties of the x-rays and gamma rays generated by the falling matter, but that's nowhere close enough to probe EH physics. – CuriousOne Sep 28 '14 at 19:08
• I understand that general relativity fails to describe the interior of a black hole Not true. It only fails to describe the singularity. – Ben Crowell Sep 28 '14 at 22:17

If you're not up to speed with general relativity this is going to be hard to explain, but I'll give it a go. The more determined reader may want to look at this PDF (just under 1MB in size) that describes the collapse in a rigorous way.

A couple of points to make before we start: you're being vague about the distinction between the singularity and the event horizon. The singularity is the point at the centre of the black hole where the curvature becomes infinite. The event horizon is the spherical surface that marks the radial distance below which light cannot escape. As you'll see, these form at different times.

The other point is that to make the calculation possible at all we have to use a simplified model. Specifically we assume the collapsing body is homogeneous (actually I see you anticipated that in your answer) and is made up of dust. In general relativity the term dust has a specific meaning - it means matter that is non-interacting (apart from gravity) and has zero pressure. This is obviously very different from the plasma found in real stars.

With the above simplifications the collapse is described by the Oppenheimer-Snyder model, and it turns out that the size of the collapsing object is described by the same equation that describes the collapse of a closed universe. This equation is called the FLRW metric, and it gives a function called the scale factor, $a(t)$, that describes the size of the ball of dust. For a closed universe the scale factor looks something like:

(image from this PDF)

A closed universe starts with a Big Bang, expands to a maximum size then recollapses in a Big Crunch. It's the recollapse, i.e. the right hand side of the graph above, that describes the collapse of the ball of dust.

The radius of the ball is proportional to $a(t)$, so the radius falls in the same way as $a(t)$ does, and the singularity forms when $a(t) = 0$ i.e. when all the matter in the ball has collapsed to zero size.

As always in GR, we need to be very careful to define what we mean by time. In the graph above the time plotted on the horizontal axis is comoving or proper time. This is the time measured by an observer inside the ball and stationary with respect to the grains of dust around them. It is not the same as the time measured by an observer outside the ball, as we'll see in a bit.

Finally, we should note that the singularity forms at the same time for every comoving observer inside the ball of dust. This is because the ball shrinks in a homogeneous way so the density is the same everywhere inside the ball. The singularity forms when the density rises to infinity (i.e. the ball radius goes to zero), and this happens everywhere inside the ball at the same time.

OK, that describes the formation of the singularity, but what about the event horizon. To find the event horizon we look at the behaviour of outgoing light rays as a function of distance from the centre of the ball. The details are somewhat technical, but when we find a radius inside which the light can't escape that's the position of the event horizon. The details are described in the paper by Luciano Rezzolla that I linked above, and glossing over the gory details the result is:

This shows time on the vertical axis (Once again this is comoving/proper time as discussed above) and the radius of the ball of dust on the horizontal axis. So as time passes we move upwards on the graph and the radius decreases.

It's obviously harder for light to escape from the centre of the ball than from the surface, so the event horizon forms initially at the centre of the ball then it expands outwards and reaches the surface when the radius of the ball has decreased to:

$$r = \frac{2GM}{c^2}$$

This distance is called the Schwarzschild radius and it's the event horizon radius for a stationary black hole of mass $M$. So at this moment the ball of dust now looks like a black hole and we can no longer see what's inside it.

However note that when the event horizon reaches the Schwarzschild radius the collapse hasn't finished and the singularity hasn't formed. It takes a bit longer for the ball to finish contracting and the singularity to form. The singularity only forms when the red line meets the vertical axis.

Finally, one last note on time.

Throughtout all the above the time I've used is proper time, $\tau$, but you and I watching the collapse from outside measure Schwarzschild coordinate time, $t$, and the two are not the same. In particular our time $t$ goes to infinity as the ball radius approaches the Schwarzschild radius $r = 2GM/c^2$. For us the part of the diagram above this point simply doesn't exist because it lies at times greater than infinity. So we never actually see the event horizon form. I won't go into this any further here because it's been discussed to death in previous questions on this site. However you might be interested to note this is one of the reasons for Stephen Hawking's claim that event horizons never form.

• Thank you very much for nice reply and the detailed explanation of the terms. In the question I probably did not express enough that I expect that the event horizon and the singularity do not form simultaneously. ------ I did not expect that the matter can really simply continue to contract to a space of zero size. Does the theory describe the zero size because it is caused just by the extremely curved space (the matter in different coordinate system will not be at a single spot)? Or could the contraction just be possibly stopped by some unknown repulsive forces stronger than gravity? – pabouk Sep 30 '14 at 18:36
• I think that there is an illogicality or something unexplained in your description: I think that the singularity should form before the red line meets the vertical line (zero size). --- Why should the singularity form only after the matter from the stellar surface collapses to the central point? Before that moment in the central point there should already be the matter which was below the stellar surface. -- Or let's see it differently: When the star starts to collapse with enough of velocity we can start to remove some matter from the stellar surface and a smaller black hole will form anyway. – pabouk Sep 30 '14 at 18:50
• @pabouk: we are using a simplified model that assumes the collapsing object is homogeneous and made of non-interacting matter i.e. dust. In this case the object remains homogeneous at all stages of the collapse, so the singularity forms at a single point in (proper) time everywhere inside the collapsing body. A real star would have some density profile i.e. it would b denser at the centre than at the surface. In this case it's likely the singularity would form while the outer layers were still at non-zero $r$. – John Rennie Sep 30 '14 at 19:48
• hey wait .. "It's obviously harder for light to escape from the centre of the ball than from the surface, so the event horizon forms initially at the centre of the ball then it expands outwards and reaches the surface" - but wait, surely inside that smaller inner ball, there is considerably less mass, so why would that small inner ball form an event horizon? – Fattie May 14 '16 at 17:18
• @JoeBlow Let's say you know the doors to a building will be locked at 5pm. And you can't break through the doors. And you want to leave the building. And you have a watch. Then you simply need to start walking to the door at an earlier time when you are in the center of the building than you do when you are only 1 meter away from the door. That's all. It literally isn't any deeper than that – Timaeus May 16 '16 at 16:24

Each collapsing star will have its own time table depending on mass. This video simulation by NASA of two subcritical mass neutron stars collapsing towards each other gives a time line:

"At 13 milliseconds, the more massive star has accumulated too much mass to support it against gravity and collapses, and a new black hole is born," NASA officials said in a statement. "The black hole's event horizon — its point of no return — is shown by the gray sphere. While most of the matter from both neutron stars will fall into the black hole, some of the less-dense, faster-moving matter manages to orbit around it, quickly forming a large and rapidly rotating torus."

