Originally Posted by

**barwick11** I wrote a long lengthy response to explain exactly what I was thinking, and found the solution in there...

Two circles of diameter X that are touching have X distance between their centers. If a third circle were placed on top of those two circles, the distance between all their centers Z would be Z = X (an equilateral triangle). The vertical distance between the centers of these circles (H) would be able to be calculated such that X^2 = H^2 + (1/2 X)^2 (in other words, A^2 + B^2 = C^2)

If the two bottom circles were separated, same story, except now it's X^2 = H^2 + (1/2 (X + Y)) ^2