I need an inversion formula with the form $f(r)=\cdots$, from this integral relation: $$g(r)=\frac{1}{2\pi}\int_0^{2\pi}d\theta\,f\left(\sqrt{r^2+r_0^2-2rr_0\cos\theta}\right)$$ where $r_0\geq0$ is a constant, $f(r)$ is a unknown real function defined on $r\geq 0$, and $g(r)$ is a given real function on $-\infty<r<\infty$.
This equation derived from a polar coordinate system like this image.
By changing of the variable: $u:=\sqrt{r^2+r_0^2-2rr_0\cos\theta}$ , I got a formula such as $$ g(r)=\frac{1}{\pi}\int_{|r-r_0|}^{|r+r_0|}du\,\frac{2uf(u)}{\sqrt{((r+r_0)^2-u^2)(u^2-(r-r_0)^2)}}. $$
I expect the inversion can be solved since this relation looks somewhat like the Abel transform formula, but it is not succesful yet.
Can someone help this transform?
If needed, $f$ and $g$ can be assumed as $(f,g<\infty),\ (f,g>0)$, differentiable, or converging fast enough when $r\rightarrow\infty$.
Thanks for your help and sorry for my halting English.