Given two possibly intersecting circles $C_1(c_1, r_1)$ resp. $C_2(c_2, r_2)$ ($c_x$ being the center and $r_x$ being the radius) and two random variables $ X_1 \in C_1 $ resp. $ X_2 \in C_2 $, what is the probability $ P(distance(X_1, X_2) < R) $ if the probability distribution of $ X_1 $ and $ X_2 $ is uniform?
What I've thought of was taking the size of the intersection $ \tilde{C}_1(c_1, r_1 + R) \cap \tilde{C}_2(c_2, r_2 + R)$ and dividing it by $ \tilde{C}_1 \cup \tilde{C}_2 $, but that doesn't account for the conditional nature of the problem.
The motivation for this problem comes from trying to estimate the probability of two people infecting each other by the covid virus based on their GPS location history.
A start:
Use basic trigonometry and the Pythagorean Theorem to calculate the distance (blue) between a point chosen from the black circle and a point chosen from the red circle. The distribution on each angle is uniform between $0 \to 2 \pi$, and the distribution of the radius is linearly proportional to the radius (up to the maximum). (Do you see why?) You'll get a (blue) distance between the two points depending upon the separation of the circle centers ($d$), the two angles, and the two radii.
Now for any criterion "infection distance" you have an implicit set of equations for $\theta_1, \theta_2, r_1, r_2$. Integrate over that set of values and divide by the integral over all possible values of those variables.