2017 AMC 12B Problems/Problem 9
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
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.