How to divide a circle with two perpendicular chords to minimize (and maximize) the following expression Consider a circle with two perpendicular chords, dividing the circle into four regions $X, Y, Z, W$(labeled):

What is the maximum and minimum possible value of 
$$\frac{A(X) + A(Z)}{A(W) + A(Y)}$$
where $A(I)$ denotes the area of $I$?
I know (instinctively) that the value will be maximum when the two chords will be the diameters of the circle, in that case, the area of the four regions will be equal and the value of the expression will be $1$. 
I don't know how to rigorously prove this, however. And I have absolutely no idea about minimizing the expression. 

When the considered quotient is not constant it has a maximum value $\mu>1$, and the minimum value is then ${1\over \mu}$. I claim that
$$\mu={\pi+2\over\pi-2}\doteq4.504\ ,\tag{1}$$
as conjectured by MvG.

Proof. I'm referring to the above figure. Since the sum of the four areas is constant the  quotient in question is maximal when
$$S(\alpha,p):={\rm area}(X)+{\rm area}(Z)\qquad\left(-{\pi\over2}\leq\alpha\leq {\pi\over2}, \quad 0\leq p\leq1\right)$$
is maximal. 
Turning the turnstile around the point $(p,0)$ and looking at the "infinitesimal sectors" so arising we see that
$$\eqalign{{\partial S\over\partial\alpha}&={1\over2}\bigl((r_1^2-r_4^2)+(r_3^2-r_2^2)\bigr)\cr &={1\over2}\bigl((r_1+r_3)^2-(r_4+r_2)^2-\bigr)+(r_2r_4-r_1r_3)\cr &=2\bigl((1-p^2\sin^2\alpha)-(1-p^2\cos^2\alpha)\bigr)+0\cr
&=2p^2(\cos^2\alpha-\sin^2\alpha)\ .  \cr}$$
From this we conclude the following: Starting at $\alpha=-{\pi\over2}$ we have $S={\pi\over2}$, then $S$ decreases for $-{\pi\over2}<\alpha<-{\pi\over4}$, from then on increases until $\alpha={\pi\over4}$, and finally decreases again to $S={\pi\over2}$ at $\alpha={\pi\over2}$. It follows that for the given $p\geq0$ the area $S$ is maximal at $\alpha={\pi\over4}$.
We now fix $\alpha={\pi\over4}$ and move the center $(p,0)$ of the turnstile from $(0,0)$ to $(1,0)$. Instead of "infinitesimal sectors" we now have "infinitesimal trapezoids" and obtain
$${\partial S\over\partial p}={1\over\sqrt{2}}((r_2+r_3)-(r_1+r_4)\bigr)>0\qquad(0<p\leq1)\ .$$
It follows that $S$ is maximal when $\alpha={\pi\over4}$ and $p=1$. In this case one has $S=1+{\pi\over2}$, which leads to the $\mu$ given in $(1)$.
