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?

By symmetry, the average distance from the boundary of $\|z\|\leq 1$  is a function of the distance from the origin. If $x\in(0,1)$ its average distance from the the boundary of the unit circle centered at the origin is given by
$$ \frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{(x-\cos\theta)^2+\sin^2\theta}\,d\theta =\frac{\sqrt{1+x^2}}{\pi}\int_{0}^{\pi}\sqrt{1-\frac{2x}{1+x^2}\cos\theta}\,d\theta$$
and by letting $\lambda=\frac{2x}{1+x^2}\in(0,1)$ we have
$$ \int_{0}^{\pi}\sqrt{1-\lambda\cos\theta}\,d\theta=\int_{0}^{\pi/2}\sqrt{1-\lambda\cos\theta}+\sqrt{1+\lambda\cos\theta}\,d\theta=\int_{0}^{1}\frac{\sqrt{1-\lambda u}+\sqrt{1+\lambda u}}{\sqrt{1-u^2}}\,du$$
clearly leading to an elliptic integral of the second kind. It is not difficult to devise tight algebraic approximations for these objects (see, for instance, the dedicated section in my notes), and according to Mathematica's notation we have that the average distance of $x\in(0,1)$ from the boundary of the unit circle centered at the origin is
$$ \frac{2}{\pi}(x+1)\cdot E\left(\frac{4x}{(x+1)^2}\right)=1+\frac{x^2}{4}+\frac{x^4}{64}+\frac{x^6}{256}+\frac{25 x^8}{16384}+\ldots $$
where all the involved coefficients of the Maclaurin series are non-negative, implying that the LHS is increasing over $(0,1)$.

Now a very tricky elementary approach. The average distance of $x\in(0,1)$ from the boundary of the unit circle is given by $\frac{1}{2\pi}$ times the perimeter of an ellipse with semiaxis lenghts $1-x$ and $1+x$. If $A,B$ are two bounded, convex sets in $\mathbb{R}^2$ and $A\subsetneq B$, then the perimeter of $A$ is less than the perimeter of $B$. This implies that the previous average distance / ellipse perimeter is an increasing function of the $x$ variable over the interval $(0,1)$.

Yet another elementary approach by convexity. The function giving the distance from a fixed point is convex and the sum of convex functions is convex. Convex functions over convex domains attain their maximum at the boundary. So we have that over $x^2+y^2\leq 1$ the average distance from $x^2+y^2=1$ is a radial and convex function. It is pretty obviously differentiable over $x^2+y^2<1$ (we already wrote an explicit integral representation) and the origin is a stationary point: in order to prove that the origin is an absolute minimum it is enough to show that the average distance is not constant over $x^2+y^2\leq 1$, and that is trivial.
