distance between centres of two overlapping congruent circles If there are two overlapping congruent circles such that the area of intersection is 10% of the area of each circle, what is the distance between their centres in terms of the radius r cm?
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.
Is it pi(rrr/5-rrrr/200)?
You assume one of the circles is xx+yy=rr while the other a circle with same radius but symmetric wrt x-axis. Assume one of the intersection points is (b,sqrt(rr-bb). Want b and then 2b is the required answer. The difference between the crcular sector  between the x-axis and the point (b,sqrt(rr-bb) and the triangle (0,0),(b,0),(b,sqrt(rr-bb)) is 1/40 .pi.rr.