Distance between 2 circles with same radius to overlap a desired percentage? I want to draw 2 circles of the same size that have a specified percentage of area overlapping.
For instance, if the overlapping percentage is 0%, the circles are next to each other and the distance between the centers is equivalent to the radius. If the overlapping percentage is 100%, they are completely on top of each other (looking like 1 circle), and the distance between the centers is 0. If the overlapping percentage is 50% they intersect so that 50% of their area is overlapping.
What is an equation that, given the percentage of area two circles should overlap, tells you the distance the centers of the circles need to be from each other?
 A: It's a straightforward but messy exercise to find the area of the "lens" formed by the intersection of the two circles.  If the radii are the same, consider a circle with center $(0,0)$ and radius 1 and another with center $(d,0)$ and radius $1$. Then the area of the lens divided by $\pi$ is the percentage you are looking for.  The circles are 
$$C_0: \ \ x^2+y^2 = 1,\ \ \ C_d: \ \ (x-d)^2 +y^2 = 1.$$
Solving for the intersection points we find
$$ P_1 = (d/2,\sqrt{4 - d^2}/2 , \ \ P_2 = (d/2,-\sqrt{4 - d^2}/2) $$
The segment $P_1 P_2$ is a chord that cuts both circles.  You can find formulas for the area between each circle and the chord -- then add them to obtain the area of the lens.
The area between the chord and $C_0$ is 
$$A_0 = \arccos(d/2)- d\sqrt{4 - d^2}/4.$$
The area between the chord and $C_d$ is identical by symmetry 
$$A_d = \arccos(d/2)- d\sqrt{4 - d^2}/4.$$
Hence the percentage of overlap is 
$$O = \frac{2 \arccos(d/2)- d\sqrt{1 - d^2/4}}{\pi}.$$
In order to find $d$ for a specified value of $O$, you have to solve a non-linear equation numerically. There is no closed form solution.
If $O = 0.5$ then $d = 0.8079455...$ approximately. 
A: I get a slightly different answer. Very sorry for not simply submitting a comment, but my neophyte status prohibits that. 
I get $$O = \frac{2 \arccos(d/2)- d\sqrt{1 - d^2/4}}{\pi}.$$
I'm sure that was just a quick fraction mess-up. Again, sorry for not just commenting.
