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2017 AMC 12B Problems/Problem 9: Difference between revisions

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


==See Also==
==See Also==
{{AMC12 box|year=2017|ab=B|before=8|num-a=10}}
{{AMC12 box|year=2017|ab=B|num-b=8|num-a=10}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 21:20, 16 February 2017

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 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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