Difference between revisions of "2017 AMC 12B Problems/Problem 6"
The 76923th (talk | contribs) (→Solution 1) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 5: | Line 5: | ||
==Solution 1== | ==Solution 1== | ||
| − | 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 <math>(4,3)</math>. The distance from <math>(0,0)</math> to <math>(4,3)</math> 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{D}</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{D}</math>. | ||
| Line 11: | Line 11: | ||
Written by: SilverLion | Written by: SilverLion | ||
| − | ==Solution 2 ( | + | ==Solution 2 (Brute Force)== |
As in solution 1, we find that the midpoint of the circle is (4,3) by finding half of the x and y coordinates of (8,6) point. We see a 3-4-5 triangle! For a point to be on the circle, it has to be 5 units away from the center. | As in solution 1, we find that the midpoint of the circle is (4,3) by finding half of the x and y coordinates of (8,6) point. We see a 3-4-5 triangle! For a point to be on the circle, it has to be 5 units away from the center. | ||
Latest revision as of 12:02, 3 July 2024
Problem
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 1
Because the two points are on a diameter, the center must be halfway between them at the point 
. The distance from to 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
Solution 2 (Brute Force)
As in solution 1, we find that the midpoint of the circle is (4,3) by finding half of the x and y coordinates of (8,6) point. We see a 3-4-5 triangle! For a point to be on the circle, it has to be 5 units away from the center.
We are able to create another 3-4-5 triangle by drawing a segment from (4,3) to (0,8). Therefore we know our answer is 
https://www.geogebra.org/geometry/xryzxpyw
-thedodecagon
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. 
