To find the desired overlap area

, recall that the area of a circle is

, and then require that

, which can immediately be solved to find that

. To find the distance

, remember that the area of a circular sector of angle

and radius

is

. (This gives the expected area for the whole circle when

.) Half of the overlap area can be found by subtracting the area of the isosceles triangle formed by the intersections of the two circles and the center of one of them

from the sector area, and then doubling gives the area of the lens in terms of

and

:

. The distance between the two centers is given by

. Setting

and numerically solving for

and

gives

and

.

The overlap area can also be found using integral calculus. For convenience, we again let the radius of the circles be 1. Then the area of each circle is

, and the area of each of the regions must be

. Let the centers of the two circles be

and

, respectively. The lens is symmetric vertically and horizontally, and one quarter of it is bounded by

,

, and the circle

. So

. Setting this equal to

and solving numerically for

again gives approximately 0.807946.

See

[1] for a solution arrived at independently.

The alert reader may have recognized the numeric value of the equal areas shown in the display, 1.570796327, as approximately

, and we have seen that this is no coincidence. But the numerical value of the distance between circle centers has no immediately obvious significance and cannot even be determined precisely, except as the solution to the transcendental equation above.

It is interesting to note that the use of

ImplicitRegion and

RegionMeasure, new functions introduced in

*Mathematica* 10, immediately gives the formula for the overlap area as a function of separation distance:

Assuming[0<d<2,

Simplify[RegionMeasure[ImplicitRegion[x^2 + y^2 <= 1 && (x-d)^2 + y^2 <= 1, {x,y}]]]]