2017 AMC 12B Problems/Problem 9

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Problem 9

A circle has center $(-10, -4)$ and has radius $13$. Another circle has center $(3, 9)$ and radius $\sqrt{65}$. The line passing through the two points of intersection of the two circles has equation $x+y=c$. What is $c$?

Solution

The equations of the two circles are $(x+10)^2+(y+4)^2=169$ and $(x-3)^2+(y-9)^2=65$. Rearrange them to $(x+10)^2+(y+4)^2-169=0$ and $(x-3)^2+(y-9)^2-65=0$, respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation $(x+10)^2+(y+4)^2-169=x-3)^2+(y-9)^2-65$. We can simplify this like follows. $(x+10)^2+(y+4)^2-169=x-3)^2+(y-9)^2-65$ \rightarrow $(x^2+20x+100)+(y^2+8y+16)-(x^2-6x+9)-(y^2-18y+81)=104$ \rightarrow $26x+26y+26=104$ \rightarrow $26x+26y=78$ \rightarrow $x+y=3$. Thus, $c = \boxed{\textbf{(D)}\ 3}$

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AMC 12 Problems and Solutions

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