Area of the gray region. How would I calculate the area of the shaded region $S_x$?
I'm trying to use similarity of triangles to find some relationship between the sides, but the equations are getting even more complicated.
In the image it seems that $r_1$ is equal to $r_2$, but it is not. And so, with $r_1 \neq r_2$, it becomes more difficult.
In Geogebra I saw that the solution holds for $r_1 \neq r_2$.
Solution: $3r_1 r_2$

 A: Hint.


Solution.

$$\left.\begin{align}
|\overline{US}| = |\overline{UV^\prime}| \implies r + s + u = w + v \\
|\overline{VR}| = |\overline{VU^\prime}| \implies r + s + v = w + u
\end{align}\right\rbrace \quad\implies\quad u = v \quad\text{and}\quad w = r + s \tag{$\star$}$$
From $(\star)$, one can derive the triangle congruences indicated in the Hint. However, that derivation seems to require a relation that leads directly to the puzzle solution, so the congruences appear to be extra work. (This makes me think that I'm missing something "obvious" about those triangles. Be that as it may ...)
The key relation comes to us courtesy of Pythagoras:
$$\begin{align}
|\overline{OU}|^2 + |\overline{OV}|^2 = |\overline{UV}|^2 &\implies
(u+r)^2 + (u+s)^2 = (r+s)^2 \\[2pt]
&\implies u^2+ur+us=rs \\[2pt]
&\implies ( u + r )( u + s ) = 2 r s \\[4pt]
&\implies |\triangle OUV| = rs
\end{align}$$
Thus, the area of the target region is $(2r)(2s) - rs = 3 rs$. $\square$ 
A: 
Area $S_x=[OFCG]-[L_1L_2O]=4r_1r_2-\tfrac14\,|L_1L_2|^2\sin2\theta$.
Let's assume without loss of generality that 
$|O_1T_1|=r_1\ge r_2=|O_2T_2|$.
Then
\begin{align} 
|PO_1|&=r_1-r_2
,\\
|O_1O_2|&=\sqrt2\,(r_1+r_2)
,\\
|T_1T_2|&=|PO_2|
.\\
\text{Let }\quad \angle O_2O_1P&=\phi
,\quad \angle O_1QP =\tfrac\pi2
.\\
\text{Note that }\quad
PO_1&\perp L_1L_2,\quad PQ\perp L_1O,\quad O_1Q\perp L_2O,\quad 
\\
\triangle PO_1Q &\sim \triangle  L_1L_2O
,\quad \angle PO_1Q =\angle L_1L_2O=\theta
.\\
\text{Then }\quad
\theta&=\tfrac34\pi-\phi
,\\
\phi&
=\arccos\left(\frac{|PO_1|}{|O_1O_2|}\right)
=\arccos\left(\frac{r_1-r_2}{\sqrt2(r_1+r_2)}\right)
,\\
|L_1T_2|=|L_1T_4|&=r_2+|L_1O|
,\\
|L_2T_1|=|L_2T_3|&=r_1+|L_2O|
,\\
|L_1T_2|+|L_2T_1|&=
|T_1T_2|+|L_1L_2|
\\
&=r_1+r_2+|L_1O|+|L_2O|
\tag{1}\label{1}
,\\
|L_1O|&=r_1+|L_1T_1|
,\\
|L_2O|&=r_2+|L_2T_2|
,\\
|L_1O|+|L_2O|
&=r_1+r_2+|L_1T_1|+|L_2T_2|
\\
&=r_1+r_2+|T_1T_2|-|L_1L_2|
\tag{2}\label{2}
.
\end{align}  
Combination of \eqref{1} and \eqref{2} results in
\begin{align} 
|L_1L_2|&=r_1+r_2
.
\end{align}
\begin{align} 
S_x&=4r_1r_2-\tfrac14\,|L_1L_2|^2\sin2\theta
\\
&=4r_1r_2-\tfrac14\,(r_1+r_2)^2
\sin(\tfrac32\pi-2\phi)
\\
&=4r_1r_2-\tfrac14\,(r_1+r_2)^2
(-\cos2\phi)
\\
&=4r_1r_2-\tfrac14\,(r_1+r_2)^2
(1-2\,\cos^2\phi)
\\
&=4r_1r_2-\tfrac14\,(r_1+r_2)^2
\left(1-\frac{(r_1-r_2)^2}{(r_1+r_2)^2}\right)
\\
&=4r_1r_2-r_1r_2=3r_1r_2
.
\end{align}
