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Difference between revisions of "2023 AMC 12B Problems/Problem 10"

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==Solution==
  
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The center of the first circle is <math>(4,0)</math>.
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The center of the second circle is <math>(0,10)</math>.
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Thus, the slope of the line that passes through these two centers is <math>- \frac{10}{4} = - \frac{5}{2}</math>.
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Because this line is the perpendicular bisector of the line that passes through two intersecting points of two circles, the slope of the latter line is <math>\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}</math>.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 18:28, 15 November 2023

Solution

The center of the first circle is $(4,0)$. The center of the second circle is $(0,10)$. Thus, the slope of the line that passes through these two centers is $- \frac{10}{4} = - \frac{5}{2}$.

Because this line is the perpendicular bisector of the line that passes through two intersecting points of two circles, the slope of the latter line is $\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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