Finding a point on a circle that makes a specific angle of incidence with a given point inside it? Suppose we have a circle of radius r centered at the origin, and we are given a point C$(x,y)$ inside the circle.

Let $\theta_i$ be the angle of incidence of the line drawn from C. If we know $\theta_i$ (say, 23º) and C, how can we find the coordinates of P$(x,y)$, the point on the circle where the angle of incidence from C is $\theta_i$?
Edit 1: Oops, forgot to draw P in the circle. It's the point where the line meets the circle.
 A: To find the points $P$ on a given circle $c$ of center $O$ and radius $R$, such that $\angle CPO=\alpha$ given, join $CO$ and from its midpoint $M$ draw the perpendicular bisector $r$. Let $A$ be a point on $r$ such that $\angle CAM=\alpha$ (that is $AM=CM/\tan\alpha$), then draw the circle $a$ having center at $A$ and passing through $O$ and $C$. The intersections between $a$ and the given circle $c$ are the required points, as well as the symmetric of those points with respect to line $OC$. In fact, if $P$ is one of such intersections, then $P$ lies on $a$ and $\angle CPO=(1/2)\angle CAO=\alpha$. Notice that we have no solutions for $OC<R\sin\alpha$ ($a$ contained in $c$), two solutions for $OC=R\sin\alpha$ ($a$ tangent to $c$) and four solutions for $OC>R\sin\alpha$ ($a$ secant to $c$).
If $x_C$, $y_C$ are the coordinates of $C$, then 
the coordinates of $A$ (or $A'$) are given by
$x_A={1\over2}(x_C\mp y_C/\tan\alpha)$, 
$y_A={1\over2}(y_C\pm x_C/\tan\alpha)$, 
and it is straightforward to get the coordinates of the intersection points:
$$
x_P={2R\sin^2\alpha\over t^2}
\left(
Rx_A\pm y_A\sqrt{{t^2\over\sin^2\alpha}-R^2}
\right),\ 
y_P={2R\sin^2\alpha\over t^2}
\left(
Ry_A\mp x_A\sqrt{{t^2\over\sin^2\alpha}-R^2}
\right),
$$
where $t=OC=\sqrt{x_C^2+y_C^2}$.
Edit from Blue. Here's a picture:

A: Assume that the center of the circle is the origin.  We have $P\cdot(P-C) = |P| |P-C| \cos\theta$, but $|P|=1$, so $P\cdot (P-C) = P\cdot P - P\cdot C$ (by linearity) $= 1-P\cdot C$ and this equation becomes $1-P\cdot C = |P-C|\cos\theta$; then squaring both sides and using $|v|^2=v\cdot v$, we get $(1-P\cdot C)^2 = \cos^2\theta (P-C)\cdot (P-C)$.  We can use linearity to expand the RHS again, getting $(1-P\cdot C)^2=\cos^2\theta(P\cdot P-2P\cdot C+C\cdot C) = \cos^2\theta (1+|C|^2-2P\cdot C)$.  This is a quadratic equation in $P\cdot C$, which you can solve to get that value (or values - I see no reason that the given point $P$ should be unique).
Finally, given the equations $P\cdot C=q$ (where $q$ is a calculated value from solving the quadratic above) and $P\cdot P=1$, there are presumably several ways to solve for $P$ but the simplest (to me) is to just set $P=(x,y)$ and note that $P\cdot C=q$ is a linear equation in $x$ and $y$, so we can use it to substitute one of $x,y$ for the other in $P\cdot P=1$ (which is just the equation $x^2+y^2=1$ of the circle), getting the final values of $x,y$.
A: *

*Fom the figure below we can calculate:


$\cos(D)=\frac{a}{d}$.....(1) 
$\cos(D+\theta)=\frac{a}{r}$ ..... (2)
(2) in (1) -> $\cos(D)=\frac{r\cos(D+\theta)}{d}$
$d=\frac{r\cos(D+\theta)}{\cos(D)}$ ........... (3)


*

*From rules of triangle we deduce the following:


