Long story short, I want to animate the rotation of an object that's based off a circle.

Given the center point of the circle, the radius, and one of the points in the arc, is it possible to find the equation to plot the 3rd point, given that the angle of the arc will be 90 degrees?

For example:

I have a circle with a radius of 10.

The center (first point) of the circle is at (3, 5). The second point is at (-7, 5).

I would like an equation to find the 3rd point. Which in this example would be at (3, 15).


Consider the vector $v=(second\ point)-(centre\ point)=(-7,5)-(3,5)=(-10,0)$ here.

Multiply it by the rotation matrix: $$R=\left [ \begin{matrix} \cos \theta& \sin \theta\\-\sin \theta & \cos \theta\end{matrix} \right ]$$ where $\theta = \pi/2$, to get: $$\left [ \begin{matrix} 0& 1\\-1 & 0\end{matrix} \right ]v = \left [ \begin{matrix} 0& 1\\-1 & 0\end{matrix} \right ]\left [ \begin{matrix} -10 \\ 0 \end{matrix} \right ] = \left [ \begin{matrix} 0 \\ 10 \end{matrix} \right]$$

Then the new point is: $$\left [ \begin{matrix} x \\ y \end{matrix} \right] -\left [ \begin{matrix} 3 \\ 5 \end{matrix} \right]=\left [ \begin{matrix} 0 \\ 10 \end{matrix} \right]$$

$$ \implies \left [ \begin{matrix} x \\ y \end{matrix} \right]=\left [ \begin{matrix} 3 \\ 15 \end{matrix} \right]$$

A similar procedure works in general. You don't even need the radius if the first point and centre are given. This gives a formula of $$(x_0+y_1-y_0, x_0-x_1+y_0)$$ for the new point, where $(x_0, y_0)$ are the coordinates of the centre and $(x_1, y_1)$ are the coordinates of the given point.

  • $\begingroup$ Thank you very much. This works perfectly. I think I was over-complicating it. $\endgroup$ – Javalsu May 6 '13 at 16:23
  • $\begingroup$ You're welcome! $\endgroup$ – Milind May 6 '13 at 16:25
  • $\begingroup$ Mind if I ask another question? Do you know how I would find the 3rd point in this chart? My animation relies on a bezier curve, so I need to figure out how to get the points for the curve as well. So I need to figure out what's the missing value in the posted pic. i.imgur.com/pLfV1lA.jpg $\endgroup$ – Javalsu May 6 '13 at 18:04
  • $\begingroup$ 3rd point as in? The next point when rotated by $\pi /2$? $\endgroup$ – Milind May 6 '13 at 18:06
  • 1
    $\begingroup$ You know the length of the bottom side is 10, and that the angle is $30^{\circ}$, so you can get the length of the perpendicular side as $10\tan 30^{\circ}=\frac{10}{\sqrt{3}}$, so the final point is $(10, 10+\frac{10}{\sqrt{3}})$. $\endgroup$ – Milind May 6 '13 at 18:12

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