Difference between revisions of "2017 AMC 12B Problems/Problem 9"
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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 | + | 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 \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</math>. Thus, <math>c = \boxed{\textbf{(D)}\ 3}</math> |
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+ | Solution by TheUltimate123 (Eric Shen) | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2017|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:21, 16 February 2017
Problem 9
A circle has center and has radius . Another circle has center and radius . The line passing through the two points of intersection of the two circles has equation . What is ?
Solution
The equations of the two circles are and . Rearrange them to and , respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation . We can simplify this like follows. . Thus,
Solution by TheUltimate123 (Eric Shen)
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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