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So I have some circles that look kinda like this:

I'm given the radius of the circle with center point A which is also the distance AB, the distance AB between the two center points on the x axis (they share same y values for the centers), and the distance CD which is the height of the shape created by the intersection.

We can also assume that despite my drawing the circle ARE perfect circles.

I'm looking for a way to find the distance BD which is also the radius of the circle centered on point B.

Thanks in advance for any help.

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"which is also the distance AC" –  Sparr Sep 21 '10 at 22:04

1 Answer 1

up vote 3 down vote accepted

First, note that segment AB bisects segment CD. Call their point of intersection E, which is also the midpoint of CD. Since you know the length of CD, you know the length of CE. The measure of angle BAC (which could also be called EAC) is $\sin^{-1}\left(\frac{CE}{AC}\right)$. Apply the Law of Cosines to triangle ABC to find BC=BD: $$BC^2=AC^2+AB^2-2\cdot AC\cdot AB\cdot\cos\left(\sin^{-1}\left(\frac{CE}{AC}\right)\right).$$ Fill in the known lengths and solve.

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As above, the first step is to solve the ACE triangle. Once you do that, you can easily calculate the coordinates of all the points. And from that, calculate the radius you need. –  phv3773 Sep 21 '10 at 20:32
thanks for you help that worked perfectly. –  AvatarOfChronos Sep 21 '10 at 21:11

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