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

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--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]
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== Solution 2 ==
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The diameter of the circle is <math>25- (-5)=30</math> so the radius is <math>15</math> and the center of the circle is on the <math>x</math>-axis at <math>x=10</math>.  The only point with <math>y=15</math> must be the point exactly above the center.
  
 
==See Also==
 
==See Also==

Latest revision as of 01:31, 28 May 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

Solution 2

The diameter of the circle is $25- (-5)=30$ so the radius is $15$ and the center of the circle is on the $x$-axis at $x=10$. The only point with $y=15$ must be the point exactly above the center.

See Also

1994 AHSME (Problems • Answer Key • Resources)
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Problem 3 		Followed by
Problem 5
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All AHSME Problems and Solutions

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