Probability that the smallest circle through two randomly selected points in a unit disk is within its interior

Question

Given two points $P$ and $Q$, chosen at random (uniformly) from the interior of a unit disk, what is the probability that the circle whose diameter is the segment $\smash[t]{\overline{PQ}}$ is entirely contained within the interior of the unit circle?

If the points within the unit circle have coordinates $P=(p_1,p_2)$ and $Q=(q_1,q_2)$, I concluded that the circle whose diameter is $\overline{PQ}$ must satisfy the inequality: $$ (p_1 + q_1)^2 + (p_2 + q_2)^2 < (p_1q_1 + p_2q_2 + 1)^2. $$

However, I don't know how I could find its probability distribution to compute $$ \mathbb{P}\bigl[(p_1+q_1)^2 + (p_2+q_2)^2 < (p_1q_1 + p_2q_2 + 1)^2\bigr]. $$

I would appreciate any suggestions or proposed solutions.

Answer 1

Fix one of the points, say $P$, & examine the collection of “largest circles through $P$”, which will be tangent to the unit circle. I claim that the locus of the centres of such circles is an ellipse with foci at $P$ and the origin, & major axis 1.

This fact follows from the fact that $OT=1$ and $PE=ET$. The line $ET$ passes through the origin as it is perpendicular to the tangent at $T$. The locus of the antipodal points $Q$ is the red ellipse of twice the size. If $OP=r$ then the area of this ellipse is $\pi\sqrt{1-r^2}$.

We have: $\mathbb{P}[r\le OP\le r+dr]=2rdr$, and $Q$ lies inside our ellipse with probability $\sqrt{1-r^2}$, so integrating our final answer is:

$$ \int_0^1 2r \sqrt{1-r^2} dr = \frac{2}{3}.$$