Horizontal distance between circle and ellipse centered at origin

Question

The figure below shows a circle of radius $r$, centered at the origin of cartesian coordinates, and an inclined ellipse also centered at the origin, with semiaxes $a$ and $b$. The area of the circle and that of the ellipse are identical, such that $r^2 = ab$.
I need to know the horizontal distance between points $P$ and $Q$, which is exactly the coordinate $x$ of point $Q$. I am interested in knowing if there is a way to relate that coordinate with the semiaxes of the ellipse. I know graphically that $a-b=x$, but I can't find a way to demonstrate it algebraically.
The angle between the $x$ coordinate axis and the mayor axis of the ellipse is unknown.
Also $Q$ is the point on the ellipse where $y$ is a maximum, equal to the radius $r$. The shape of the ellipse will vary according to its inclination, such that when the inclination angle approaches zero, $a$ increases and $b$ decreases, while the relationship $y=r$ always is maintained

Answer 1

Let $O=(0,0)$ and $A$ the intersection of the ellipse with positive $x$-axis. Then $OA$ and $OQ$ are, by definition, conjugate semi-diameters of the ellipse, and thus have the following nice properties: $$ area_{AOQ}={1\over2}ab \quad\text{and}\quad OA^2+OQ^2=a^2+b^2. $$ The first equality implies $OA=r$. Substituting this and $OQ^2=x^2+r^2$ into the second equality yields: $$ x^2=a^2+b^2-2r^2=a^2+b^2-2ab=(a-b)^2. $$