Say that these distances we are looking for are counterclockwise distances.
I only know how to find it, if one of the sets has at most one element:
Let the sets be X = {$x_1$, $x_2$..., $x_n$}, Y = {y}, then dist($x_1$,y) = $d_1$, dist($x_2$,y) = $d_2$..., dist($x_n$,y) = $d_n$ are iid uniform random variables along the length of the circle. Let E be the minimum.Then:
$f_E$(x) = P(E<x) = 1 - P(E>x) = 1 - P($d_1$<x)P($d_2$<x)...P($d_n$<x) = 1 - $(1-x)^n$
The catch, as far as I can tell, is that while distances from one point are independent, distances from two and more are not (let $y_1$, $y_2$ $\in$ Y, x $\in$ X, then dist($y_1$,x) and dist($y_2$,x) always differ by dist($y_1$, $y_2$))
Notation. The unit circle can be denoted by $T$ for torus. The $n$-fold Cartesian product will be written as $T_n$. An interval $B\subset T$ is considered to be a one-dimensional ball whose center is its midpoint.
Problem Statement.
Suppose first that $Y =(y_1, y_2)\in T_2$ has only 2 points and $X=(x_1, x_2, \ldots, x_n) \in T_n$ has $n$ points.
Pick randomly chosen $X$ and $Y$. Let $D(X,Y)$ denote the minimum distance between all pairwise choices $(x_i, y_j)$.
We wish to determine the distribution of this random variable, which is defined as $f(d):={\mathbb E}( D(X,Y)<d)$ where expectation is taken over the product space $X\in T_n, Y\in T_2$.
Solution.
Note that $D(X,Y)<d\iff $ some ball of radius $d$ centered at some $y$ contains some member of $X$. The complementary event that is easier to analyze is that neither ball of radius $d$ centered at either point in $Y$ contains any member of $X$. That is, the union of the two balls $B_1$ (centered at $y_1$) and $B_2$ (centered at $y_2$) is a set completely free of any members of $X$. The probability that all $n$ random independent choices of $X\in T_n$ satisfy this condition is $\mu^n$ where $\mu = m(B_1\cup B_2)$ is the measure of the union of the two balls. By inclusion/exclusion this is $\mu= m(B_1) + m(B_2) - m(B_1\cap B_2)$ where each ball has measure $2d$. Thus $$(1) \mu= 4d -m(B_1 \cap B_2)$$ Clearly the last term depends only on the relative separation $s=|y_1- y_2|$ between the centers of the balls and the widths of the balls. When $s=0$ (concentric balls), the measure is $2d$. As the balls slide apart as we increase $s$, the measure of the overlap declines linearly until it becomes zero when $s=2d$. Thus we see that $\mu(B_1\cap B_2)$ is an explicit piecewise-linear function of $s=|y_1-y_2|$ that changes behavior when $s=2d$ (hence has to be split into cases, unfortunately too tedious to typeset).
Now just substitute (1) into $$(2) f(d) = \int _{(y_1, y_2)\in T_2} (1- (\mu(y_1, y_2))^n ) dy_1 dy_2$$
Higher Dimensions
When we add more points to $Y$ the calculations become much worse. The evaluation of the measure of intersection of e.g. three randomly centered balls $m(B_1\cap B_2\cap B_3)$ that all have equal radius $d$ can ultimately be expressed as the volume of the intersection of two sets: a 3D cube centered at a random point $(y_1, y_2,y_3)$ defined by $\{(u_1, u_2, u_3) : |u_i- y_i|<d\}$ and a slab in space defined by $-d\leq u_1+u_2+u_3 <d$. This volume can thus be interpreted as the convolution of the characteristic function of a cube and a slab. It is a piece-wise polynomial function of the coordinates $(y_1,y_2, y_3)$. One way to obtain it is to use a multidimensional Laplace transform and its inverse.
A clever method for representing this piece-wise polynomial function using spline notation is given in Barrow, D. L., and P. W. Smith. “Spline Notation Applied to a Volume Problem.” The American Mathematical Monthly, vol. 86, no. 1, 1979, pp. 50–51. JSTOR, https://doi.org/10.2307/2320304.