Angle of point on one circle to match view from another circle 
This should be a simple geometry problem, but I can't seem to find a simple answer.
 A: Let $\alpha =\text{ASM}$, $\beta=\text{ASF}$, $\gamma =\text{ASE}$
Taking S as origin. 
The point M has the coordinates $(x_3,y_3)= (r_3 \cos \alpha, r_3 \sin \alpha)$
The point F has the coordinates $(x_2,y_2)= (r_2 \cos \beta, r_2 \sin \beta)$
The point E has the coordinates $(x_1,y_1)= (r_1 \cos \gamma, r_1 \sin \gamma)$
The line between M and E is given by 
$$y=\frac{x_3-x}{x_3-x_1}y_1 + \frac{x-x_1}{x_3-x_1}y_3
=\frac{x_3y_1-x_1 y_3}{x_3-x_1}
 + \frac{y_3-y_1}{x_3-x_1}x
$$
Denote the slope $m=\frac{y_3-y_1}{x_3-x_1}$ and the shift
$b=\frac{x_3y_1-x_1 y_3}{x_3-x_1}$
We looking to describe F in terms of this line, so we ask what point $(x,y)$ on this line satisfies $x^2+y^2 =r^2$ that is we looking for a positive 
$(x,y)$
satisfying
$$x^2+y^2=x^2+\left(\frac{x_3-x}{x_3-x_1}y_1 + \frac{x-x_1}{x_3-x_1}y_3\right)^2=r^2$$
and solve it for $x$, that is the following quadratic equation
$$
x^2 +2 \frac{b m}{1+ m^2} - \frac{r^2-b^2}{1+ m^2}x=0
$$
to get
$$
x_{1,2}=-\frac{b m}{1+ m^2} \pm \sqrt{\frac{b^2 m^2}{(1+ m^2)^2}+\frac{r^2-b^2}{1+ m^2}}
$$
and choose the positive solution, i.e. the one with $+$ in $\pm$ and denote it $x^+$.
Finaly, 
$$\beta =\arctan\frac{y(x^+)}{x^+}=
\arctan\frac{\frac{x_3y_1-x_1 y_3}{x_3-x_1}
 + \frac{y_3-y_1}{x_3-x_1}x^+}{x^+}
$$
Switching back to radius's and angles will give you the expression you looking for.
