# Distance of centers of two unit circles with area of intersection $1/2$ area of one.

In a psychology 101 text book it said on the margins that the perfect relationship between two people is when the amount of things they have it common exactly equals the amount of qualities that are unique to each individual. It showed two circles of equal radius that have an intersection area equal to half the area of one circle.

What is this distance if both circles are unit circles?

The area of the segment is $$A=1/2(a-\sin(a))$$ and the intersection is twice this the area of intersection is: $$a-\sin(a)$$ Given a unit circle area is $$\pi$$, the angle for the solution is: $$\pi/2 = a-\sin(a)$$. The distance between the two circles in then: $$D = 2\times \text{cos}(a)$$

And this is where I'm stuck.

Wolfram Alpha gives the unique solution of equation $$a-\sin a=\dfrac{\pi}{2}$$ as $$a\approx 2.30988\approx 132.35^\circ$$
Notice, $$a$$ is the aperture angle exerted by the common chord at the center of any of intersecting unit circles. Therefore the distance between the centers of intersecting unit circles is $$D=2\cos \frac a2\approx 2\cos\left(\frac{132.35^\circ}{2}\right)\approx \color{blue}{0.807946842 \ \text{unit}}$$