Thread:

1. help with points on a circle

If I know a coordinate of a point (red dot on image) of a circle with radius x and i want to rotate it y degrees to the right, how do I get the coordinate of the new point (blue dot on image) ?

thanks
Attached Thumbnails

2. Re: help with points on a circle

As you are seeking a pair of co-ordinates $\left(x_2,\;y_2\right)$

you could set up a pair of equations in 2 unknowns.

If the circle centre is the origin, then

$\left(x_2\right)^2+\left(y_2 \right)^2=\left(x_1 \right)^2+\left(y_1 \right)^2=r^2$

where the other co-ordinates are known.

Then the area of the triangle between the origin and the 2 points is

$0.5r^2sin\theta$

and so, your 2nd equation is, using the area of a triangle for which one vertex is the origin

$0.5r^2sin\theta=0.5|x_1y_2-x_2y_1|$

However, a trigonometric solution will be much simpler.

If you label the acute angle between the negative part of the x-axis, the origin and the red dot "A",
and the obtuse angle between the positive part of the x-axis and the blue dot "C",
then use the given co-ordinates to calculate the circle radius and the angle "A",
then calculate "C" and use

$x=rcosC,\;\;y=rsinC$

3. Re: help with points on a circle

Originally Posted by Sneaky
If I know a coordinate of a point (red dot on image) of a circle with radius x and i want to rotate it y degrees to the right, how do I get the coordinate of the new point (blue dot on image) ?

thanks
p-radius(not x, for this matter)

x=pcos (a)
y=psin(a)

you have red point, (x1,y1) put it above, find angle a.

blue point is:

x=pcos (a-b)
y=psin(a-b)

I think that would work...

4. Re: help with points on a circle

Originally Posted by Sneaky
If I know a coordinate of a point (red dot on image) of a circle with radius x and i want to rotate it y degrees to the right, how do I get the coordinate of the new point (blue dot on image) ?
Let's change some of the notation here. Say that the the coordinates of the red point are $(x,y)$ then the coordinates of the blue point $\left( {x',y'} \right)$ can be found using:
$\begin{array}{*{20}c}{x' = x\cos ( - \phi ) - y\sin ( - \phi )} \\ {y' = x\sin ( - \phi ) + y\cos ( - \phi )} \\ \end{array}$

NOTE. I have changed the notation. $\phi$ is the angle you have called $y$. Also note that you said that you know $\color{red}(x,y)$
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8888199329376221, "perplexity_flag": "middle"}