Difference between revisions of "2017 AMC 12B Problems/Problem 9"
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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> | <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 1== |
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>. | ||
Revision as of 10:22, 17 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 1
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)
Solution 2: Shortcut with right triangles
Note the specificity of the radii, and , and that specificity is often deliberately added to simplify the solution to a problem.
One may recognize as the hypotenuse of the 
right triangle and as the hypotenuse of the right triangle with legs 
and 
. 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 
units to the right of and 
units above the center of the first circle, we find the point 
, which is in fact unit to the left of and units below the center of the second circle at 
.
Plugging 
into 
gives us 
.
A similar solution uses the other intersection point, 
.
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. 
