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

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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==See Also==
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{{AMC12 box|year=2023|ab=B|num-b=9|num-a=11}}
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{{MAA Notice}}

Revision as of 20:27, 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)

See Also

2023 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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