So I have some circles that look kind of 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.
I'm looking for a way to find the distance $BD$ which is also the radius of the circle centered at the point $B$.
Best Answer
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.