1954 AHSME Problems/Problem 9

Problem

A point $P$ is outside a circle and is $13$ inches from the center. A secant from $P$ cuts the circle at $Q$ and $R$ so that the external segment of the secant $PQ$ is $9$ inches and $QR$ is $7$ inches. The radius of the circle is:

$\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7"$

Solution

Using the Secant-Secant Power Theorem, you can get $9(16)=(13-r)(13+r)$ , where $r$ is the radius of the given circle. Solving the equation, you get a quadratic: $r^2-25$ . A radius cannot be negative so the answer is $\boxed{\textbf{(C) }5"}$

1954 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1954 AHSME Problems/Problem 9

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