Polar integration seems to be the best method for creating a general formula for this.
Re-expressing both circles as polar yields this:
$r1 = 1$ and $r2 = cos\theta+\sqrt(a^2-sin^2\theta)$ based off of the general formula for a polar circle, centered at $(1, 0)$ with radius $a$. Really all we need here are the points of intersection which can easily be solved by setting each equation equal and following algebraic and trigonometric rules
$$cos\theta+\sqrt(a^2-sin^2\theta) = 1$$
$$a^2-sin^2\theta=1+cos^2\theta-2cos\theta$$
Replacing $sin^2$ and $cos^2$ with 1 and solving for theta yields:
$$\theta=\pm arcos(a^2-2/-2)$$
From here it should be noted that we must add $\pi$ to the resulting angle because our result accounts for the part of the circle outside of the central unit circle. So our angles from which we base our integral are
$$\theta = \pi \pm arcos(a^2-2/-2)$$
At this point I believe you can evaluate this regularly
$$.5\int_{\theta_1}^{\theta_2} 1^2-[cos\theta+\sqrt({a^2-sin^2\theta})]^2 \,d\theta$$
There may be some errors I missed in my answer however this general method should work
Computing the Density for One Disk
Without loss of generalization, let the radius of the larger circle be $1$ and the radius of the smaller circle be $s$.
The probability of a point at a distance $r$ from the center of the circle of radius $1$ being in a circle of radius $s$, whose center is uniformly distributed in a circle of radius $1-s$ concentric with the circle of radius $1$, is the ratio of the area of the intersection of a circle of radius $1-s$ with a circle of radius $s$ whose centers are at a distance $r$ divided by the area of the circle of radius $1-s$.
Using the Law of Cosines and Heron's formula, we get
$$
\rho(s,r)=\left\{\begin{array}{}
1&\text{if }r\le2s-1\\[12pt]
\frac{s^2}{(1-s)^2}&\text{if }r\le1-2s\\
\frac{s^2\cos^{-1}\left(\frac{r^2+2s-1}{2rs}\right)+(1-s)^2\cos^{-1}\left(\frac{r^2-2s+1}{2r(1-s)}\right)-\frac12\sqrt{\left(1-r^2\right)\left(r^2-(2s-1)^2\right)}}{\pi(1-s)^2}&\text{if }r\gt|1-2s|
\end{array}\right.
$$
Computing the Density for Multiple Disks
Each of the disks is independent of the others, so the density for the intersection of $n$ disks is $\rho(s,r)^n$. Thus, the expected overlap of all $n$ disks is
$$
2\int_0^1\rho(s,r)^n\,r\,\mathrm{d}r
$$
If $0\le r\le2s-1$, then $\rho(s,r)=1$; otherwise, $\rho(s,r)\lt1$. Therefore,
$$
\begin{align}
\lim_{n\to\infty}2\int_0^1\rho(s,r)^n\,r\,\mathrm{d}r
&=2\int_0^{2s-1}r\,\mathrm{d}r\\
&=(2s-1)^2
\end{align}
$$
Since $s=\sqrt{\frac\omega\Omega}$, this verifies the formula guessed when $\frac\omega\Omega\ge\frac14$. When $\frac\omega\Omega\lt\frac14$, the limit of the expected values is $0$.
Explanation of the Claim in the Second Paragraph
Symmetry says that the density will be radially symmetric, so we will evaluate at the point $(r,0)$.
Consider
$$
\frac1{\pi(1-s)^2}\iint\overbrace{[|t|\lt1-s]}^{t\in B(1-s,0)}\,\overbrace{[|x-t|\lt s]}^{x\in B(s,t)}\,\mathrm{d}t\,\delta(x-(r,0))\,\mathrm{d}x
$$
Integrating in $t$ first, we get the average of a disk of radius $s$ whose center has been placed uniformly in the disk of radius $1-s$ centered at the origin. Integrating in $x$, against the delta function, we get evaluation of the resultant average at the point $(r,0)$. This gives the average inclusion probability at $(r,0)$.
The substitution $t\mapsto x-t$ yields
$$
\frac1{\pi(1-s)^2}\iint[|x-t|\lt1-s]\,[|t|\lt s]\,\mathrm{d}t\,\delta(x-(r,0))\,\mathrm{d}x
$$
and now we can apply Fubini to change the order of integration
$$
\frac1{\pi(1-s)^2}\iint[|x-t|\lt1-s]\,\delta(x-(r,0))\,\mathrm{d}x\,[|t|\lt s]\,\mathrm{d}t
$$
Integrating in $x$ yields
$$
\frac1{\pi(1-s)^2}\int[|(r,0)-t|\lt1-s]\,[|t|\lt s]\,\mathrm{d}t
$$
which is the area of the intersection of a disk of radius $s$ and a disk of radius $1-s$ whose centers are separated by a distance $r$ divided by area of a disk of radius $1-s$.
Best Answer
Here is one possible interpretation of this problem, and method of calculating an answer.
Let $D(r)$ denote the disk of radius $r$ centered at $(0,0)$ in the plane. Pick a random vector $X$ uniformly from $D(15)$ and independently pick a random vector $Y$ uniformly from $D(20)$. What is sought is the probability that $X+Y\in D(10).$
That is, the event that origin $(0,0)$ is within distance $10$ of the point $X+Y$ can be expressed either by stating that $(0,0)$ is in the $10$-disk centered at $X+Y$ or, equivalently, that $X+Y$ is in the $10$-disk centered at $(0,0)$.
The chance of this can (in principle) be worked out by writing the density function of $\|X+Y\|$ in terms of Bessel functions, exploiting the rotational symmetry of the density functions of $X$, of $Y$, and of $X+Y$.