https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_9&feed=atom&action=history1954 AHSME Problems/Problem 9 - Revision history2024-03-29T15:25:37ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_9&diff=117892&oldid=prevRayfish at 00:39, 18 February 20202020-02-18T00:39:59Z<p></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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></div></td></tr>
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</table>Rayfishhttps://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_9&diff=95574&oldid=prevSkyraptor79: 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..."2018-06-23T19:48:33Z<p>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..."</p>
<p><b>New page</b></p><div>== Problem==<br />
<br />
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 <math>R</math> <br />
so that the external segment of the secant <math>PQ</math> is <math>9</math> inches and <math>QR</math> is <math>7</math> inches. The radius of the circle is: <br />
<br />
<math>\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7" </math><br />
<br />
<br />
==Solution==<br />
<br />
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></div>Skyraptor79