Assume WLOG that the problem has been re-scaled so that the cirle has radius 1. Also note that by double symmetry we need only consider 1/4 of the figure. Then we have:
$$
f(x)=\sqrt{1-x^2}\\
g(x)=a\cdot f(x/b)
$$
where $a>1>b$ and $x\geq 0$. The situation looks like this:
The point of intersection will be found via:
$$
f(x)=g(x)\\
\iff\\
\sqrt{1-x^2}=a\cdot\sqrt{1-x^2/b^2}\\
\iff\\
1-x^2=a^2(1-x^2/b^2)\\
\iff\\
(1-a^2/b^2)x^2=1-a^2\\
\iff\\
(b^2-a^2)x^2=b^2(1-a^2)\\
\iff\\
x^2=\frac{b^2(1-a^2)}{b^2-a^2}\\
\iff\\
x=\sqrt{\frac{b^2(1-a^2)}{b^2-a^2}}
$$
and then the area of the overlap can be found as:
$$
x_0=\sqrt{\frac{b^2(1-a^2)}{b^2-a^2}}
\int_0^{x_0}f(x)\ dx
+
\int_{x_0}^b g(x)\ dx\\
=\\
F(x_0)-F(0)+G(b)-G(x_0)
$$
Note how $F(0)=0$ and $G(b)=ab\cdot\frac\pi 4$. Hence the overlap can be reduced to:
$$
F(x_0)+ab\cdot\frac\pi 4-G(x_0)
$$
Finally, we can use the primitive functions:
$$
F(x)=\int \sqrt{1-x^2}\ dx=\frac12\left(x\cdot f(x)+\arcsin(x)\right)
$$
and
$$
G(x)=\int a\cdot f(x/b)\ dx=a\cdot b\cdot F(x/b)
$$
Final summary
Given a circle with radius $r$ and an ellipse with axes $A,B$ compute:
$$
\begin{align}
a&=A/r\\
b&=B/r\\
x_0&=\sqrt{\frac{b^2(1-a^2)}{b^2-a^2}}\\
ellipse&=ab\cdot\frac\pi 4\\
F(x_0)&=\frac12\left(x_0\cdot\sqrt{1-x_0^2}+\arcsin(x_0)\right)\\
x_1&=x_0/b\\
G(x_0)&=a\cdot b\cdot\frac12\left(x_1\cdot\sqrt{1-x_1^2}+\arcsin(x_1)\right)\\
overlap&=F(x_0)+ellipse-G(x_0)\\
ratio&=overlap/ellipse
\end{align}
$$
NOTE: here $overlap$ and $ellipse$ values are really only $1/4r^2$ of the actual areas, but for the ratio this will not matter. Otherwise multiply by $4r^2$ if you need actual areas.
