I have been thinking about how to approach this problem for a while, and it still seems a little obscure to me. I will really appreciate a detailed answer with relevant Proof. I try to clearly describe it below:
Part A:
In Figure (a) (See below), $X$ is uniformly distributed in the smaller circle. I understand that the pdf is $f(x)= \dfrac{2 x}{r^2}$. Let's call this Case 1.
Now, let's say we modify this distribution, which we call "new distribution" with a "tuning factor," i.e for $0 \leq f < 2$, the pdf becomes
$$\frac{(2- f) r^{1-f}}{R^{2-f}}.$$
Note that with $f=0$ we have the same pdf as in Case 1.
Part B
In Figure (b), $X$ obeys a uniform distribution in the shaded region and $S$ is the area of the shaded region. $\ell(r)$ is the length of any arc $BPQ$. Points $A$ and $B$ are the centers of the intersecting circles with the same radii, respectively.
Distance between point $A$ to $X$ obeys $0 < r \leq R$.
I know the pdf of $r$, i.e. $f(r)$, is given by $\dfrac{\ell(r)}{S}$, which I clearly understand from this question.
My Question:
How does the pdf $f(r)$ change with the new distribution in Part A? In other words, what is the new pdf given the shaded area in Figure (b) and its expectation?
$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$Here is an answer based on one possible interpretation of the question. Denote the tuning factor by $c$ instead.
First, reverse-engineering from the modified density of $r$ in Case 1, the density of the point $X$ can be assumed as$$ f_1(x_1, x_2) = \begin{cases} \dfrac{2 - c}{2π R^{2 - c}} \dfrac{1}{(x_1^2 + x_2^2)^{\frac{c}{2}}}; & x_1^2 + x_2^2 < R^2\\ 0; & \text{otherwise} \end{cases} $$
Now for Case 2, suppose the origin is $A$. Assume that $X$ has the above density conditioned to the shaded area $S$. Because in Case 1,$$ p := P(X \in S) = \iint\limits_{(x_1 - R)^2 + x_2^2 < R^2} f_1(x_1, x_2) \,\d x_1\d x_2, $$ then the density of $X$ in Case 2 is$$ f_2(x_1, x_2) = \begin{cases} \dfrac{1}{p} f_1(x_1, x_2); & (x_1 - R)^2 + x_2^2 < R^2\\ 0; & \text{otherwise} \end{cases} $$ Thus the distribution of $r$ in Case 2 is\begin{align*} &\peq F_r(r_0) = P(\|X\| \leqslant r_0)\\ &= \iint\limits_{x_1^2 + x_2^2 \leqslant r_0^2} f_2(x_1, x_2) \,\d x_1\d x_2 = \iint\limits_{\substack{r \leqslant r_0\\-π < θ \leqslant π}} f_2(r \cos θ, r \sin θ) r\,\d r\d θ\\ &= \int_0^{r_0} r\,\d r \int_{-π}^π f_2(r \cos θ, r \sin θ) \,\d θ = \int_0^{r_0} r\,\d r \int\limits_{r < 2R \cos θ} f_1(r \cos θ, r \sin θ) \,\d θ\\ &= \frac{1}{p} \int_0^{r_0} r\,\d r \int_{-\arccos \frac{r}{2R}}^{\arccos \frac{r}{2R}} f_1(r \cos θ, r \sin θ) \,\d θ = \frac{1}{p} \int_0^{r_0} r\,\d r \int_{-\arccos \frac{r}{2R}}^{\arccos \frac{r}{2R}} \frac{2 - c}{2π R^{2 - c}} \frac{1}{r^c} \,\d θ\\ &= \frac{1}{p} \int_0^{r_0} r · 2\arccos \frac{r}{2R} · \frac{2 - c}{2π R^{2 - c}} \frac{1}{r^c} \,\d r = \frac{2}{p} · \frac{2 - c}{2π R^{2 - c}} \int_0^{r_0} r^{1 - c} \arccos \frac{r}{2R} \,\d r, \end{align*} which implies the density is$$ f_r(r_0) = \frac{1}{p} · \frac{2 - c}{π R^{2 - c}} · r_0^{1 - c} \arccos \frac{r_0}{2R}. \quad \forall 0 < r_0 < R $$
For the degenerate case $c = 0$, since by definition,$$ p = P(X \in S) = \iint\limits_{(x_1 - R)^2 + x_2^2 < R^2} f_1(x_1, x_2) \,\d x_1\d x_2 = \iint\limits_{(x_1 - R)^2 + x_2^2 < R^2} \frac{1}{πR^2} \,\d x_1\d x_2 = \frac{|S|}{πR^2}, $$ then$$ f_r(r_0) = \frac{1}{p} · \frac{2}{π R^2} · r_0 \arccos \frac{r_0}{2R} = \frac{2r_0}{|S|} \arccos \frac{r_0}{2R}. $$
For $c = 0.5$,$$ f_r(r) = \frac{1}{p} · \frac{1.5}{π R^{1.5}} · \sqrt{r_0} \arccos \frac{r_0}{2R}. \quad \forall 0 < r_0 < R $$ Here is a figure for $f_r$ and $F_r$: