What is the average thickness of an eccentric annulus?

Question

I have a small circle of radius $r$ entirely contained in a large circle of radius $R\gt r$, these two circles are not concentric and so they define an eccentric annulus.

The center of the small circle is in $\left(0,0\right)$; a point $A=\left(r\cos\left(\alpha\right),r\sin\left(\alpha\right)\right)$ is on the small circle; $A$ and the center of the small circle define a line; this line intersects the large circle in two points, the farthest point from the small circle is $B$ and it is depicted in the following figure; I define the thickness $t$ as a function of the angle $\alpha$ as the length of the line segment $\overline{AB}$. What is the average value of $\overline{AB}$ when $\alpha$ goes from $0$ to $2\pi$?

If I had to guess, I would say the average value is $R-r$, i.e. the thickness of an annulus.

Using the formulae¹ I derived an (horrible) expression for $t(\alpha)$ but I have no idea on how to integrate it in order to compute its mean value.

¹Weisstein, Eric W. "Circle-Line Intersection." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Circle-LineIntersection.html

Answer 1

Hint:

WLOG, $R=1$. In polar coordinates, the equation of a circle centered at $(k,0)$ is

$$(\rho\cos\theta-k)^2+\rho\sin^2\theta=1$$

or

$$\rho=k\cos\theta+\sqrt{1-k^2\sin^2\theta}.$$

So you need to compute the half of

$$\int_0^{2\pi}(k\cos\theta+\sqrt{1-k^2\sin^2\theta}-r)^2\,d\theta=\\ \int_0^{2\pi}(k^2\cos^2\theta+1-k^2\sin^2\theta+r^2+2k\cos\theta\sqrt{1-k^2\sin^2\theta}-2r\sqrt{1-k^2\sin^2\theta}-2rk\cos\theta)\,d\theta$$ where $r$ is the inner radius.

Unfortunately, all the terms are integrable but $2r\sqrt{1-k^2\sin^2\theta}$ which can only be solved via a complete elliptic integral of the second kind $E(k)$, and this is a sure sign that no simpler solution is possible.