2023 AMC 12B Problems/Problem 10: Difference between revisions

Revision as of 02:09, 16 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 1

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)

Solution 2 (Coordinate Geometry)

The first circle can be written as $(x-4)^2 + y^2 = 4^2$ we'll call this equation $(1)$ The second can we writen as $x^2 + (y-10)^2 = 10^2$ , we'll call this equation $(2)$

Expanding $(1)$ : $\begin{align*} x^2 -8x + 4^2 + y^2 &= 4^2 \\ x^2 - 8x + y^2 &= 0 \end{align*}$ Exapnding $(2)$ $\begin{align*} x^2 + y^2 -20y + 10^2 = 10^2\\ x^2 + y^2 - 20y = 0 \end{align*}$

Now we can set the equations equal to eachother: $\begin{align*} x^2 - 8x + y^2 &= x^2 + y^2 - 20y \\ \frac{8}{20}x &= y \\ \frac{2}{5}x &= y \end{align*}$ This is in slope intercept form therefore the slope is $\boxed{\textbf{(E) } \frac{2}{5}}$ .

~luckuso

Video Solution 1 by OmegaLearn

https://youtu.be/IUB6r1iNgpU

2023 AMC 12B (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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 ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
+==Video Solution 1 by OmegaLearn==
+https://youtu.be/IUB6r1iNgpU
 ==See Also==
 {{AMC12 box|year=2023|ab=B|num-b=9|num-a=11}}
 {{MAA Notice}}

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