Difference between revisions of "1994 AHSME Problems/Problem 4"

(Solution)
(Solution)
Line 7: Line 7:
  
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 +
 +
==See Also==
 +
 +
{{AHSME box|year=1994|num-b=3|num-a=5}}
 +
{{MAA Notice}}

Revision as of 17:26, 9 January 2021

Problem

In the $xy$-plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$

$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20$

Solution

We see that the center of this circle is at $\left(\frac{-5+25}{2},0\right)=(10,0)$. The radius is $\frac{30}{2}=15$. So the equation of this circle is \[(x-10)^2+y^2=225.\] Substituting $y=15$ yields $(x-10)^2=0$ so $x=\boxed{\textbf{(A) }10}$.

--Solution by TheMaskedMagician

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png