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Difference between revisions of "2017 AMC 12B Problems/Problem 6"

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==Solution 1==
 
==Solution 1==
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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>.  
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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>.

Latest revision as of 13:02, 3 July 2024

Problem

The circle having $(0,0)$ and $(8,6)$ as the endpoints of a diameter intersects the $x$-axis at a second point. What is the $x$-coordinate of this point?

$\textbf{(A) } 4\sqrt{2} \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 5\sqrt{2} \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

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 $(x-4)^2+(y-3)^2=25$.

To find the x-intercept, y must be 0, so $(x-4)^2+(0-3)^2=25$, so $(x-4)^2=16$, $x-4=4$, $x=8, \boxed{D}$.

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 $\boxed{D}$

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

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