Point of reflection on a circle I am trying to find the point on a circle where a light source is reflected to the focal point of a camera...
So I have the coordinates of the light source, the focal point and the centre of a circle. I believe the light will reflect at an angle equal to the incident angle. I have tried forming equations with this information but I always seem to have too many unknowns...
Diagram of problem
Thanks for any help!
 A: Define $\tan A={y_1-y_p\over x_1-x_p}\ ;\ \tan B={y_0-y_p\over x_0-x_p}\ \&\ \tan C={y_2-y_p\over x_2-x_p}$
Since incident angle equals reflected angle,at $(x_p,y_p)$, so $$\tan (A-B)=\tan (B-C)\implies {\tan A-\tan B\over 1+\tan A\tan B}={\tan B-\tan C\over 1+\tan B\tan C}$$
Moreover, since $x_p,y_p$ must be on a circle, you must know its radius$(=r)$: there goes another equation $$(x_0-x_p)^2+(y_0-y_p)^2=r^2$$
$2$ equation and $2$ unknowns ,BINGO!
A: To make things easier, I limit all the points to the first quadrant and I also assume that the mirror surface is the horizontal tangent line y = k. Our target is to find k.

Let R be the reflection of Q about the line y = k. Because of the angle of incidence = angle of reflection, we have two sets of equal angles (red and green). Then, $R = (x_2, 2k – y_2)$.
Equation of PR is $L :\dfrac {y – y_1}{x – x_1} = \dfrac {(2k – y_2) – y_1}{x_2 – x_1}$.
Solving L and y = k, we will get the co-ordinates of O which is precisely $(x_0, k)$. Then we have one equation and only one unknown.
Note:- Special treatment will be needed in cases like $x_1 = x_2$ etc.
