# Thread:

1. ## circle

Let C be the circle in R² having the point (h, k) and (0, 1) as diameter. Prove that this circle intersects the x-axis if and only if h²-4k≥0 and in this case the two intercepts are the roots of the equation x²-hx+k=0.

2. First, it doesn't make sense. I am guessing the diameter is the line segment between (h,k) and (0,1)

Second, isn't this just an algebra problem?

Center of Circle

$\left(\frac{h}{2},\frac{k+1}{2}\right)$

Radius of Circle

$\sqrt{(h-0)^{2}+(k-1)^{2}}$

Where's the tricky part?

3. ## confused

i cant even understand the question is that the question says that find the radius/

4. Originally Posted by apple2009
Let C be the circle in R² having the point (h, k) and (0, 1) as diameter.
As TKHunny says, two "points" do not form a diameter- you mean the line segment having those points as endpoints.

Prove that this circle intersects the x-axis if and only if h²-4k≥0 and in this case the two intercepts are the roots of the equation x²-hx+k=0.
The center of a circle is the midpoint of any diameter so this circle has $\left(\frac{h+0}{2}, \frac{k+1}{2}\right)= \left(\frac{h}{2}, \frac{k+1}{2}\right)$ as center.

The radius of a circle is the distance from the center to any point on the circumference. Since $\left(\frac{h}{2}, \frac{k+1}{2}\right)$ is the center and (0, 0) is on the circumference, the radius is $\sqrt{\frac{h^2}{4}+ \frac{(k+1)^2}{4}}$

The equation of a circle with center at (a, b) and radius r is $(x- a)^2+ (y- b)^2= r^2$ so the equation of this circle is $\left(x- \frac{h}{2}\right)^2+ \left(y- \frac{k+1}{2}\right)^2= \frac{h^2+ (k+1)^2}{4}$.

The first thing we can do is multiply both sides of that by 4, leaving $(2x- h)^2+ (2y- (k+1))^2= h^2+ (k+1)^2$.

The circle "intersects the x-axis" where y= 0. Put y= 0 into that equation and solve for x. That will be a quadratic equation so it will have two distinct solutions if and only its discrimant is 0. (The "discriminant" of the quadrtic equation, $ax^2+ bx+ c= 0$ is $b^2- 4ac$.)

5. ## thanks

thank you very much

6. There is a circle equation formed from 2 end points of a diameter but it does not seem to be regularly taught.
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