What's the distance between two circumcenters This problem was came to a Facebook post of mathigon ......the problem seems trickier than I expected...can you help me to find the distance??
 A: 
Let $\angle CAB = \angle ABC = \angle BAD = \angle DBA = \theta$
Area of the $\bigtriangleup CAD = \frac{1}{2} 1^2 \sin 2 \theta = \frac{1}{2} \sin 2 \theta$
Area of the sector $CAD$ (consider the right side of the circle on the left) $ = \frac{1}{2} 1^2 2\theta = \frac{1}{2} (2 \theta)$
So the common area between the circles $ = 2 (\frac{1}{2} (2 \theta) - \frac{1}{2} \sin 2 \theta)$
However, it is given that the common area between the circles $= \frac{\pi}{2}$
Hence we need to solve the following equation for $\theta$
$$2 \left(\frac{1}{2} \left(2 \theta \right) - \frac{1}{2} \sin 2 \theta \right) = \frac{\pi}{2}$$
Solving, $\theta = 1.15494$
Finally length of $AB$ is $2 \cos \theta = 0.80794684217$
A: The area of the lens formed by two overlapping circle of radius $r$ and at distance $d$ apart is
given by https://mathworld.wolfram.com/Lens.html
$$Area= \pi r^2-2r^2 \tan^{-1}\frac d{\sqrt{4r^2 -d^2}}-\frac12d \sqrt{4r^2 -d^2}
$$
Set $Area =\frac\pi2$ and $r=1$ to obtain numerically $d= 0.8079$.
