Length of intersecting chords inside circle, given angle, vertex and circle radius Let's assume a circle with the known radius of R and origin O. We know the position of the intersection of two lines within that circle: vector A. Those lines both terminate at the perimeter of the circle and we know the value of the angle a between those lines, for simplicity let's assume AB is at a right angle (Y-up). How does one determine the length of line CA or just the position of C (Note that it is not necessarily at the same angle as AO).
I think this one is hard to explain without a picture so either bare with the crappy ASCII art.
          B  
   C..|''''..
  .'\ |      '.              B
 :   \|        :      C      |
:     A         :      \     |
:       O       :       \  a |
:               :        \.''|
 :             :          \  |
  '.         .'   CA = ?   \ |
    ''.....''               \|
                             A

Or see this image.

I will implement it with programming / scripting but need to figure out which method to use, I could "project" point A onto the circle if there's no better option.
Thanks in advance!
 A: @WW1's comment pointed me in the right direction (I would upvote if I had the rep):

*

*Get the angle $\angle CAO$, use law of sines to get the length of AC.

The triangle CAO is a SSA triangle, so using the law of sines could give ambiguous results but the code I've implemented seems solid regardless of input. I don't know why but I guess that's another question in its own right. Here's a short gif-animation of the implementation.
My original question was a bit vague, so I'll just walk through the steps I used.

*

*The direction from A to C is known unit vector $\hat{AC}$

*We also know the values for unit vector $\hat{AO}$, and magnitude $|AO|$

*Magnitude $|CO| = \text{radius}$

*Get the (shortest) angle difference using the dot product:
$$\angle CAO = \arccos(\hat{AC} \cdot \hat{AO})$$

*Get remaining angles of triangle CAO:
$$\angle ACO = \arcsin(\frac{|AO| * \sin(\angle CAO)}{|CO|})$$
$$\angle AOC = 180^{\circ} - \angle ACO - \angle CAO$$

*Get distance from A to C:
$$|AC| = \frac{\sin(\angle AOC) * |CO|}{\sin(\angle CAO)}$$
A: 
Draw a radius from O parallel with AB. For point $A (x_A, y_A)$ We have:
$m=\tan(\widehat{AOD})=\dfrac {x_A}{y_A}$
$$ b=\widehat{AOD}=\tan^{-1}\left(\dfrac{x_A}{y_A}\right)$$
If $\widehat {BAC}=a$ is given the we have:
$c=\widehat {CAE}= a-b$
and:
$$OA=\sqrt{x_A^2+y_A^2}$$
finally:
$$CA\approx \frac{R-OA}{\cos c}$$
A: 
Following formula can be also used to determine the length of $AC$ (see $\mathrm{Fig.\space 1}$)
$$AC=\sqrt{r^2-\left(x_A^2+y_A^2\right)\sin^2\left(\omega+\tan^{-1}\left(\dfrac{x_A}{y_A}\right)\right)}-\sqrt{x_A^2+y_A^2}\cos \left(\omega+\tan^{-1}\left(\dfrac{x_A}{y_A}\right)\right)$$
This formula gives the length of $AC$ for a point $A$ defined anywhere in the interior of a circle and two points $B$ and $C$ lying on its circumference making $\measuredangle BAC=\omega\enspace$ when measured counterclockwise from $B$ to $C$. $\mathrm{Fig.\space 1}$ to $\mathrm{Fig.\space 4\enspace}$ illustrate how $\angle BAC$ is defined. Furthermore, $x_A$ and $y_A$ are not signless lengths, but signed $x$- and $y$-Cartesian coordinates of $A$

