Prove the center of the unit circle has the highest average ray length? If the average ray length is the average distance of the segments from a point inside the circle to points evenly distributed on the boundary:
Prove the center of the unit circle has the highest average ray length.
My Attempt
Convert the circle $x^2+y^2=1$ into polar coordinates so the distances are evenly distrbuted. Since distance $r$ is centered at the origin we must move point $(u,v)$ in the circle to the origin.
$$(x+u)^2+(y+v)^2=1$$
$$(r\cos(\theta)+u)^2+(r\sin(\theta)+v)^2=1$$
$$r^2+2r(u\cos(\theta)+v\sin(\theta))+u^2+v^2-1$$
Solving for $r$ and simplifying give us
$$r=-\left(u\cos(\theta)+v\sin(\theta)\right)\pm\sqrt{(u\cos(\theta)+v\sin(\theta))^2-(u^2+v^2-1)}$$
Since $r$ must be positive, we get the average radius is 
$$\frac{1}{2\pi}\int_{0}^{2\pi}\left|-\left(u\cos(\theta)+v\sin(\theta)\right)\pm\sqrt{(u\cos(\theta)+v\sin(\theta))^2-(u^2+v^2-1)}\right| d\theta$$
Then solve the integral and find the maximum in terms of $(u,v)$ 
The problem is I'm not sure if the integral is solvable. Is there another way of approaching this problem?

There is a much more obvious way of answering this question.  Take any point in the interior of the circle, pass a chord through that point. The average of the 2 rays that make up this chord is 1/2 the length of the chord.  All chords except the diameter are shorter than the diameter.  For the center of the circle, all chords are diameters.  For all other points inside the circle, the chords (other than the diameter) are shorter, therefore the average is less than the average for the center.
