Given two separate annulus with centers $[C_1,C_2]$ and their corresponding radii being $[R_1,r_1]$ and $[R_2,r_2]$ respectively, larger radius being $R$. There are methods to look at whether they are overlapping or not. If they overlap how to find the overlapping area. Is there any close form expression for it? equation of first annulus
$r_1^2<(x-x_{c1})^2 + (y-y_{c1})^2 <R_1^2$
equation of second annulus
$r_2^2<(x-x_{c2})^2 + (y-y_{c2})^2 <R_2^2$
If there are three circles overlapping, we get overlapping area as solution to set of 4 in-equations by least squares approach(Though we don't end up getting total area but, area within tangents of circles ).
Just for fun I tried the area bound by $4<(x+1)^2+(y+1)^2<9$ and $1<(x-2)^2+(y-3)^2<25$. This turned out to be very ugly and long (below). In general probably there one can find a formula ... \begin{align} Area&=\frac{1}{2} \left[8 \sqrt{6}-3 \sqrt{91}-4 \pi +25 \sin ^{-1}\left(\frac{2}{625} \left(63 \sqrt{6}-142\right)\right)-9 \sin ^{-1}\left(\frac{1}{50} \left(9-4 \sqrt{91}\right)\right)+25 \sin ^{-1}\left(\frac{3}{250} \left(4 \sqrt{91}-41\right)\right)+9 \sin ^{-1}\left(\frac{1}{50} \left(9+4 \sqrt{91}\right)\right)-8 \cos ^{-1}\left(\frac{1}{25} \left(3+8 \sqrt{6}\right)\right)+18 \cos ^{-1}\left(\frac{1}{50} \left(9+4 \sqrt{91}\right)\right)+4 \tan ^{-1}\left(\frac{24}{7}\right)+25 \csc ^{-1}(5)+25 \csc ^{-1}\left(\frac{1250}{69 \sqrt{91}-164 \sqrt{6}}\right)\right] \end{align}