Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
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>.  
  
−
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{B}</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>.
  
 
Written by: SilverLion
 
Written by: SilverLion

Revision as of 16:42, 20 January 2018

Problem 6

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

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_6&oldid=89895"