1-the big triangle:
$\frac{r}{\sin(e)}=\frac{B}{\sin\theta}$
$\sin(e)=\frac{r\sin\theta}{B}$.....(4)
$\frac{C}{sin(Y+y)}=\frac{C}{\sin(2\pi-e-\theta)}=\frac{C}{\sin(-e-\theta)}=-\frac{C}{\sin(e)\cos\theta+\cos(e)\sin\theta}=\frac{B}{\sin\theta}$
$C=-B\sin(e)cot\theta-B\cos(e)$....(5)
2- from right triangle:
$\frac{A}{\sin(e)}=\frac{B}{\sin(\pi-t)}$
$A\sin(t)=B\sin(e)$,
from (4): $A\sin(t)=B\frac{r\sin\theta}{B}=r\sin\theta$....(6)
$\frac{A}{\sin(e)}=\frac{C-d}{\sin(Y)}$
$\frac{A\sin(Y)}{C-d}=\sin(e)$
From (5), $\frac{A\sin(Y}{C-d}=\frac{A\sin(Y)}{-B\sin(e)\cot\theta-B\cos(e)-d}=\sin(e)$,
using (4), 
$A\frac{sin(-Y)}{B\frac{r\sin\theta}{B}\cot\theta+B\sqrt{1-\frac{r^2\sin^2\theta}{B^2}}+d}=\frac{r\sin\theta}{B}$
...... (7)
3- From left triangle:
$\frac{A}{\sin(\theta)}=\frac{d}{\sin(y)}$
$\frac{A*\sin(y)}{d}=\sin\theta$ ..... (8)




*

*Knowing that $\sin(y)=\frac{a}{r}=\cos(D+\theta)$ ....(9)


(3),(9) in (8) -> $\frac{A*\cos(D+\theta)}{\frac{r\cos(D+\theta)}{\cos(D)}}=\sin\theta$
$\frac{A*\cos(D)}{r}=\sin\theta$ .... (10)
(3) in (7) ,
$A\frac{sin(-Y)}{B\frac{r\sin\theta}{B}\cot\theta+B\sqrt{1-\frac{r^2\sin^2\theta}{B^2}}+\frac{r\cos(D+\theta)}{\cos(D)}}=\frac{r\sin\theta}{B}$
...... (11)
(10) in (11)
$\frac{r\sin\theta}{\cos(D)} \frac{sin(-Y)}{B\frac{r\sin\theta}{B}\cot\theta+B\sqrt{1-\frac{r^2\sin^2\theta}{B^2}}+\frac{r\cos(D+\theta)}{\cos(D)}}=\frac{r\sin\theta}{B}$
$ {B}\frac{sin(-Y)}{{\cos(D)}{r\cos\theta}+{\cos(D)}\sqrt{B^2-{r^2\sin^2\theta}}+{r\cos(D+\theta)}}=1$
$\cos^2 D + \frac{{\left(\cos D\, \left(2\, r\, \mathrm{\cos}\!\left(\mathrm{\theta}\right) + \sqrt{B^2 - r^2\, {\mathrm{\sin}\!\left(\mathrm{\theta}\right)}^2}\right) - B\, \mathrm{\sin}\!\left(- Y\right)\right)}^2}{r^2\, {\mathrm{\sin}\!\left(\mathrm{\theta}\right)}^2} - 1=0$...(12)
with:
$x=cos(D)$
$a1=2*r*\cos(\theta)+\sqrt{B^2-(r^2*\sin(\theta)^2)}$
$a2=(B)*\sin(-Y)$
$a3=(r*\sin(\theta))^2$
solutions:
$x_1=\frac{\mathrm{a1}\, \mathrm{a2} + \mathrm{a3}\, \sqrt{\frac{{\mathrm{a1}}^2 - {\mathrm{a2}}^2 + \mathrm{a3}}{\mathrm{a3}}}}{{\mathrm{a1}}^2 + \mathrm{a3}}$
$x_2=\frac{\mathrm{a1}\, \mathrm{a2} - \mathrm{a3}\, \sqrt{\frac{{\mathrm{a1}}^2 - {\mathrm{a2}}^2 + \mathrm{a3}}{\mathrm{a3}}}}{{\mathrm{a1}}^2 + \mathrm{a3}}$
A: Let $Q$ be the point of incidence. We can solve the triangle OCQ with sides $r$ and $|OC|$ and the angle $\theta$ by the sine law and find the angle $\angle{COP}$.
$$\sin(\angle OCQ)=\frac r{|OC|}\sin(\theta)\\
\angle COQ=\pi-\angle OCQ-\theta.$$
This gives the polar angle of $Q$ (knowing that of $C$).
The requested point $P$ has a polar angle equal to that of $Q$ minus the angle at the apex of an isosceles triangle, $\pi-2\theta$.
