Difference between revisions of "2017 AMC 12B Problems/Problem 9"
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==Problem 9== | ==Problem 9== | ||
A circle has center <math>(-10, -4)</math> and has radius <math>13</math>. Another circle has center <math>(3, 9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x+y=c</math>. What is <math>c</math>? | A circle has center <math>(-10, -4)</math> and has radius <math>13</math>. Another circle has center <math>(3, 9)</math> and radius <math>\sqrt{65}</math>. The line passing through the two points of intersection of the two circles has equation <math>x+y=c</math>. What is <math>c</math>? | ||
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| + | <math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}</math> | ||
==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 the following. <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{( | + | 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 the following. <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{(A)}\ 3}</math>. |
Solution by TheUltimate123 (Eric Shen) | Solution by TheUltimate123 (Eric Shen) | ||
Revision as of 21:32, 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 the following. 
. 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
