Difference between revisions of "2017 AMC 12B Problems/Problem 6"
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Because the two points are on a diameter, the center must be halfway between them at the point (4,3). The distance from (0,0) to (4,3) is 5 so the circle has radius 5. Thus, the equation of the circle is <math>(x-4)^2+(y-3)^2=25</math>. | Because the two points are on a diameter, the center must be halfway between them at the point (4,3). The distance from (0,0) to (4,3) is 5 so the circle has radius 5. Thus, the equation of the circle is <math>(x-4)^2+(y-3)^2=25</math>. | ||
| − | To find the x-intercept, y must be 0, so <math>(x-4)^2+(0-3)^2=25</math>, so <math>(x-4)^2=16</math>, <math>x-4=4</math>, <math>x=8, boxed{B}</math>. | + | To find the x-intercept, y must be 0, so <math>(x-4)^2+(0-3)^2=25</math>, so <math>(x-4)^2=16</math>, <math>x-4=4</math>, <math>x=8, \boxed{B}</math>. |
Written by: SilverLion | Written by: SilverLion | ||
Revision as of 11:41, 12 January 2018
Problem 6
The circle having 
and 
as the endpoints of a diameter intersects the 
-axis at a second point. What is the -coordinate of this point?
Solution
Because the two points are on a diameter, the center must be halfway between them at the point (4,3). The distance from (0,0) to (4,3) is 5 so the circle has radius 5. Thus, the equation of the circle is 
.
To find the x-intercept, y must be 0, so 
, so 
, 
, 
.
Written by: SilverLion
See Also
| 2017 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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. 
