Expected area of the intersection of two circles

Let $\vec{x}_1$ and $\vec{x}_2$ be the two points. Let $r = |\vec{x}_1 - \vec{x}_2|$ be the distance between them. By elementary geometry, if you draw two circle of radius $r$ using these two points as center, the area of their intersection is given by $(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})r^2$. Notice the picking of two points are independent, we have: $$E\left[ \vec{x}_1 \cdot \vec{x}_2 \right] = E\left[\vec{x}_1\right] \cdot E\left[\vec{x}_2\right] = \vec{0} \cdot \vec{0} = 0$$ This implies $$E\left[|\vec{x}_1 - \vec{x}_2|^2\right] = E\left[|\vec{x}_1|^2 + |\vec{x}_2|^2\right] = 2\frac{\int_0^R r^3 dr}{\int_0^R rdr} = R^2$$

As a result, the expected area of the intersection is $(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})R^2$.

Update for those who are curious

Let $\mathscr{C}$ be the set of events such that the intersection contains the origin, then: $$\begin{align} \operatorname{Prob}\left[\,\mathscr{C} \right] &= \frac{2\pi + 3\sqrt{3}}{6\pi}\\ E\left[\,|\vec{x}_1 - \vec{x}_2|^2 : \mathscr{C}\right] &= \frac{20\pi + 21\sqrt{3}}{6(2\pi + 3\sqrt{3})} \end{align}$$ and the expected area of intersection conditional to containing the center is given by: $$\frac{(4\pi - 3\sqrt{3})(20\pi + 21\sqrt{3})}{36(2\pi + 3\sqrt{3})}$$

To evaluate $E\left[ \varphi(\vec{x}_1,\vec{x}_2) ) : \mathscr{C} \right]$ for any function $\varphi( \vec{x}_1, \vec{x}_2 )$ which is symmetric and rotational invariant w.r.t its argument, you need to compute an integral of the from:

$$\int_{\frac{\pi}{3}}^{\pi} \frac{d\theta}{\pi} \left[2\int_{0}^{R} \frac{2udu}{R^2} \left( \int_{\alpha(\theta)u}^{u} \frac{2vdv}{R^2} \phi( \vec{x}_1, \vec{x}_2 ) \right) \right] $$

where $u \ge v$ are $|\vec{x}_1|$ and $|\vec{x}_2|$ sorted in descending order. $\theta$ is the angle between $\vec{x}_1$ and $\vec{x}_2$. The mysterious $\alpha(\theta)$ is $\max(2\cos(\theta),0)$ for $\theta \in [\frac{\pi}{3},\pi]$.

The integral is a big mess and I need a computer algebra system to crank that out. I won't provide more details on this part not relevant to the main answer.