Let $L=(-1,0),R=(1,0)$, $O=(0,0)$, the center, and $M$ the random point on the circle. WLOG assume $M$ is on the upper semicircle. Note the picking the point uniformly is equivalent to sampling $\theta =\angle ROM$ uniformly over $[0,\pi]$.
Write $X$ for the length of $LM$ and $Y$ for the length of $RM$.
- Observation: $\angle LMR$ is a right angle. Indeed, let $\alpha = \angle RLM$ and $\beta = \angle LRM$. Since the triangles $MOL$ and $MOR$ are each an isosceles, it follows that $\angle LMR=\alpha+\beta$. But then
$$180^{\circ} =
\angle RLM+\angle LRM+ \angle LMR=\alpha +\beta + (\alpha+\beta) = 2 (\alpha+\beta).$$
- Corollary: $XY$ is twice the area of the triangle $LMR$.
- Twice area of $LMR$ is also given by the length of its base $LR$ times its height. This is easy to compute. Base is $LR$ and has length $2$, and the height is equal to $\sin \theta$. Thus,
$$ E[XY ] =2 E[ \sin \theta ] =2 \frac{1}{\pi} \int_0^{\pi}\sin (\theta)d \theta = \frac{4}{\pi}.$$
- Next find $E[X]$ and $E[Y]$. Because of symmetry, $E[X]=E[Y]$. By looking at the triangle $MOR$, we see that $Y= 2 \sin (\theta/2)$. Therefore,
$$ E [ Y ] = \frac{2}{\pi} \int_0^\pi \sin (\theta /2) d \theta = \frac{4}{\pi}\int_0^{\pi/2}\sin (t) dt = \frac 4 \pi. $$
- Final calculation:
$$ \mbox{Cov} (X,Y) = E [XY] - E[X] E[Y] = \frac{4}{\pi} (1- \frac{4}{\pi}).$$
