What is the distance between the centers? Two unit circles overlap in such a way that exactly half the area of each circle is within the overlap. How far apart are the centers of the two circles?
 A: I suggest you position the circles so that their centers are on the $y$-axis at points $(0,a)$ and $(0,-a)$. Then you can find the area of their overlap by an integral - you just need to determine the $x$-limits of the integral and the two functions $y_1(x)$ and $y_2(x)$ bounding the region. The area of the region will depend on $a$. You want to know when the area is exactly equal to half the area of the unit circle (i.e. $\frac{\pi}{2}$), so set this equal to $\frac{\pi}{2}$ and solve for $a$. Then you can find the distance between the centers of the circles. 

A: 
If the area of overlap is $\frac 12$ the area of each circle then the red area is $\frac 14$ the area of the circle.
Lets assume that the radius of these circles is 1 
The area of the red region plus the green region is the section of a circle with angle $A = \frac 12 A$ (with $A$ measured in radians)
The area of the green triangle is $\frac 12 \sin A$
$\frac 12 A - \frac 12 \sin A  =  \frac \pi4\\
A - \sin A  =  \frac \pi2$
$A\approx 2.31$ 
the distance between the centers is $2 \cos \frac 12 A \approx 0.81$
If the radius is not $1\approx 0.81 r$
A: 
One half of the sought area
is found as a difference between the area
of the circular sector $\smallfrown O_1BA$
and the area of $\triangle AO_1B$,
so we have an equation
in terms of the angle $\phi=\angle AO_1B$
\begin{align}
2(S_{\smallfrown O_1BA}-
S_{\triangle AO_1B})&=\tfrac12\cdot\pi\cdot 1^2
,\\
\phi-\sin\phi &= \tfrac\pi2
,
\end{align}
which after the substitution $\phi=\mathbf{d}+\tfrac\pi2$
transforms to a well-known equation
\begin{align}
\cos\mathbf{d}&=\mathbf{d}
,
\end{align}
which has a numeric solution
$\mathbf{d}\approx 0.73908513321516$,
known as
the Dottie number,
A003957.
Then the distance $|O_1O_2|$ between the centers of the circles
is found as
\begin{align}
2\cos\tfrac\phi2&=
2\cos(\tfrac{\mathbf{d}}2+\tfrac\pi4)
=\sqrt2(\cos\tfrac{\mathbf{d}}2-\sin\tfrac{\mathbf{d}}2)
\\&=
\sqrt{1+{\mathbf{d}}}-\sqrt{1-{\mathbf{d}}}
=\sqrt{2-2\sqrt{1-\mathbf{d}^2}}
\approx 0.807945506599
,
\end{align}
which is also known as
A133741,
"decimal expansion of offset at which two unit disks overlap by half each's area".
