Problem:
Circle C1 has its center O lying on circle C2. The two circles meet at X and Y. Point Z in the exterior of C1 lies on circle C2 and XZ=13,OZ=11, and YZ=7. What is the radius of circle C1 ?
Answer Choices:
A. 5
B. 26
C. 33
D. 27
E. 30 Solution:
Let r be the radius of C1. Because OX=OY=r, it follows that ∠OZY=∠XZO. Applying the Law of Cosines to triangles XZO and OZY gives
Let P be the point on XZ such that ZP=ZY=7. Because OZ is the bisector of ∠XZY, it follows that △OPZ≅△OYZ. Therefore OP=OY=r and thus P is on C1. By the Power of a Point Theorem, 13⋅7=ZX⋅ZP=OZ2−r2=112−r2. Solving for r2 gives r2=30 and so r=30.
