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Difference between revisions of "2017 AMC 12B Problems/Problem 9"

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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>.
 
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>.
  
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Solution by TheUltimate123 (Eric Shen)
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Solution by TheUltimate123
  
 
==Solution 2: Shortcut with right triangles==
 
==Solution 2: Shortcut with right triangles==

Revision as of 19:00, 4 February 2019

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 $c$?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}$

Solution 1

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 the following. $(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65 \rightarrow (x^2+20x+100)+(y^2+8y+16)-(x^2-6x+9)-(y^2-18y+81)=104 \rightarrow 26x+26y+26=104 \rightarrow 26x+26y=78 \rightarrow x+y=3$. Thus, $c = \boxed{\textbf{(A)}\ 3}$.

Solution by TheUltimate123

Solution 2: Shortcut with right triangles

Note the specificity of the radii, $13$ and $\sqrt{65}$, and that specificity is often deliberately added to simplify the solution to a problem.

One may recognize $13$ as the hypotenuse of the $\text{5-12-13}$ right triangle and $\sqrt{65}$ as the hypotenuse of the right triangle with legs $1$ and $8$. We can suppose that the legs of these triangles connect the circles' centers to their intersection along the gridlines of the plane.

If we suspect that one of the intersections lies $12$ units to the right of and $5$ units above the center of the first circle, we find the point $(-10 + 12,-4 + 5) = (2,1)$, which is in fact $1$ unit to the left of and $8$ units below the center of the second circle at $(3,9)$.

Plugging $(2,1)$ into $x + y$ gives us $c = 2 + 1 = \boxed{\textbf{(A) } 3}$.

A similar solution uses the other intersection point, $(-5,8)$.


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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