2017 AMC 12B Problems/Problem 6

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 (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 $\boxed{D}$

https://www.geogebra.org/geometry/xryzxpyw

-thedodecagon

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.

Page

Toolbox

Search

2017 AMC 12B Problems/Problem 6

Contents

Problem

Solution 1

Solution 2 (no calculation)

See Also