Let $O$ and $O'$ be the centers and $P$ and $Q$ be the intersection points of both circles.
First, let's calculate the area of the circular segment $PQ$. We need the angle $\alpha=\angle POQ$.
Call $d=OO'/2$ and $r$ the radius of the circles. Then
$$\alpha=2\arccos\frac dr$$
The area of the circular segment is obtained substracting the triangle $OPQ$ to the circular sector $OPQ$, and is
$$\frac{\alpha r^2}{2}-d\sqrt{r^2-d^2}$$
being $\alpha$ expressed in radians (this makes the formulae clearer).
The area of intersection is twice the circular segment's. Therefore, you must solve this for $d$:
$$r^2\arccos\frac dr-d\sqrt{r^2-d^2}=0.05\pi r^2$$
I'd say that this is impossible to solve by algebraic methods, but any math program can do it with numerical methods, given the radius.
The equation becomes a bit cleaner if you call $k=d/r$ the ratio between $d$ and $r$. Dividing the equation by $r^2$ yields:
$$\arccos k-k\sqrt{1-k^2}=0.05\pi$$
Sadly it still can't be solved by algebraic methods.
PS: Don't forget that the arccos must be in radians, and that $d$ is the distance between centers halved.
Since the radius increases at a constant rate relative to the change of the angle $\theta$ the polar equation would be $r = r_1 + \frac{r_2-r_1}{\theta_1 - \theta_0}\theta$ where $r_2 > r_1$. To find the distance you need the integral $\int_{\theta_0}^{\theta_1}\sqrt{1+(\frac{\text{d}r}{\text{d}\theta})^2}\text{d}\theta$, where $\frac{\text{d}r}{\text{d}\theta}=\frac{r_2-r_1}{\theta_1 - \theta_0}$. Which would yield a length of $(\theta_1 - \theta_0)\sqrt{1 + (\frac{r_2-r_1}{\theta_1 - \theta_0})^2}$. Even simpler it would be $\sqrt{(\theta_1 - \theta_0)^2 + (r_2-r_1)^2}$.
Best Answer
Suppose length of the arc is l, radius of circles is r, then $r\theta = l$ where $\theta$ (in radian) is angle subtended by the arc. So $\theta = l/r$. Then the distance between center of the two circles is $2r\cos\dfrac{\theta}{2}$