For these distances and these masses,

It starts off with two neutron stars — the city-size, dense remnants of a violent supernova explosion — separated by about 11 miles (18 kilometers), NASA officials said. One object contains about 1.7 times the mass of our sun, while the other weighs in at 1.4 solar masses.

the order of time for an event horizon to form is 13 milliseconds. What happens within is not accessible to mathematical simulation. The formation of the event horizon during the collapse of a star implies a black hole "singularity", as in the video, but singularities could exist without an event horizon.

In general relativity, a naked singularity is a gravitational singularity without an event horizon. In a black hole, the singularity is completely enclosed by a boundary known as the event horizon, inside which the gravitational force of the singularity is strong enough so that light cannot escape. Hence, objects inside the event horizon—including the singularity itself—cannot be directly observed. A naked singularity, by contrast, is observable from the outside.

• It is the existence of the even horizon that defines the black hole "singularity". No, the event horizon and the singularity are two different things. – Ben Crowell Sep 28 '14 at 22:19
• Thank you for the interesting article and video but I agree with Ben Crowell that event horizon is not enough for the singularity to exist. – pabouk Sep 30 '14 at 18:17


text:
# Families of Lines

## Key Questions

• A family of lines is a set of lines that have something in common with each other.

In brief, we can have two types of families of lines:

1. one where the slope is the same $\to$ keep the slope unchanged and vary the $y$-intercept $\to$ parallel lines

2. one where the $y$-intercept is the same $\to$ keep the $y$-intercept unchanged and vary the slope $\to$ concurrent lines

• All lines following a specific criteria are called a family of lines.

For eg.
If the criteria is passing through a point (0,0)
Then all possible lines passing through that point are called the family of lines.

If the criteria is have a slope of -1,
Then all possible lines have slope = -1 are its family of lines.

The idea is similar to human families where all our relatives belong to the same family (the criteria) and therefore everyone is a family member.

I hope this helps.


text:
# Compressive time-lapse seismic monitoring of carbon storage and sequestration with the joint recovery model

 Title Compressive time-lapse seismic monitoring of carbon storage and sequestration with the joint recovery model Publication Type Conference Year of Publication 2021 Authors Ziyi Yin, Mathias Louboutin, Felix J. Herrmann Conference Name SEG Technical Program Expanded Abstracts Month 09 Keywords CCS, Compressive Sensing, Imaging, JRM, marine, SEG, time-lapse Abstract Time-lapse seismic monitoring of carbon storage and sequestration is often challenging because the time-lapse signature of the growth of CO$_2$ plumes is weak in amplitude and therefore difficult to detect seismically. This situation is compounded by the fact that the surveys are often coarsely sampled and not replicated to reduce costs. As a result, images obtained for different vintages (baseline and monitor surveys) often contain artifacts that may be attributed wrongly to time-lapse changes. To address these issues, we propose to invert the baseline and monitor surveys jointly. By using the joint recovery model, we exploit information shared between multiple time-lapse surveys. Contrary to other time-lapse methods, our approach does not rely on replicating the surveys to detect time-lapse changes. To illustrate this advantage, we present a numerical sensitivity study where CO$_2$ is injected in a realistic synthetic model. This model is representative of the geology in the southeast of the North Sea, an area currently considered for carbon sequestration. Our example demonstrates that the joint recovery model improves the quality of time-lapse images allowing us to monitor the CO$_2$ plume seismically. Notes (IMAGE, Denver) URL https://slim.gatech.edu/Publications/Public/Conferences/SEG/2021/yin2021SEGcts/yin2021SEGcts.html DOI 10.1190/segam2021-3569087.1 Presentation URL2 Software Citation Key yin2021SEGcts


text:
TORONTO KIDS COMPUTER CLUB | Tuesday Python Homework 18:30 – 22.12.13.
21556

# Tuesday Python Homework 18:30 – 22.12.13.

## 14 Dec Tuesday Python Homework 18:30 – 22.12.13.

Question 1:
Being able to send encoded messages during World War II was very important to the Allies. The messages were always sent after being encoded with a known password. Having a fixed password was of course insecure, thus there was a need to change it frequently. However, a mechanism was necessary to send the new password. One of the mathematicians working in the cryptographic team had a clever idea that was to send the password hidden within the message itself. The interesting point was that the receiver of the message only had to know the size of the password and then search for the password within the received text.

A password with size N can be found by searching the text for the most frequent substring with N characters. After finding the password, all the substrings that coincide with the password are removed from the encoded text. Now, the password can be used to decode the message.

Your mission has been simplified as you are only requested to write a program that, given the size of the password and the encoded message, determines the password following the strategy given above.

To illustrate your task, consider the following example in which the password size is three (N = 3) and the text message is just ‘baababacb’. The password would then be aba because this is the substring with size 3 that appears most often in the whole text (it appears twice) while the other six different substrings appear only once (baa ; aab ; bab ; bac ; acb).

Input
The input file contains several test cases, each of them consists of one line with the size of the password, 0 < N ≤ 10, followed by the text representing the encoded message. To simplify things, you can assume that the text only includes lower case letters.

Output
For each test case, your program should print as output a line with the password string.

Sample Input
3 baababacb

Sample Output
aba

Question 2:
Ken is working for the secret agency. He is going to send a secret message that encodes UPPERCASE words by shifting their letters forward.

Shifting a letter by S positions means to go forward S letters in the alphabet. For example, shifting B by S = 3 positions gives E. However, sometimes this makes us go past Z, the last letter of the alphabet. Whenever this happens we wrap around, treating A as the letter that follows Z. For example, shifting Z by S = 2 positions gives B.

Ken’s code depends on a parameter K and also varies depending on the position of each letter in the word. For the letter at position P, they use the shift value of S = 2*K+3*P.

For example, here is how “FIND” is encoded when K = 3. The first letter F has a shift value of S = 2 × 3 + 3 × 1 = 9; it wraps around and becomes the letter O. The second letter, I, has S = 2 × 3 + 3 × 2 = 12 and becomes U. The last two letters become C and V. So Ken sends the secret message: OUCV

Write a program to encode messages sent by Ken.

