Probability distribution or the distance between two points in $n$-dimensional Euclidean space after a random perturbation of one point Take two points, $p_0$ and $p_k$, in $n$-dimensional Euclidean space, where $d(p_0,p_k)$ is the distance between the points.  Now, draw an $n$-sphere of radius $r$ centered on $p_0$ and uniformly select a new point, $p_1$, in the volume of the sphere, and a new point $p_2$ on the surface of the sphere.
Let $h_1 = d(p_0,p_k) - d(p_1,p_k)$, and $h_2 = d(p_0,p_k) - d(p_2,p_k)$, represent the difference in the distance from $p_0$ to $p_k$ if $p_0$ is moved to $p_1$ or $p_2$, respectively.  To clarify, if we blow up another $n$-sphere, $S$, about $p_k$ until $p_0$ "touches" its contour, $h_1$ / $h_2$ will be positive if $p_1$ / $p_2$ is inside $S$, zero if $p_1$ / $p_2$ are on the contour of $S$, and negative if $p_1$ / $p_2$ fall outside $S$.
What probability distribution and expectation do we have for $h_1$ and $h_2$?
Update [2/15/2014] :: Thanks to Bjørn Kjos-Hanssen's efforts, we have a nice exact expression for the $n = 2$ case for $h_1$ (which implies that we have expectation --- $\int_{1-r}^{1+r} \space R \times f(R) \space dR$).  At this point, I think its probably wise to restrict the focus or scope of this question to the $n = 2$ case.  Can a PDF for $h_2$ (where we select points along the contour of the circle) be derived in a similar manner?
 A: Let us consider the case $n=2$. Assume $p_k$ is at the origin and $p_0$ is at the point $(1,0)$ on the $x$-axis, and that $r<1$.
Distribution of $h_1$
By subtracting a constant it suffices to find the distribution of $d:=d(p_1,0)$.
The density for $d$, $f_d(R)$, is proportional to $R\Theta$ where $\Theta$ is length of the interval of angles $\theta$ for which the point $(R\cos\theta,R\sin\theta)$ is within $r$ of $(1,0)$. Now, calculation shows that
$$
|(R\cos\theta,R\sin\theta)-(1,0)| < r
$$
is equivalent to
$$
|\theta| < \arccos\left(\frac{R^2+1-r^2}{2R}\right)
$$
so $f_d(R)$ is proportional to
$$
2R\arccos\left(\frac{R^2+1-r^2}{2R}\right),\quad R\in [1-r,1+r].
$$
and $f_d(R)=0$ for $R\not\in [1-r,1+r]$.
For geometric reasons (and one could also check it analytically) the integral of the given expression over the given interval is $\pi r^2$. So

$$ f(R)=\frac{2R\arccos\left(\frac{R^2+1-r^2}{2R}\right)}{\pi
 r^2},\quad 1-r\le R\le 1+r. $$

Distribution of $h_2$
The squared distance from $(1,0)+r(\cos\theta,\sin\theta)$ to the origin is
$$
D^2 : = (r\cos\theta+1)^2+(r\sin\theta)^2 = r^2 + 1 + 2r\cos\theta$$
and the probability of picking a point $R$ or more away from the origin is then $1/\pi$ times $\theta_R>0$ where $\theta=\theta_R$ makes $D^2=R^2$. Namely
$$
\theta_R = \arccos\left(\frac{R^2-r^2-1}{2r}\right)
$$
Then the density is
$$ f(R)= \frac{-1}{\pi} \frac{d}{dR}
 \arccos\left(\frac{R^2-r^2-1}{2r}\right);
$$

$$
 f(R)=\frac1{\pi}\frac{1}{\sqrt{1-u(R)^2}} \frac{R}{r},\quad 1-r\le R\le
 1+r, $$ where $u(R)=\frac{R^2-r^2-1}{2r}$.

