Difference between revisions of "2017 AMC 12B Problems/Problem 6"
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==Solution== | ==Solution== | ||
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| − | Because the two points are a diameter the center must be half way between them at the point (4,3). The distance from (0,0) to (4,3) is 5 therefore the circle has a radius of 5. The equation of the circle can thus be written as (x-4)^2+(y-3)^2=25. To find the x intercept y must be 0, so (x-4)^2+(0-3)^2=25. Then (x-4)^2=16. x-4=4. x=8 | + | Because the two points are a diameter the center must be half way between them at the point (4,3). The distance from (0,0) to (4,3) is 5 therefore the circle has a radius of 5. The equation of the circle can thus be written as (x-4)^2+(y-3)^2=25. To find the x intercept y must be 0, so (x-4)^2+(0-3)^2=25. |
| + | Then (x-4)^2=16. | ||
| + | x-4=4. | ||
| + | x=8 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=5|num-a=7}} | {{AMC12 box|year=2017|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:34, 16 February 2017
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
Solution 1 Because the two points are a diameter the center must be half way between them at the point (4,3). The distance from (0,0) to (4,3) is 5 therefore the circle has a radius of 5. The equation of the circle can thus be written as (x-4)^2+(y-3)^2=25. To find the x intercept y must be 0, so (x-4)^2+(0-3)^2=25. Then (x-4)^2=16. x-4=4. x=8
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. 