Input Specification
The input will be two lines. The first line will contain the positive integer K (K < 10), which is used to compute the shift value. The second line of input will be the original word, which will be a sequence of uppercase characters of length at most 20.

Output Specification
The output will be the encoded word of uppercase letters.

Sample Input 1
3
FIND

Output for Sample Input 1
OUCV

Sample Input 2
5
MISSILE

Output for Sample Input 2
ZYLOHNJ

Question 3:

Today in gym class, your class decided to play dodgeball, and Jack is selected as one of the team captains! N students are lined up in a row, waiting to be put on teams. Jack is allowed to pick multiple students as long as they are adjacent to one another, and the first letter of each of their names are the same (case insensitive).

As an assistant to Jack, help him decide what his first choice should be by writing a program to determine all the possible choices he could make!

Input Specification
The first line contains the integer N (1N105), representing the total number of students.

The second line contains N space separated strings, each representing the name of the ith student in line. Each name will be no longer than 20 characters and only contain letters from the English alphabet.

Output Specification
Output one integer, representing the total number of choices Jack can make.

Sample Input
5
Sarah Timmy Turner Betty Bob

Sample Output
7

Explanation for Sample Input
The possible groups of people that aurpine could choose are:
Sarah
Timmy
Turner
Betty
Bob
Timmy, Turner
Betty, Bob


text:
# Acid bases help

1. Feb 6, 2005

### physicsss

can some tell me how to write net ionic equations of acid base recations in the cases of a weak base reacts with a strong acid, a strong base reacting with a weak acid, and weak base reacting with weak acid? How do you cancel things out?

Also, does NH3+HCl--->NH4Cl or NH3+HCl--->NH4+Cl-?

2. Feb 6, 2005

### chem_tr

Calcium hydroxide is a weak base than NaOH, let's try this with HCl:

$$Ca(OH)_2_{(aq)}+2HCl_{(aq)} \rightarrow CaCl_2_{(aq)} + H_2O_{(l)}$$

Here, you can show the ones with (aq) in ionized form, too.

About your question with ammonium chloride, ammonium does not produce water when neutralized. So either write it as (aq) or ionized form.

3. Feb 6, 2005

### physicsss

so both are correct?

4. Feb 6, 2005

### chem_tr

Yes, depending on your need.


text:
# Identity for simple 1D random walk

The question is to find a purely probabilistic proof of the following identity, valid for every integer $n\geqslant1$, where $(S_n)_{n\geqslant0}$ denotes a standard simple random walk:

$$E[(S_n)^2;S_{2n}=0]=\frac{n}2\,P[S_{2n-2}=0].$$

Standard simple random walk is defined as $S_0=0$ and $S_n=\sum\limits_{k=1}^nX_k$ for every $n\geqslant1$, where $(X_k)_{k\geqslant1}$ is an i.i.d. sequence such that $P[X_k=+1]=P[X_k=-1]=\frac12$.

Of course, the RHS of the identity is $$\frac{n}{2^{2n-1}}\,{2n-2\choose n-1}.$$ For a combinatorial proof, see this MSE question and its comments.

For an accessible introduction to the subject, see the corresponding chapter of the Chance project.

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Here, are you using $\mathbb{E}[X; A]$ to denote the integral taken only over the subset $A$? – Nick Peterson Jul 14 '13 at 13:38
@nrpeterson Yes, $E[X;A]=E[X\mathbf 1_A]$. – Did Jul 14 '13 at 21:28
This is an interesting question... I'll do some thinking on it. – Nick Peterson Jul 14 '13 at 21:36

I don't know if what I will write is a "purely probabilistic proof" as the question requests, or a combinatorial proof, but Did should decide that. At the end I do use combinatorial identities (UPDATE 12-1-2014: an alternative final step of the proof has been found that does not use the identities. The initial final step is preserved at the end, in grey).

The LHS of the identity can be re-written as

$$E[(S_n)^2;S_{2n}=0] = E\left[\left(\sum_{i=1}^nX_i\right)^2 ; S_{2n}=0\right]=E\left[\sum_{i=1}^nX_i^2 + \sum_{i\neq j}X_iX_j ; S_{2n}=0\right]$$ hence $$E[(S_n)^2;S_{2n}=0]=E\left[\sum_{i=1}^nX_i^2 ; S_{2n}=0\right]+ E\left[\sum_{i\neq j}X_iX_j ; S_{2n}=0\right]$$

But $X_i^2=1$ almost surely for every $i$. Likewise, the distribution of $(X_i,X_j)$ conditionally on $S_{2n}=0$ does not depend on $i\ne j$ since the $X_i$s are i.i.d and $S_{2n}$ is a symmetric function of $(X_i)$. So we arrive at

$$E[(S_n)^2;S_{2n}=0] = nP[S_{2n}=0] +n(n-1)E\left(X_{2n-1}X_{2n}; S_{2n}=0\right)$$

The random variable $X_{2n-1}X_{2n}$ takes on the values $\{-1,1\}$. Let $q \equiv P[X_{2n-1}X_{2n}=-1; S_{2n}=0]$. Then $P[X_{2n-1}X_{2n}=1; S_{2n}=0]=P[S_{2n}=0]-q$ hence $$E\left(X_{2n-1}X_{2n};S_{2n}=0\right) = -1\cdot q + 1\cdot(P[S_{2n}=0]-q) = P[S_{2n}=0]-2q$$

Note that if $X_{2n-1}X_{2n}=-1$ then $X_{2n-1}+X_{2n} = 0$ and $S_{2n-2}=S_{2n}$, therefore $$q= P[S_{2n-2}=0,S_{2n}=0]= P[S_{2n-2}=0,X_{2n-1}+X_{2n} = 0]= \frac 12 P[S_{2n-2}=0]$$ the last equality being due to the fact that $S_{2n-2}$ and $X_{2n-1}+X_{2n}$ are independent and that $P[X_{2n-1}+X_{2n} = 0]= \frac 12$. So $$E\left(X_{2n-1}X_{2n}; S_{2n}=0\right) =P[S_{2n}=0]-P[S_{2n-2}=0]$$

