Geometry question: ray paths and circles I was working on a problem and used the image below to make an argument regarding an effective line-of-sight (from one of my papers).  My question below is more of an intellectual curiosity since the paper was already published (I am a physicist, so if my mathematical lingo is sloppy, I will entertain criticism if witty).
Suppose one moved along either of the red arrows (or equally the obstacle moved over you) at a velocity $\mathbf{V}_{o}$, then we could say that $L_{s}$ = $\lvert \mathbf{V}_{o} \rvert t$.  Using some simple geometric rules, we can also show that:
$$
L_{s} = 2 \left( a^{2} - b^{2} \right)^{1/2}
$$
where $a$ is the radius of the circle and $b$ is the impact parameter (see figure).  I was also able to find an analytic solution for $L_{a}$ = $L_{a}\left( L_{s},a,\theta \right)$ for the top trajectory, given by:
$$
L_{a} = -\frac{ L_{s} }{ 2 } \pm \frac{ \left\{ \left[ 2 \ a \ \cos \theta + \sqrt{4 \ a^{2} - L_{s}^{2} } \right]^{2} + L_{s}^{2} \sin^{2} \theta \right\}^{1/2} }{ 2 \sin \theta }
$$
This is the length of the trajectory over which one would be in the effective "shadow" of this obstacle (don't worry about what SLAMS stands for… unless you are interested in nonlinear solitons).  
However, this solution only works for the top trajectory.  Therefore, I have two questions:


*

*Is there an analytic solution for the lower trajectory? 

*If so, how would I go about finding it [I used power of point etc. for the top part but couldn't convince myself it would work for the bottom part]?



Edits
In the figure below, you see two red arrows.  These correspond to two possible trajectories through the obstacle.  One could imagine sitting still while the obstacle moved over you at a velocity $\mathbf{V}_{o}$ (or you move through the obstacle at that velocity, it doesn't matter).  Therefore, $L_{s}$ is the distance traversed while inside the obstacle.
The distance $L_{a}$ shown in blue is the distance between the point of exit from the obstacle to the point where the green arrows no longer intersect with the obstacle.  Meaning, $L_{a}$ defines the distance along the red arrow one must traverse before one can observe something along a green arrow line of sight that does not intersect the obstacle.
In the current figure, the $L_{a}$ is only shown for the top trajectory.  The corresponding $L_{a}$ for the bottom trajectory with this specific $\theta$ and $b$ would be very short (actually, just to the right of the blue vertical dashed line is where it would end and just to the left is where it would start).
 A: Your formula for the shadowed length of the upper trajectory simplifies to
$$L_a=\frac{a+b\cos\theta}{\sin\theta}-\sqrt{a^2-b^2}.$$
Similarly, the shadowed length of the lower trajectory is
$$\frac{a-b\cos\theta}{\sin\theta}-\sqrt{a^2-b^2}.$$
You can derive this in the straightforward though tedious way as follows:

The marked angles are all equal to $\theta$, and $AB=b$ and $AC=a$ are the radii of your two circles. So
$$|AE|=\frac b{\cos\theta},\quad |BE|=\frac{b\sin\theta}{\cos\theta},\tag{$\triangle ABE$}$$
meaning
$$\begin{gather}
|EC|=|AC|-|AE|=a-\frac b{\cos\theta},\\
|ED|=\frac{|EC|}{\sin\theta}=\frac a{\sin\theta}-\frac b{\sin\theta\cos\theta},\tag{$\triangle ECD$}
\end{gather}$$
and finally
$$|BD|=|BE|+|ED|=\frac{b\sin\theta}{\cos\theta}+\frac a{\sin\theta}-\frac b{\sin\theta\cos\theta}=\text{(lots of algebra...)}=\frac{a-b\cos\theta}{\sin\theta}.$$
Your shadowed length is the part of $BD$ outside the grey region. But you already know that the part inside has length $\sqrt{a^2-b^2}$. The result follows.
A quicker derivation is possible if you draw a rectangle whose one side is $AC$ and whose other side, say $FG$, passes through $B$. Then $|FG|=a$ but $|FB|=b\cos\theta$, so $|BG|=a-b\cos\theta$, and from the right triangle $BGD$ you can conclude that $|BD|=(a-b\cos\theta)/\sin\theta$.
