2023 AMC 12B Problems/Problem 10

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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2023 AMC 12B Problems/Problem 10

Solution