SOLUTION: two circles intersect and have a common chord that is 16cm long . the centers are 21cm apart . if the radius of one circle is 10cm ,find the radius of the other circle

Question 344990: two circles intersect and have a common chord that is 16cm long . the centers are 21cm apart . if the radius of one circle is 10cm ,find the radius of the other circle
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!

First we will assume the smaller circle is the one with radius 10.
We will let the radius of the larger circle be r. We draw in a
radius of each circle (in green):

Then AC = 10, AB=21, BC = r, and CE = 8 since it is half of
the common chord CD, which is 16.

Triangle ACE is a right triangle, so we can find AE by the Pythagorean
theorem:

(We can see here that the larger circle could not have been the
one with radius 10, for that would have made BE = 6 which would have
been shorter than AE.)
Therefore since AB = 21 and AE = 6
AB = AE + EB
21 = 6 + EB
15 = EB

Triangle CEB is a right triangle, so we can find r by
the Pythagorean theorem

Edwin
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