Inserting into the main expression and multiplying out we have

$$E[(S_n)^2;S_{2n}=0] = nP[S_{2n}=0] +n(n-1)\Big (P[S_{2n}=0] - P[S_{2n-2}=0]\Big)$$

$$=n^2P[S_{2n}=0] - n(n-1)P[S_{2n-2}=0] \tag{A}$$

Guided by what we want to prove, the issue now is to express $P[S_{2n}=0]$ in terms of $P[S_{2n-2}=0]$. We note that the event $\{S_{2n}=0\}$ can only occur and so has strictly positive probability, if and only if one of the following events occur:

1) $E_0 \equiv [\{S_{2n-2}=0\}\; \cap \{X_{2n-1}+X_{2n}= 0\}]$
2) $E_1 \equiv [\{S_{2n-2}=2\}\; \cap \{X_{2n-1}+X_{2n}= -2\}]$
3) $E_2 \equiv [\{S_{2n-2}=-2\}\; \cap \{X_{2n-1}+X_{2n}= 2\}]$

The events inside the brackets, if we look at them as elements of a $3\times 2$ matrix, are (pardon my language) ... "horizontally independent and vertically mutually exclusive". Combined these imply that $$P[S_{2n}=0] = P[E_0 \cup E_1 \cup E_2] = P[E_0] + P[E_1]+P[E_2]$$

while the "horizontal" independence implies that the joint probabilities can be separated as the product of the probabilities of the two events involved in each case. More over, by symmetry we have that $P[S_{2n-2}=-2]= P[S_{2n-2}=2]$, while
$P[X_{2n-1}+X_{2n}= 0]= 1/2$ , $P[X_{2n-1}+X_{2n}= -2]= 1/4$ , $P[X_{2n-1}+X_{2n}= 2]= 1/4$. Using all these results we arrive at

$$P[S_{2n}=0] = \frac 12 P[S_{2n-2}=0] + \frac 12 P[S_{2n-2}=2]$$

Inserting this into equation $[A]$ we obtain

$$E[(S_n)^2;S_{2n}=0] = n^2\left( \frac 12 P[S_{2n-2}=0] + \frac 12 P[S_{2n-2}=2]\right) - n(n-1)P[S_{2n-2}=0]$$

$$\Rightarrow E[(S_n)^2;S_{2n}=0] = \frac n2 (2-n)P[S_{2n-2}=0] + \frac {n^2}2P[S_{2n-2}=2] \tag{B}$$

Equation $[B]$ is an improvement over eq. $[A]$ because, through probabilistic arguments, only $S_{2n-2}$ is now present in the RHS. Now, if we denote by $Y_k$ a binomial random variable with probability of success $1/2$, $Y_k = Bin(k,1/2)$, then, the probabilities related to $S_{2n-2}$ can be expressed in terms of $Y_{2n-2}$,

$$P[S_{2n-2}=0] = P[Y_{2n-2}=n-1],\;\; P[S_{2n-2}=2] = P[Y_{2n-2}=n]$$

It is then a natural step to use the probability generating function of $Y_{2n-2}$, which is

$$G_Y(z) = \left (\frac 12 +\frac 12z\right)^{2n-2}$$ while we have

$$P[S_{2n-2}=0] = P[Y_{2n-2}=n-1] = \frac 1 {(n-1)!}G_Y^{(n-1)}(0)$$ $$P[S_{2n-2}=2] = P[Y_{2n-2}=n] = \frac 1 {n!}G_Y^{(n)}(0)$$

where the superscript parentheses denote the order of derivative taken. But

$$G_Y^{(n)}(0) = (n-1)G_Y^{(n-1)}(0) = (n-1)\cdot[(n-1)!]P[Y_{2n-2}=n-1]$$

and so

$$P[S_{2n-2}=2] = P[Y_{2n-2}=n] = \frac {n-1}{n} P[Y_{2n-2}=n-1] = \frac {n-1}{n}P[S_{2n-2}=0]$$

Inserting this into equation $[B]$ we obtain

$$E[(S_n)^2;S_{2n}=0] = \frac n2 (2-n)P[S_{2n-2}=0] + \frac {n^2}2\frac {n-1}{n}P[S_{2n-2}=0]$$

$$\Rightarrow \frac n2P[S_{2n-2}=0] \cdot (2-n+n-1) = \frac n2P[S_{2n-2}=0]$$

which is what we wanted to prove. One could argue that the use of the probability generating function is too "algebraic" a step, and that in general, the moment we enter binomial distribution territory, combinatorics rage in the background -but at least now they rage only in the background.

--

And don't forget, on the side, rearranging the relations above we can obtain the conditional expectation

$$E[(S_n)^2 \mid S_{2n}=0] = \frac {n^2}{2n-1}$$

INITIAL final step of the proof (using combinatorial identities)

We have arrived at equation $[A]$.

Using $Y_k$ to denote a binomial random variable with probability of success $1/2$, $Y_k = B(k,1/2)$, the probabilities related to $S_n$ can be expressed as probabilities of binomial random variables, $$P[S_{2n}=0] = P[Y_{2n}=n],\qquad P[S_{2n-2}=0] =P[Y_{2n-2}=n-1]$$

Since we want at the end to have only $P[S_{2n-2}=0]$ present, we must somehow express $P[Y_{2n}=n]$ in terms of $P[Y_{2n-2}=n-1]$, and here is where I use the combinatorial identities, since one can arrive through them at

$$P[Y_{2n}=n] = \frac 1{2^{2n}}{2n \choose n} = \frac 12 \frac {2n-1}{n}\frac 1{2^{2n-2}}{2n-2 \choose n-1} = \frac {2n-1}{2n}P[Y_{2n-2}=n-1]$$

Reverting back to $S$- notation and inserting in the main expression, we have

$$E[(S_n)^2;S_{2n}=0] = n^2\frac {2n-1}{2n}P[S_{2n-2}=0] - n(n-1)P[S_{2n-2}=0]$$

$$=\frac{n}2\,P[S_{2n-2}=0]\Big (2n-1 - 2(n-1)\Big) = \frac{n}2\,P[S_{2n-2}=0]$$

which is what we wanted to prove. Maybe there are other ways to relate $P[S_{2n}=0]$ and $P[S_{2n-2}=0]$, but they currently escape me.

