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Difference between revisions of "1954 AHSME Problems/Problem 9"

(Created page with "== Problem== A point <math>P</math> is outside a circle and is <math>13</math> inches from the center. A secant from <math>P</math> cuts the circle at <math>Q</math> and <mat...")
 
 
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Using the Secant-Secant Power Theorem, you can get <math>9(16)=(13-r)(13+r)</math>, where <math>r</math> is the radius of the given circle. Solving the equation, you get a quadratic: <math>r^2-25</math>. A radius cannot be negative so the answer is <math>\boxed{\textbf{(C) }5"}</math>
 
Using the Secant-Secant Power Theorem, you can get <math>9(16)=(13-r)(13+r)</math>, where <math>r</math> is the radius of the given circle. Solving the equation, you get a quadratic: <math>r^2-25</math>. A radius cannot be negative so the answer is <math>\boxed{\textbf{(C) }5"}</math>
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==See Also==
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{{AHSME 50p box|year=1954|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 19:39, 17 February 2020

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 $\boxed{\textbf{(C) }5"}$

See Also

1954 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AHSME Problems and Solutions


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