Consider two circles $c_1$ and $c_2$ both of radius $r$ located in 2-D plane such that the distance between their centers is $r$. Assume a point is randomly and uniformly chosen within their intersection. What is $E[x]$, the expected value of the point's distance to the center of $c_1$ (or $c_2$ equivalently)?
I assume the exact computation might be too involved. Can you think of an easy approximation? For example, I am under the impression that $E[x] = \Theta(r)$, but I'd also like to get a sense of the constants involved in this equality. Any ideas?
Assume that $r=1$. Since you want uniform distribution, you integrate over the area. Assume the first circle is centered at $(0,0)$, the second at $(1,0)$. The lens-shaped intersection is described by
$$\lvert y\rvert \le\begin{cases} \sqrt{1-(1-x)^2} = \sqrt{2x-x^2} & \text{for }0\le x\le\tfrac12 \\ \sqrt{1-x^2} & \text{for }\tfrac12\le x\le1 \end{cases}$$
You can integrate over the upper half of this area ($y\ge 0$) only, since the lower half will be identical. Use one integral to compute the distance, and the other to compute the area only, then divide them by one another and you have the expected distance.
\begin{align*} A &= \int_0^{\frac12}\int_0^{\sqrt{2x-x^2}} \sqrt{x^2+y^2} \;\mathrm dy\,\mathrm dx + \int_{\frac12}^1\int_0^{\sqrt{1-x^2}} \sqrt{x^2+y^2} \;\mathrm dy\,\mathrm dx \\ B &= 2\cdot \int_0^{\frac12}\int_0^{\sqrt{2x-x^2}} \mathrm dy\,\mathrm dx = 2\cdot \int_0^{\frac12}\sqrt{2x-x^2}\;\mathrm dx \\ E[x] &= \frac AB \end{align*}
Numerical computation (using sage, since this seems to exceed the powers Wolfram Alpha grants to anonymous users) tells me that
$$ E[x] \approx 0.64279153263859578984682902596387936174378729 $$
and randomized experiments support that figure.
If you have $r\neq1$, all lengths scale by that factor $r$, so the expected distance has to scale by that same amount. Therefore in general you'll have
$$ E[x] \approx 0.64279153 \cdot r $$