-
This is excellent. I rewrote marginally the post to avoid the conditioning/deconditioning steps, which (I think) were unnecessary. +1 of course. If no further answer is posted, I plan to "accept" this one in a while. – Did Jan 9 '14 at 8:32
Thank you. I do appreciate also the re-writing, I used the conditional framework only because I am a bit more familiar with it compared to the "censored" one - but it is now evident it wasn't necessary. – Alecos Papadopoulos Jan 9 '14 at 9:25
@Did An ex post bounty award is a very gratifying gesture, although you had already clearly stated your appreciation for this effort of mine. Thank you Did. – Alecos Papadopoulos Jan 11 '14 at 19:43
The bounty is well earned--although, to be honest, and as you stress yourself, the step from $P[S_{2n-2}=n-1]$ to $P[S_{2n}=n]$ uses a not "purely probabilistic" workaround. One can complete this step using the generating function of the first return time to $0$ and such, but the ways I see to do so are still all, at the moment, slightly artificial. – Did Jan 12 '14 at 16:09
@Did I added an alternative final step of the proof, a bit more probabilistic I think. – Alecos Papadopoulos Jan 13 '14 at 0:31


text:
From SU(1,1) to the Lorentz group

From the two-dimensional disk we are moving to three-dimensional space-time. We will meet Einstein-Poincare-Minkowski special relativity, though in a baby version, with x and y, but without z in space. It is not too bad, because the famous Lorentz transformations, with length contraction and time dilation happen already in two-dimensional space-time, with x and t alone. We will discover Lorentz transformations today. First in disguise, but then we will unmask them.
[latexpage]
First we recall, from The disk and the hyperbolic model, the relation between the coordinates $(x,y)$ on the Poincare disk $x^2+y^2<1$, and $(X,Y,T)$ on the unit hyperboloid $T^2-X^2-Y^2=1.$ [caption id="attachment_3054" align="aligncenter" width="640"] Space-time hyperboloid and the Poincare disk models[/caption]

\begin{eqnarray}
X&=&\frac{2x}{1-x^2-y^2},\\
Y&=&\frac{2y}{1-x^2-y^2},\\
T&=&\frac{1+x^2+y^2}{1-x^2-y^2}.
\label{eq:XYT1}\end{eqnarray}

\begin{eqnarray}
x&=&\frac{X}{1+T},\\
y&=&\frac{Y}{1+T}\label{eq:Xx1}.
\end{eqnarray}

We have the group SU(1,1) acting on the disk with fractional linear transformations. With $z=x+iy$ and $A$ in SU(1,1)
$$A=\begin{bmatrix}\lambda&\mu\\ \nu&\rho\end{bmatrix},$$
the fractional linear action is
$$A:z\mapsto z_1=\frac{\rho z+\nu}{\mu z+\lambda}.$$
By the way, we know from previous notes that $A$ is in SU(1,1) if and only if
$$\nu=\bar{\mu},\,\rho=\bar{\lambda},\quad|\lambda|^2-|\mu|^2=1.$$

Having the new point on the disk, with coordinates $(x_1,y_1)$ we can use Eq. (\ref{eq:XYT1}) to calculate the new space-time point coordinates $(X_1,Y_1,T_1).$ This is what we will do now. We will see that even if $z_1$ depends on $z$ in a nonlinear way, the space-time coordinates transform linearly. We will calculate the transformation matrix $L(A),$ and express it in terms of $\lambda$ and $\mu.$ We will also check that this is a matrix in the group SO(1,2).

The program above involves algebraic calculations. Doing them by hand is not a good idea. Let me recall a quote from Gottfried Leibniz, who, according to Wikipedia

He became one of the most prolific inventors in the field of mechanical calculators. While working on adding automatic multiplication and division to Pascal’s calculator, he was the first to describe a pinwheel calculator in 1685[13] and invented the Leibniz wheel, used in the arithmometer, the first mass-produced mechanical calculator. He also refined the binary number system, which is the foundation of virtually all digital computers.

I used Mathematica as my machine. The same calculations can be certainly done with Maple, or with free software like Reduce or Maxima. For those interested, the code that I used, and the results can be reviewed as a separate HTML document: From SU(1,1) to Lorentz.

Here I will provide only the results. It is important to notice that while the matrix $A$ has complex entries, the matrix $L(A)$ is real. The entries of $L(A)$ depend on real and imaginary parts of $\lambda$ and $\mu$

$$\lambda=\lambda_r+i\lambda_i,\, \mu=\mu_r+i\mu_i.$$

Here is the calculated result for $L(A)$:

L(A)=\begin{bmatrix}
-\lambda_i^2+\lambda_r^2-\mu_i^2+\mu_r^2 & 2 \lambda_i \lambda_r-2 \mu_i \mu_r & 2 \lambda_r \mu_r-2 \lambda_i \mu_i \\
-2 \lambda_i \lambda_r-2 \mu_i \mu_r & -\lambda_i^2+\lambda_r^2+\mu_i^2-\mu_r^2 & -2 \lambda_r \mu_i-2 \lambda_i \mu_r \\
2 \lambda_i \mu_i+2 \lambda_r \mu_r & 2 \lambda_i \mu_r-2 \lambda_r \mu_i & \lambda_i^2+\lambda_r^2+\mu_i^2+\mu_r^2
\label{eq:a2l}\end{bmatrix}

In From SU(1,1) to Lorentz it is first verified that the matrix $L(A)$ is of determinant 1. Then it is verified that it preserves the Minkowski space-time metric. With $G$ defined as
$$G=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1\end{bmatrix}$$
we have
$$L(A)GL(A)^T=G.$$
Since $L(A)_{3,3}= \lambda_i^2+\lambda_r^2+\mu_i^2+\mu_r^2 =1+2|\mu|^2\geq 1>0,$ the transformation $L(A)$ preserves the time direction. Thus $L(A)$ is an element of the proper Lorentz group $\mathrm{SO}^{+}(1,2)$.

Remark: Of course we could have chosen $G$ with $(+1,+1,-1)$ on the diagonal. We would have the group SO(2,1), and we would write the hyperboloid as $X^2+Y^2-T^2=-1.$ It is a question of convention.

