Two points $A$ and $B$ are chosen uniformly at random from the interior of a circle $X_1$. Let $X_2$ be the circle whose diameter is the segment $AB$. What is the probability that $X_2$ is contained entirely inside $X_1$?
Progress: Not much luck with random integrating.
For points $P$ and $Q$ in the plane, let $\Gamma(P,Q)$ denote the circle with segment $PQ$ as its diameter.
We first prove a lemma.
Lemma: Let $\Gamma$ be a circle, $P$ a point inside $\Gamma$, and $P'$ the reflection of $P$ about the center of $\Gamma$. Denote by $\mathcal{L}$ the locus of points $Q$ inside $\Gamma$ such that $\Gamma(P,Q)$ is internally tangent to $\Gamma$. Then $\mathcal{L}$ is the unique ellipse with foci $P$ and $P'$ that is internally tangent to $\Gamma$.
Proof: Suppose $Q$ is inside $\Gamma$ such that $\Gamma(P,Q)$ internally tangent. Let $A$ be the point of tangency, $O$ the center of $\Gamma$, and $O_1$ the center of $\Gamma(P,Q)$. Note that $PQ = 2PO_1$, and $OO_1 = AO - AO_1 = AO - PO_1$. Since $PO = P'O$ and $PO_1 = QO_1$, it follows that $\triangle PO_1 O \sim \triangle PQP'$. Thus $P'Q = 2OO_1 = 2(AO - PO_1)$. Then $PQ + P'Q = 2AO$, which is a constant.
Assume that $X_1$ is of radius 1, and suppose $A$ is distance $x$ from the center of the circle. By the lemma, the probability that $B$ is chosen such that $\Gamma(A,B) \subset \Gamma$ is exactly $(\pi \sqrt{1-x^2})/\pi = \sqrt{1-x^2}$. Thus, the answer is $$\int_{0}^{1} x \sqrt{1-x^2} \, dx = \frac{1}{3}$$
