2012 AMC 12A Problem 16

Question

Circle

C_1

has its center

O

lying on circle

C_2.

The two circles meet at

X

and

Y.

Point

Z

in the exterior of

C_1

lies on circle

C_2

and

XZ = 13,

OZ = 11,

and

YZ = 7.

What is the radius of circle

C_1?

5

\sqrt{26}

3\sqrt{3}

2\sqrt{7}

\sqrt{30}

Let

r

be the radius, so

OX = OY = r

; these equal chords of

C_2

subtend equal angles at

Z

Apply the Law of Cosines to triangles

XZO

and

YZO

and set the two expressions for

\cos\angle Z

equal

Let

r

be the radius of

C_1,

so

OX = OY = r.

These are equal chords of

C_2,

so they subtend equal angles at

Z:

\angle XZO = \angle OZY.

Applying the Law of Cosines to triangles

XZO

and

YZO,

\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

r^2 = 30,

so

r = \sqrt{30}.

Thus, the correct answer is E.

Problem 16 in other years