In SU(1,1) straight lines on the disk we considered three one-parameter subgroups of SU(1,1):

\begin{eqnarray}
X_1&=&\begin{bmatrix}0&i\\-i&0\end{bmatrix},\\
X_2&=&\begin{bmatrix}0&1\\1&0\end{bmatrix},\\
X_3&=&\begin{bmatrix}i&0\\0&-i\end{bmatrix}.
\end{eqnarray}
\begin{eqnarray}
A_1(t)&=&
\exp(tX_1)=\begin{bmatrix}\cosh(t)&i\sinh(t)\\-i\sinh(t)&\cosh(t)\end{bmatrix},\\
A_2(t)&=&
\exp(tX_2)=\begin{bmatrix}\cosh(t)&\sinh(t)\\ \sinh(t)&\cosh(t)\end{bmatrix},\\
A_3(t)&=&
\exp(tX_3)=\begin{bmatrix}e^{it}&0\\ 0&e^{-it}\end{bmatrix}.
\end{eqnarray}

We can now use Eq. (\ref{eq:a2l}) in order to see which space-time transformations they implement. Again I calculated it obeying Leibniz and using a machine (see From SU(1,1) to Lorentz).

Here are the results of the machine work:

L_1(t)=L(A_1(t))=\begin{bmatrix}
1 & 0 & 0 \\
0 & \cosh (2 t) & -\sinh (2 t) \\
0 & -\sinh (2 t) & \cosh (2 t)\end{bmatrix},

L_2(t)=L(A_2(t))=\begin{bmatrix}
\cosh (2 t) & 0 & \sinh (2 t) \\
0 & 1 & 0 \\
\sinh (2 t) & 0 & \cosh (2 t)
\end{bmatrix},
L_3(\phi)=L(A_3(\phi))=\begin{bmatrix}
\cos (2 \phi ) & \sin (2 \phi ) & 0 \\
-\sin (2 \phi ) & \cos (2 \phi ) & 0 \\
0 & 0 & 1 \\
\end{bmatrix},
The third family is a simple Euclidean rotation in the $(X,Y)$ plane. That is why I denoted the parameter with the letter $\phi.$ In order to “decode” the first two one-parameter subgroups it is convenient to introduce new variable $v$ and set $2t=\mathrm{arctanh}(v).$ The group property $L(t_1)L(t_2)=L(t_1+t_2)$ is then lost, but the matrices become evidently those of special Lorentz transformations, $L_1(v)$ transforming $Y$ and $T$, leaving $X$ unchanged, and $L_2(v)$ transforming $(X,T)$ and leaving $Y$ unchanged (though with a different sign of $v$). Taking into account the identities
\begin{eqnarray}
\cosh(\mathrm{arctanh} (v))&=&\frac{1}{\sqrt{1-v^2}},\\
\sinh(\mathrm{arctanh} (v))&=&\frac{v}{\sqrt{1-v^2}}
\end{eqnarray}
we get
L_1(v)=\begin{bmatrix}
1 & 0 & 0 \\
0 & \frac{1}{\sqrt{1-v^2}} & -\frac{v}{\sqrt{1-v^2}} \\
0 & -\frac{v}{\sqrt{1-v^2}} & \frac{1}{\sqrt{1-v^2}}\end{bmatrix},

L_2(v)=\begin{bmatrix}
\frac{1}{\sqrt{1-v^2}} & 0 & \frac{v}{\sqrt{1-v^2}} \\
0 & 1 & 0 \\
\frac{v}{\sqrt{1-v^2}} & 0 & \frac{1}{\sqrt{1-v^2}}
\end{bmatrix},

In the following posts we will use the relativistic Minkowski space distance on the hyperboloid for finding the distance formula on the Poincare disk.

When China Rules the World

The title of the post can be found on Wikipedia with additional information. But this post is more personal. Indeed China rulez! Few weeks ago I received the following email:

Quantum World-2017

Time: 16th-18th October, 2017

Place: Changsha, Hunan Province, China

On behalf of the organizing committee of CQW-2017, we sent you a letter a few days ago invited you to join us and give a speech at first Annual Conference of Quantum World (CQW-2017), which will be held on 16th-18th October, in Changsha, Hunan Province, China. It seems you have not received that letter yet, so I am writing again to extend our sincere invitation. As we have learnt your valuable contribution to Asymptotic Formula for Quantum Harmonic Oscillator Tunneling Probabilities…, we believe your inspirational speech and participation will highlight this congress a lot!

Under the theme “From E=MC2 to Quantum Industry”, the first Annual Conference of Quantum World (CQW-2017) aims at 200+ oral presentations in Quantum Physics and Mechanics, Quantum Information Science, Quantum Chemistry, Quantum Optics, Quantum Materials, and Quantum System, Quantum Engineering and Application, which cover hot topics with both theoretical and experimental contributions.

The conference venue Changsha, as the capital of Hu’nan province, is a beautiful, creative, historical and cultural city with comfortable climate, unique scenery and convenient transportation. It will give you a special experience on the colligation and integration of the Huxiang Culture with the modern civilization.

Worth mentioning that partial scientific program with speakers’ profile and excellent speech titles has been updated on website, kindly click here to view and give us your valuable advice.

Sincerely yours,

CQW-2017 Organizing Committee

Of course I was surprised, because I do not expect anybody but few experts in the whole world care about my paper. Looks to me like a huge conference industry.

In my blog post Lorentz transformation from an elementary point of view – from blogging to science publishing I wrote about a paper that came as the result of blogging. This is the second paper that I wrote together with prof. Jerzy Szulga, a continuation of our previous paper, A Comment on “On the Rotation Matrix in Minkowski Space-time” by Ozdemir and Erdogdu, http://arxiv.org/abs/1412.5581, Reports on Mathematical Physics, 74(1), 2014, 39-44,
DOI: 10.1016/S0034-4877(14)60056-2. Today I received from my coauthor a message with a copy of another invitation from China:

Dear Dr. ….,

I’m writing to follow-up my last invitation as below, would you please give me a tentative reply? Thank you very much. I apologize for the inconvenience if the letter disturbed you more than once.

It is our great pleasure and privilege to welcome you to join the 8th World Gene Convention-2017, which will take place in Macao, China during November 13-15, 2017. We would like to welcome you to be the chair/speaker in Theme 902: Agriculture, Food and Plant Biotechnology while presenting about A Comment on “On the Rotation Matrix in Minkowski Space-Time” by Ozdemir and Erdogdu…….

