Basically, I know that for a uniformly distributed points, r within the intersection $XQYD$ in Figure 1, the pdf is given by $$f(r) = \frac{l(r)}{A}$$ where A is the area of the intersection region and $l(r)$ is the length of $ACB$ and $\theta$ is given by $\cos^{-1}\left(\frac{r}{2R}\right)$ from Figure 2.
In case someone might find this useful
$$l(r) = 2r \cos^{-1}\left(\frac{r}{2R}\right) $$ while $$A = \left( \frac{2\pi}{3}-\frac{\sqrt{3}}{2} \right) $$
Question: Now suppose this changes to a non-uniform way-point spatial distribution. In this distribution, points are non-uniformly distributed around an arbitrary 'center' with a PDF
\begin{equation} f\left ({ r}\right )=\frac {1}{R^{2}}\left ({\frac {-4 \, r^{3}}{R^{2}}+4\, r}\right ) , \end{equation}
How do one go about modifying the PDF of $r$ i.e conditioning on the PDF above on the intersection of two equal circles?
In case it might be of interest, please note that to understand the pdf for the uniform distribution I have asked this question before.
