Angle and circle intersection, find the circular segment area Playing Kerbal Space Program, I found myself wondering about what a satellite would see of a planet depending on its field of view and its altitude. I tried attacking the problem from various angles but I always end up with a missing piece of information. 
Speaking a bit more formally, I have a circle Planet with a radius r. I have a point Sat which is at a distance Orb from the center of Planet (for our case, Orb > r). The line Sat-Center(Planet) bisects A-Sat-B, with A and B being points on the circle.
I know Orb, r and the angle A-Sat-B, what ways do I have to calculate A-B? (See below for a diagram.)
From there I think the Circular segment will let me find the information I need but I'm stumped on the calculation.

 A: So you have the angle $\theta$, the planet radius $r$ and the orbital distance ${\rm Orb}$

the solution arrives from two trigonometry  equations:
$$ \left. 
\begin{aligned} {\rm Orb} & = r \cos \frac{\psi}{2} + \ell \cos \frac{\theta}{2} \\
 r \sin \frac{\psi}{2} & = \ell \sin \frac{\theta}{2} \end{aligned} 
\right\} 
\begin{aligned}  \frac{\psi}{2} & = \frac{\pi-\theta}{2} - \cos^{-1} \left(  \frac{{\rm Orb}}{r}\sin \frac{\theta}{2}\right) 
\end{aligned}
 $$
With $h = r \sin \frac{\psi}{2} $  and ${\bf AB} = 2 h$
For example, when $r=1$, ${\rm Orb}=3.5$ and $\theta=25°$ then

$$ \frac{\psi}{2} = \frac{180°-25°}{2} - \cos^{-1} \left(  \frac{3.5}{1}\sin \frac{25°}{2}\right) =  36.74768°$$  which is same as measured $\beta$ angle in diagram above.
Then $h = 1 \sin (36.74768°) = 0.5982922 $ which also matches diagram above.
Also we can check $\ell = \frac{ r \sin \frac{\psi}{2} }{\sin\frac{\theta}{2}} = { r \sin(0.6413681) }{\sin(12.5°)} = 2.764245$
Diagram is made with GeoGebra 4
A: 
$\text{The answer can be as explained below -  }\\$
$\text{Join $A$ and $B$, Let $C$ be centre,}\\$
$\text{Now join $AC$ and $BC$, get $\theta$, the $\angle$ that lines $AC$ and $BC$ make at the centre,}\\$
$\text{find the length of the chord $AB$, which will be $\frac{1}{2}$ the $diameter$ if angle is $\frac {\pi}{2}$,}\\$
$\text{depending on the $\angle \theta $, whether it is $\pi$, $\frac{\pi}{2}$ respectively, and so on.}\\$
$\text{Now when you know $AB$,}\\$
$\text{You can also check (A-SAT) $or$ (SAT-B), by geometry finding the 2 equal}\\$
$\text{sides of a triangle by pythagorous theorem when base and height are known,}\\$
$\text{$or$ The Arc length $AB$ $=$ $r$ $\cdot$ $\theta$ & the }\\$
$$\text{$"Darkened Circular Segment Area"$} = \left[\frac {r^2}{2} \cdot \left(\frac{\pi \cdot \theta}{180} - \sin \theta \right)\right]$$