If the suggested thematic session is not your current focused core, you may look through the whole sessions and transfer another one that fit your interest (more info about the program is available athttp://www.bitcongress.com/wgc2017/ProgramLayout.asp

Under our SAB members’ contributions and endeavor, BIT’s 7th World Gene Convention-2016 (WGC-2016), successfully held in Shanghai, China during November 13-15, 2016. Totally, there were nearly 300 world-renowned experts, professors, laboratory principals, project leaders and representatives of well-known enterprises attended the WGC-2016. Participants from the international enterprises, academic and research institutions enjoyed the three days scientific program. Depending on the warmly support and good suggestions from all of the participants, we are confident in organizing WGC-2017 which would be better and more successful than WGC-2016.

WGC-2017 features a very strong technical program, mainly focused on: breakthroughs in gene, advances genomics & genetics, new research of DNA and RNA, focus on basic research, the frontier research of life sciences, new biotherapy discovery, emerging areas for medicine applications, robust technology development, and cutting-edge Biotechnology. It aims to provide a platform for all experts from academia, industry and national labs to discuss latest hot researches and achievements. Attendees will hear world-class speakers discussing the challenges and opportunities facing the gene, biotechnology and life sciences field. The business & academic experts who are from home and abroad will give excellent speeches.

In addition to the dynamic scientific program, you will benefit from the wonderful experience in Macao, China. Macao is an international free port. It’s famous for light industry, tourism, and hotel. Macao is also one of the most developed and richest regions in the world, this is a city of amazing and fascinating cultural wealth. The unique blend of European and Oriental cultures existing here creates a pleasurable and laid back atmosphere in a truly beautiful city. We hope you will enjoy your stay in this beautiful city with all its feature, beauty, architecture and hospitality!

We expect your precious comments or suggestions; also your reference to other speakers will be highly appreciated. We look forward to receiving your replies on the following questions:

1. What is the title of your speech?

2. Do you have any suggestions about our program?

We look forward to see you in Macao in 2017 for this influential event.

Sincerely yours,

Ms. Teresa Xiao

Organizing Committee of WGC-2017

It is only Chinese people can have such a broad and brave imagination and vision of the future. To connect Lorentz and spin groups of matrices with genes and agriculture – it is a real feat.

Elliptic m-deformed relativity

[latexpage]
According to Wikipedia special relativity theory was originally proposed in 1905 by Albert Einstein. But Wikipedia is not always the best source of information. For instance Wikipedia has a section about “Causality and prohibition of motion faster than light“. Quite often we can read sentences like that one:

” Since the moving clouds travel slightly slower than the speed of light, they do not actually violate Einstein’s theory of relativity which sets light as the speed limit.”

It continues to be alleged that superluminal inﬂuences of any sort would be inconsistent with special relativity for the following three reasons: (i) they would imply the existence of a ‘distinguished’ frame; (ii) they would allow the detection of absolute motion; and (iii) they would violate the relativity of simultaneity. This paper shows that the ﬁrst two objections rest upon very elementary misunderstandings of Minkowski geometry and lingering Newtonian intuitions about instantaneity. The third objection has a basis, but rather than invalidating the notion of faster-than-light inﬂuences it points the way to more general conceptions of simultaneity that could allow for quantum nonlocality in a natural way.

The point is that very often physicists do not think. They repeat what someone told them, or what they read, without much thinking. To quote from “Superluminal motions?A bird-eye view of the experimental situation“, Found.Phys.31:1119-1135,2001, by Erasmo Recami

… Still in pre-relativistic times, one meets various related works, from those by J.J.Thomson to the papers by the great A.Sommerfeld. With Special Relativity, however, since 1905 the conviction spread over that the speed c of light in vacuum was the upper limit of any possible speed. For instance, R.C.Tolman in 1917 believed to have shown by his “paradox” that the existence of particles endowed with speeds larger than c would have allowed sending information into the past. Such a conviction blocked for more than half a century (aside from an isolated paper (1922) by the Italian mathematician G.Somigliana) any research about Superluminal speeds.

Science is not free from “religious wars”. But that is not the subject of my post today. My post is about a certain curious observation that gave me some idea, and I do not know whether this idea is new, or it already occurred to someone else before. And I do not care, because the idea may be not crazy enough to be worth of discussing. Nevertheless it fits the subject of discussion in my recent series, so I will tell it to you now, and, perhaps, ask some questions.

In Special relativity we have a strange formula for addition of velocities (here we will discuss only velocities in one space dimension):

Q & A: Relativistic velocity addition

To simplify the notation I will assume that $c=1,$ or, if you wish, I will understand my velocity $\beta$ as the quotient $u/c$ etc.
The relativistic addition of velocities is sometimes denoted as $u\oplus v$
$$\beta\oplus \beta’=\frac{\beta+\beta’}{1+\beta\beta’}.\label{eq:op}$$
John Baez, whom we know from my previous posts, has a web page on How Do You Add Velocities in Special Relativity? There he notices the well know fact that the relativistic addition of velocities is essentially the same as for hyperbolic tangent, where we have
$$\tanh (x+y)=\frac{\tanh x +\tanh y}{1+\tanh x\tanh y}.\label{eq:th}$$

One of the consequences of the above addition formula is that if, say $\beta=0.9$ and $\beta’=0.9$ then $\beta\oplus \beta’=0.994475.$
Your spaceship moves with respect to the Sun with velocity that of 90% of the speed of light, and you send from it, in the direction of its flight, a missile traveling with respect to the spaceship with another 90% speed of light, and yet, with respect to the Sun the missile has the speed of 99% of the speed of light, rather than 180% as we would expect from naive addition.

