Difference between revisions of "2017 AMC 12B Problems/Problem 6"
m (→Solution) |
Silverlion (talk | contribs) (→Solution) |
||
| Line 3: | Line 3: | ||
==Solution== | ==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 <math>(x-4)^2+(y-3)^2=25</math>. | |
| − | Because the two points are a diameter the center must be | + | |
| − | + | 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</math>. | |
| − | x-4=4 | + | |
| − | x=8 | + | Written by: SilverLion |
==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 10:01, 17 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
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. 
