Please help me to find an equation to find the 3rd point in an arc. 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).
 A: 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.
