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Revision as of 08:57, 4 July 2013

Problem

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 $C_1$?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ \sqrt{26}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 2\sqrt{7}\qquad\textbf{(E)}\ \sqrt{30}$

Solution

Solution 1

Let $r$ denote the radius of circle $C_1$. Note that quadrilateral $ZYOX$ is cyclic. By Ptolemy's Theorem, we have $11XY=13r+7r$ and $XY=20r/11$. Let t be the measure of angle $YOX$. Since $YO=OX=r$, the law of cosines on triangle $YOX$ gives us $\cos t =-79/121$. Again since $ZYOX$ is cyclic, the measure of angle $YZX=180-t$. We apply the law of cosines to triangle $ZYX$ so that $XY^2=7^2+13^2-2(7)(13)\cos(180-t)$. Since $\cos(180-t)=-\cos t=79/121$ we obtain $XY^2=12000/121$. But$XY^2=400r^2/121$ so that $r=\sqrt{30}$. $\boxed{E}$.

Solution 2

Let us call the $r$ the radius of circle $C_1$, and $R$ the radius of $C_2$. Consider $\triangle OZX$ and $\triangle OZY$. Both of these triangles have the same circumcircle ($C_2$). From the Extended Law of Sines, we see that $\frac{r}{\sin{\angle{OZY}}} = \frac{r}{\sin{\angle{OZX}}}= 2R$. Therefore, $\angle{OZY} \cong \angle{OZX}$. We will now apply the Law of Cosines to $\triangle OZX$ and $\triangle OZY$ and get the equations

$r^2 = 13^2 + 11^2 - 2 \cdot 13 \cdot 11 \cdot \cos{\angle{OZX}}$,

$r^2 = 11^2 + 7^2 - 2 \cdot 11 \cdot 7 \cdot \cos{\angle{OZY}}$,

respectively. Because $\angle{OZY} \cong \angle{OZX}$, this is a system of two equations and two variables. Solving for $r$ gives $r = \sqrt{30}$. $\boxed{E}$.

Solution 3

Let $r$ denote the radius of circle $C_1$. Note that quadrilateral $ZYOX$ is cyclic. By Ptolemy's Theorem, we have $11XY=13r+7r$ and $XY=20r/11$. Consider isosceles triangle $XOY$. Pulling an altitude to $XY$ from $O$, we obtain $\cos(\angle{OXY}) = \frac{10}{11}$. Since quadrilateral $ZYOX$ is cyclic, we have $\angle{OXY}=\angle{OZY}$, so $\cos(\angle{OXY}) = \cos(\angle{OZY})$. Applying the Law of Cosines to triangle $OZY$, we obtain $\frac{10}{11} = \frac{7^2+11^2-r^2}{2(7)(11)}$. Solving gives $r=\sqrt{30}$. $\boxed{E}$.

-Solution by thecmd999

Solution 4

Let $P = XY \cap OZ$. Consider an inversion about $C_1 \implies C_2 \to XY, Z \to P$. So, $OP \cdot OZ = r^2 \implies OP = r^2/11 \implies PZ = \dfrac{121 - r^2}{11}$. Using $\triangle YPZ \sim OXZ \implies r = \sqrt{30} \implies \boxed{E}$.


-Solution by IDMasterz

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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