Now, in the recent series of posts we were discussing elliptic functions, and in particular Jacobi sinus function $\sn(u,m).$ We know that for the parameter $m=1$ we have $\sn(u,m)=\tanh u.$ We also have addition formula for $\sn(u,m)$. It is thus natural to ask how would special relativity look like when the formula (\ref{eq:op}) is replaced by one derived from the addition formula for $\sn(u,m)$ given in the post Elliptic addition theorem:

\mathrm{sn} (u+v,m)=\frac{\mathrm{sn}(u,m)\mathrm{cn}(v,m)\mathrm{dn}(v,m)+\mathrm{sn}(v,m)\mathrm{cn}(u,m)\mathrm{dn}(u,m)}{1-m\,\mathrm{sn}^2(u,m)\,\mathrm{sn}^2(v,m)}.
We can set $\beta=\sn(u,m),\, \beta’=\sn(v,m),$ then $\cn(u,m)=\sqrt{1-\beta^2},\, \dn(u,m)=\sqrt{1-m\beta^2},$ $\cn(v,m)=\sqrt{1-\beta’^2},\, \dn(v,m)=\sqrt{1-m\beta’^2},$ and the new, proposed addition formula, involving parameter $m$ not ncessrily equal to 1, reads:
\beta\oplus_m\beta’=\frac{\beta\sqrt{1-\beta’^2}\sqrt{1-m\beta’^2}+\beta’\sqrt{1-\beta^2}\sqrt{1-m\beta^2}}{1-m\beta^2\beta’^2}.
That is my candidate for the m-deformed relativity. How it compares with the non-deformed (that is “standard”) relativity? It looks weird.
Assume our space-ship travels with the speed 90% of the speed of light. Assume $m=0.9$, and assume we shoot a missile from our ship, in the direction of its motion. What will be the speed of the missile? Here are the plots:

The blue curve is the special relativity. The $0.9\oplus \beta’$ speed always increases, though slower and slower as $\beta’$ approaches 1. But the m-deformed relativity, represented by the red curve is even crazier. If the missile is shot with a speed over a certain value, it starts to move slower with respect to the Sun.

Is that crazy enough to have a chance to be useful?

Can these elliptically deformed addition formulas be included in some geometrical setting? Will it follow from some algebra involving a generalization of the Lorentz group? I do not know.


text:
# Definition:Elementary Abelian P-Group

Let $G$ be an abelian group.
Then $G$ is an elementary abelian $p$-group if and only if all its non-trivial elements are of order $p$.


text:
# [NTG-context] Asymptote or Metapost and ConTeXt parameters.

Sat Dec 24 04:05:21 CET 2016

```On Sat, 24 Dec 2016, Jaroslav Hajtmar wrote:

> It seems now, that everything at this time works as intended. Many
> thanks for your support. Below is a minimal example. Attached is a PDF
> sample result.

Nice to know that it works.

> I tried also to experiment with the choice "object = no", as suggested
> by Hans, but without success.

Note that it is "object=no" (without spaces around equal to)

> I would asking if you make any changes in filter module?
> Or have I misunderstood Your opinion?

I will add the code between \unprotect ... \protect to the filter module after
I have tested it a bit more thoroughly.


text:
Area of a circumcenter triangle equals area of medial triangle

Let $X$, $Y$, $Z$ be the midpoints of sides $BC$, $AC$, $AB$ respectively in triangle $ABC$. Let $O_{A}$, $O_{B}$, and $O_{C}$ be the circumcenters of triangles $AZX$, $BXY$, and $CYZ$ respectively. Prove that the area of triangle $O_{A} O_{B} O_{C}$ equals the area of triangle $XYZ$.

Basically, I've done a horrific coordinate bash, and the statement is definitely true. But I'd be very interested in seeing a purely Euclidean solution.

• What is the source of this problem? – Ahaan S. Rungta Dec 20 '14 at 5:51
• This is something I conjectured was true, then confirmed via coordinates, then could not solve with pure geometry. – Sameer Kailasa Dec 20 '14 at 15:39
• Then why does this have the contest-math tag? Is this from an ongoing contest? I'm afraid nobody will want to help you without further details. – Ahaan S. Rungta Dec 20 '14 at 15:58
• It's not from an ongoing contest. The contest-math tag has description "problems from or inspired by mathematics competitions." This is certainly inspired by olympiad geometry problems. – Sameer Kailasa Dec 20 '14 at 15:59
• Would you like to show us your work in “co-ordinate bash” supporting your claim is definitely true? – Mick Dec 21 '14 at 2:57

We assign coordinates $A(0,0)$, $B(b,0)$, $C(c,d)$ to the vertices of triangle $ABC$. This gives $X = ((b+c)/2, d/2)$, $Y = (c/2, d/2)$, $Z = (b/2, 0)$. $O_A$ has coordinates corresponding to the intersection of the perpendicular bisector of $AZ$ and $AX$. These bisectors have equations $x = b/4$ and $$y = -\frac{b+c}{d} \left(x-\frac{b+c}{2} \right)+ \frac{d}{4}$$ Solving the system gives $$O_{A} = \left(\frac{b}{4}, \frac{d}{4} - \frac{bc+c^2}{4d} \right)$$ Similarly, we compute $O_{B}$ and $O_{C}$: $$O_{B} = \left(\frac{2c+b}{4}, \frac{3bc+d^2-c^2-2b^2}{4d} \right)$$ $$O_{C} = \left(\frac{b^2 + 2(c^2 + d^2)}{4b}, -\frac{b^2 c+ (2c-3b)(c^2 + d^2)}{4bd} \right)$$ Plugging these in to the shoelace formula and simplifying with a CAS yields $$[O_{A} O_{B} O_{C}] = \frac{1}{4} \left(\frac{1}{2} bd\right) = \frac{1}{4} [ABC] = [XYZ]$$ as desired.


text:
# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Corollary 4.5.8.14. Let $f: A \hookrightarrow B$ be a right anodyne morphism of simplicial sets. Then the induced map

$\theta : B \coprod _{A} (A \diamond \Delta ^0) \hookrightarrow B \diamond \Delta ^0$

is a categorical equivalence of simplicial sets.

Proof. Proposition 4.3.6.4 guarantees that the natural map $B \coprod _{A} A^{\triangleright } \hookrightarrow B^{\triangleright }$ is inner anodyne, and therefore a categorical equivalence (Corollary 4.5.3.14). Using Proposition 4.5.4.11, we conclude that the diagram

$\xymatrix@R =50pt@C=50pt{ A \ar [r] \ar [d]^{f} & A^{\triangleright } \ar [d]^{f^{\triangleright }} \\ B \ar [r] & B^{\triangleright } }$

is categorical pushout square. It then follows from Theorem 4.5.8.8 and Proposition 4.5.4.9 that the equivalent diagram

$\xymatrix@R =50pt@C=50pt{ A \ar [r] \ar [d]^{f} & A \diamond \Delta ^0 \ar [d]^{f \diamond \operatorname{id}_{\Delta ^{0}}} \\ B \ar [r] & B \diamond \Delta ^{0} }$

is also categorical pushout square, so that $\theta$ is a categorical equivalence by virtue of Proposition 4.5.4.11. $\square$