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

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==Problem==
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In the <math>xy</math>-plane, a circle of radius <math>4</math> with center on the positive <math>x</math>-axis is tangent to the <math>y</math>-axis at the origin, and a circle with radius <math>10</math> with center on the positive <math>y</math>-axis is tangent to the <math>x</math>-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?
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<math>\textbf{(A)}\ \dfrac{2}{7} \qquad\textbf{(B)}\ \dfrac{3}{7}  \qquad\textbf{(C)}\ \dfrac{2}{\sqrt{29}}  \qquad\textbf{(D)}\ \dfrac{1}{\sqrt{29}}  \qquad\textbf{(E)}\ \dfrac{2}{5}</math>
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==Solution==
 
==Solution==
  

Revision as of 20:37, 15 November 2023

Problem

In the $xy$-plane, a circle of radius $4$ with center on the positive $x$-axis is tangent to the $y$-axis at the origin, and a circle with radius $10$ with center on the positive $y$-axis is tangent to the $x$-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?

$\textbf{(A)}\ \dfrac{2}{7} \qquad\textbf{(B)}\ \dfrac{3}{7}  \qquad\textbf{(C)}\ \dfrac{2}{\sqrt{29}}  \qquad\textbf{(D)}\ \dfrac{1}{\sqrt{29}}  \qquad\textbf{(E)}\ \dfrac{2}{5}$

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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