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2012 AMC 12A Problem 16

Problem 16 of 25IntermediateGeometry

Circle C1C_1 has its center OO lying on circle C2.C_2. The two circles meet at XX and Y.Y. Point ZZ in the exterior of C1C_1 lies on circle C2C_2 and XZ=13,XZ = 13, OZ=11,OZ = 11, and YZ=7.YZ = 7. What is the radius of circle C1?C_1?

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Solution

Let rr be the radius of C1,C_1, so OX=OY=r.OX = OY = r. These are equal chords of C2,C_2, so they subtend equal angles at Z:Z: XZO=OZY.\angle XZO = \angle OZY. Applying the Law of Cosines to triangles XZOXZO and YZO,YZO, 132+112r221311=72+112r22711. \begin{aligned} &\frac{13^2 + 11^2 - r^2}{2 \cdot 13 \cdot 11} \\ &= \frac{7^2 + 11^2 - r^2}{2 \cdot 7 \cdot 11}. \end{aligned} Clearing denominators and solving gives r2=30,r^2 = 30, so r=30.r = \sqrt{30}. Thus, the correct answer is E.

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Concepts: law of cosines · chord · circle

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