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

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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