# What is the average thickness of an eccentric annulus?

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

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.