If radii of three concentric circles are related as $r_1(r_2+r_3)+r_2(r_3+r_1)+r_3(r_1+r_2)=118$ and $$\sum_{cyclic}\frac{r_1^2+r_2^2}{r_1r_2}=\frac{44}{5}$$ then area of enclosed region between any two circles can be
(A)$21\pi$
(B)$45\pi$
(C)$32\pi$
(D)$19\pi$
I tried eliminating $r_1$
$r_1=\frac{59-r_2r_3}{r_2+r_3}$
The second condition may be written as
$r_1(\frac{r_2+r_3}{r_2r_3})+\frac{1}{r_1}(r_2+r_3)+\frac{(r_2+r_3)^2-2r_2r_3}{r_2r_3}=\frac{44}{5}$
Clearly we are looking for something like $r_3^2-r_2^2$ but calculation appears very tricky. Could there be a geometrical way to go about it
Easily, $r_1=\frac{59-r_2r_3}{r_2+r_3}$. But I found the second equation as $$\frac{(r_2+r_3)^2}{59-r_2r_3}+\frac{(r_2+r_3)^2+59}{r_2r_3}=\frac{59}{5}.$$ This can be written as $$(r_2+r_3)^2=\frac{1}{5}(59-r_2r_3)(r_2r_3-5).\tag{*}$$ Assuming $r_3>r_2$, the positive integer solution of $(*)$ is $(r_3,r_2,r_1)=(7,5,2)$. So, the possible difference for squares are $45, 24$ and $21$. Is there a double answers? A and B seem to be the answers. There are also non-integer solutions.
https://www.wolframalpha.com/input?i=%28x%2By%29%5E2%3D%281%2F5%29%2859-xy%29%28xy-5%29+and+x%5E2-y%5E2%3D21 https://www.wolframalpha.com/input?i=%28x%2By%29%5E2%3D%281%2F5%29%2859-xy%29%28xy-5%29+and+x%5E2-y%5E2%3D45