Let $C_0$ be a circle centered on the origin, and $C_1$ a circle centered on $(1,0,0)$, center distance of $1$.
Q1. If both $C_0$ and $C_1$ are randomly oriented and have the same radius $r > \frac{1}{2}$, what is the probability $P(r)$ that the circles are linked?
This is in some sense a straighforward question. Although "straighforward," I am not finding it an easy computation. (I would like an exact expression, although numerical approximations are welcomed.)
It may help to start with a simpler question:
Q2. Assume $C_0$ is fixed in the $xy$-plane, and only $C_1$'s orientation is random. What is the probability $P'(r)$ that the circles are linked?
The orientation of $C_1$ is determined by a unit normal vector $\hat{n}$ whose tip is on a unit sphere $S$. The challenge in Q2 is to work out the region $R$ on $S$ that leads to linking. The probability $P'(r)$ is then the area of $R$ divided by $4 \pi$.
Even Q2 seems tricky. If anyone can see considerations that simplify the calculations, I would appreciate hearing of them—Thanks!
For Q2, the probability should be just: $$\frac{2}{\pi}\arccos\frac{1}{2r}.$$ Consider the plane $\pi$ where $C_0$ lies. $C_1$ intersect such a plane in two antipodal points $P_1,P_2$ with respect to $(1,0,0)$. If a point between $P_1$ and $P_2$ lies inside the disk that $C_0$ cuts on $\pi$ the two circles are linked, otherwise they are not. But the distribution of the antipodal points $P_1,P_2$ is uniform on a circle that lies on $\pi$, with center $(1,0,0)$ and radius $r$. Hence we just need to compute the amplitude of the angle depicted in the following figure:
and divide it by $\pi$ to get our answer. For Q1, consider $C_0$ as a fixed circle lying in the plane $\pi$ and let $\Gamma$ be the intersection between $\pi$ and the sphere $S$ concentric with $C_1$ having radius $r$. Let $\pi_1$ be a plane parallel to $\pi$ through the center of $S$. $C_1$ intersects $\Gamma$ only if the angle $\theta$ between $\pi_1$ and the plane containing $C_1$ is big enough. In such a case, $C_0$ and $C_1$ are linked only if $C_1$ intersects $\Gamma$ in a point that lies inside $C_0$. In this case, our probability depends on two (Euler) angles $(\theta,\phi)$ that are uniformly distributed over $[0,\pi]\times[0,2\pi]$.
Given that $d$ is the distance of the center of $S$ from the plane $\pi$ where $C_0$ lies, then $\sqrt{1-d^2}$ is the distance between the center of $C_0$ and the center of $\Gamma$, while $\sqrt{r^2-d^2}$ is the radius of $\Gamma$. The probability depending on the azimutal angle $\theta$ is $\frac{2}{\pi}\arccos\frac{d}{r}$, while the probability depending on $\phi$ can be found by resolving a triangle with side lengths $(r,\sqrt{r^2-d^2},\sqrt{1-d^2})$ through the Cosine theorem. Putting all together, the final probability is:
$$\frac{4}{\pi^2}\cdot\left(\arccos\frac{d}{r}\right)\cdot\left(\arccos\frac{1-2d^2}{2\sqrt{(r^2-d^2)(1-d^2)}}\right).$$