Area of circles: represent $x$ in terms of $r_1$ and $r_2$ See the image. Area of green and red regions are equal. Can you represent $x=|O_2D|$ in terms of $r_1$ and $r_2$ for $r_1> r_2$ ?

Edit: The point $O_1$ does not enter in the region of small circle.
 A: Consider this pink rhombus:

If we split the green area into two sections (P and Q) along line M, we can find each of their areas in terms of the radius and angle of one of the circles, by subtracting the area of the triangle formed by the two radii and M from the area of the whole sector. (I'm going to use big $R$ for $r_1$ and little $r$ for $r_2$.)
$$ Area\space of\space P = \frac{\theta}{2\pi}{\pi{r^2}} - r^2\cos{\frac{\theta}{2}}\sin{\frac{\theta}{2}}   $$
$$ Area\space of\space Q = \frac{\phi}{2\pi}{\pi{R^2}} - R^2\cos{\frac{\phi}{2}}\sin{\frac{\phi}{2}}   $$
$$Area \space of\space Q + \space Area \space of \space P = \frac{\phi}{2\pi}{\pi{R^2}} - R^2\cos{\frac{\phi}{2}}\sin{\frac{\phi}{2}} + \frac{\theta}{2\pi}{\pi{r^2}} - r^2\cos{\frac{\theta}{2}}\sin{\frac{\theta}{2}} = \frac{1}{2}\pi {r^2}\tag{1}$$
That's four variables. We can relate $\theta$ to $\phi$ by expressing the length of $M$ in terms of each.
$$ 2r\sin{\frac{\theta}{2}} = M = 2R\sin{\frac{\phi}{2}}$$
$$ \phi = 2\arcsin{(\frac{r}{R}\cdot\sin{\frac{\theta}{2}})}\tag{2}$$
Pluging $(2)$ into $(1)$ should be sufficient to express $\theta$ in terms of $r$ and $R$.
Now, $x$ is $R$ less the length of the rhombus.
$$ x = R - r\cos{\frac{\theta}{2}} - R\cos{\frac{\phi}{2}} $$
and the rest is algebra.
