Is there a simple formula for this simple question about a circle? What is the average distance of the points within a circle of radius $r$ from a point a distance $d$ from the centre of the circle (with $d>r$, though a general solution without this constraint would be nice)?
The question arose as an operational research simplification of a real problem in telecoms networks and is easy to approximate for any particular case. But several of my colleagues thought that such a simple problem should have a simple formula as the solution, but our combined brains never found one despite titanic effort. It looks like it might involve calculus.
I'm interested in both how to approach the problem, but also a final algebraic solution that could be used in a spreadsheet (that is, I don't want to have to integrate anything).

I guess it involves calculus. Let $(x,y)$ be a point within the circle of radius $R$ and $(d,0)$ the coordinates of the point a distance $d$ away from the origin (because of the symmetry we can choose it to lie on the $x$-axis). Then the distance between the two points is given by $$\ell = \sqrt{(x-d)^2 + y^2}.$$
Averaging over the circle is best done in polar coordinates with $x=r \cos \phi$ and $y=r \sin\phi$. We have
$$\begin{align}
\langle \ell \rangle &= \frac{1}{\pi R^2} \int_0^R  dr \int_0^{2\pi} d\phi\, r  \sqrt{(r\cos \phi -d)^2 + r^2\sin^2 \phi}\\
&= \frac{1}{\pi R^2} \int_0^R dr \int_0^{2\pi} d\phi\, r \sqrt{r^2 + d^2 -2 d r\cos\phi}.
\end{align}$$
I am not sure it the integral has a simple analytic solution. Thus, I calculate it in three simple limits.
(a) $d\gg R$:
we can expand the $\sqrt{}$ and have
$$\langle \ell \rangle = \frac{1}{\pi R^2} \int_0^R dr \int_0^{2\pi} d\phi\, [r d- r^2 \cos \phi + \frac{r^3}{2d} \sin^2 \phi] = d + \frac{R^2}{8d}$$
(b) for $d \approx R$, we have 
$$\langle \ell \rangle = \frac{1}{\pi R^2} \int_0^R dr \int_0^{2\pi} d\phi\,  r^2 \sqrt{2 (1-\cos \phi)}  = \frac{8 R}{3\pi}$$
(c) for $d\ll R$ [joriki's comment] 
$$\langle \ell \rangle = \frac{1}{\pi R^2} \int_0^R dr \int_0^{2\pi} d\phi\,  \left[r^2 -rd\cos\phi + \frac{d^2}{2} \cos^2\phi\right]    = \frac{2 R}{3} + \frac{d^2}{2R}$$
