Trigonometry of an hyperbolic Omega Triangle I was puzzeling with the trigonometry of an Omega Triangle ( a triangle / biangle  in hyperbolic geometry where one of the vertices is an Ideal point see https://en.wikipedia.org/wiki/Hyperbolic_triangle#Ideal_vertices )
And using the Poincare disk model ( https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model ) 
Translated the Omega triangle to the following problem in Euclidean geometry:
(the angle $\angle AOB $ in one of the angles of the Omega triangle the other angle is $\angle CAO - 90 ^{\circ} $ )
Given:


*

*the unit circle (circle centered around $ (0 . 0)$ and radius 1 )

*a point $A = (a,0) $ with $ 0 < a < 1 $

*a point $ B$ on the arc of the unit circle with $0^\circ < \angle AOB < 90^\circ $

*A circle $c$ with centre $C$ such that $c$ is orthogonal to the unit circle and passes trough $A$ and $B$ 
Question:


*

*What is $ \angle CAO$ as function of  $ \angle AOB $ and $ a$ ?

 A: the cosine of angle $CAO=$
$$\frac{\left(a^2-1\right)\sin  \theta }{1+a^2-2a \cos  \theta }$$
where $\theta$ is angle $AOB$
Outline of a proof usin trig
Let $OB=1, OA=a, AB=b, BC=CA=c$
angles in degrees: $CAO=x, AOB=\theta, OBA=\beta,CAB=ABC=90-\beta$
Law of Sines: $\frac{\sin  \beta }{a}=\frac{\sin  \theta }{b}$
Calculating $\text{OC}^2$ in two different ways we get
$$a^2+c^2-2a c \cos  x=c^2+1$$
and therefore $2a c \cos  x=a^2-1$
On the other hand 
$$\frac{a \sin  \theta }{b}=\sin  \beta =\cos  (90-\beta )=\frac{b}{2c}$$
$$2a c \sin  \theta =b^2$$
$$\cos  x = \frac{\left(a^2-1\right)\sin  \theta }{b^2}=\frac{\left(a^2-1\right)\sin  \theta }{1+a^2-2a \cos  \theta }$$
Some interesting special cases:
$$a=\cos  \theta , x=\theta +\pi /2$$
$$a=2 \cos  \theta , x = 3\theta -\pi /2$$
$$\theta =\pi /2, x=\arccos  \left(\frac{a^2-1}{a^2+1}\right)$$
another special case
$$\cos  \theta  = \frac{2a}{a^2+1},x=\pi $$
A: HINT.-Here an answer put in equations (perhaps there may be another easier one). Do not solve the system nor calculate the angle between the straight lines $CA$ and $OA$ (I leave this as an exercise for the OP).
In order to know angle $\angle {CAO}$ one can find out the two points $B=(x_1,y_1)$ and $C=(x_2,y_2)$ so there are four unknowns, therefore one needs four (independent) equations.
First equation.- $B$ is in the unit circle: $$x_1^2+y_1^2=1\qquad (1)$$
Second equation.-$C$ is in the line segment bisector of $\overline{AB}$. The midpoint of $\overline{AB}$ being $M=(\frac {x_1+a}{2},\frac {y_1}{2})$ and the slope of $\overline{AB}$ being $\frac{y_1}{x_1-a}$ we have
$$\frac{2y_2-y_1}{2x_2-x_1}=\frac{-x_1+a}{y_1}\iff 2y_1y_2+2(x_1-a)x_2+ax_1=1\qquad (2)$$
Third equation.- $|\overline{CB}|=|\overline{CA}|$ (radius of the circle $c$).
$$(x_2-x_1)^2+(y_2-y_1)^2= (x_2-a)^2+y_2^2\iff 2(a-x_1)x_2-2y_1y_2+1=-2ax_2+a^2\qquad (3)$$
Fourth equation.-Line $CB$ is tangent to the unit circle.
$$\frac{y_1-y_2}{x_1-x_2}=\frac{-x_1}{y_1}\iff x_1x_2+y_1y_2=1\qquad (4)$$
