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.
The key point to a proof is that if you have three non-collinear points, they determine a unique circle. (So two distinct circles can intersect in at most two points.) You can prove this by construction: The center of the circle will be the intersection of the perpendicular bisectors of the segments joining pairs of the points.
Best Answer
Label the center of the first circle $C$ and the center of the second circle $C'$. Label one of the points of intersection of the two circles $A$ and the other $B$. Let the radius of the circles be $r>0$. It should be clear that the following lengths are all equal to $r$. $AC$, $AC'$, $BC$, $BC'$, $CC'$. With a simple application of Pythagoras' Theorem, we get that the length of the line segment $AB$ is $\sqrt{3}r$.
With some basic trigonometry, we find the angles $\angle ACB=\angle AC'B=\dfrac{2\pi}{3}$. So, the area of one half of the intersection is the area of a circular segment with angle $\theta=\dfrac{2\pi}{3}$ and radius $r$, which gives an area of $\dfrac{r^2}{2}(\theta-\sin\theta)=\dfrac{r^2}{2}\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right)$ and so the area of the entire intersection is twice this. This gives an area of $$r^2\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right).$$