Difference between revisions of "2017 AMC 12B Problems/Problem 6"
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<math> \textbf{(A) } 4\sqrt{2} \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 5\sqrt{2} \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2} </math> | <math> \textbf{(A) } 4\sqrt{2} \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 5\sqrt{2} \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2} </math> | ||
| − | ==Solution== | + | ==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 (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>. | ||
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Written by: SilverLion | Written by: SilverLion | ||
| + | |||
| + | ==Solution 2 (no calculation)== | ||
| + | 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 <math>\boxed{D}</math> | ||
| + | |||
| + | https://www.geogebra.org/geometry/xryzxpyw | ||
| + | |||
| + | -thedodecagon | ||
==See Also== | ==See Also== | ||
Revision as of 20:50, 21 January 2021
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 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 
.
To find the x-intercept, y must be 0, so 
, so 
, 
, 
.
Written by: SilverLion
Solution 2 (no calculation)
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. 
