1994 AHSME Problems/Problem 4

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.

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1994 AHSME Problems/Problem 4

Contents

Problem

Solution

Solution 2

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