For reference: Two circles whose radii measure R are situated in such a way that the distance between their centers measures R. In the region common to both circles is inscribed a square , calculate the length of the side of the square.(Answer: $\frac{R}{2}(\sqrt7-1)$) *Solution by geometry, not by analytics.
My progress:
$GF = l\\AI = JB = x\\
MF = \frac{DF}{2} \\
DF = l\sqrt2 \therefore MF = \frac{l\sqrt2}{2}\\
\triangle JFM: (R-x)^2+(\frac{l}{2})^2 = (\frac{l\sqrt2}{2})^2\implies\\
(\frac{R}{2}-x)^2 = \frac{l^2}{2}-\frac{l^2}{4} \implies (\frac{R}{2}-x)^2 = \frac{l^2}{4}$
needed to find the value of x
Use the triangle $AFJ$ where we know the length of $AF=R$. Using your notation, we can write $FJ=\frac{l}{2}$. Since $AB=R$ and $MJ=\frac{l}{2}$ we can find that $AJ=AM + MJ = \frac{R}{2} + \frac{l}{2}$. Now we can use the abc formula to solve for $l$.