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

Revision as of 15:11, 28 June 2014

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

@@ Line 4: / Line 4: @@
 <math> \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20 </math>
 ==Solution==
+We see that the center of this circle is at <math>\left(\frac{-5+25}{2},0\right)=(10,0)</math>. The radius is <math>\frac{30}{2}=15</math>. So the equation of this circle is <cmath>(x-10)^2+y^2=225.</cmath> Substituting <math>y=15</math> yields <math>(x-10)^2=0</math> so <math>x=\boxed{\textbf{(A) }10}</math>.
+--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]

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

Revision as of 15:11, 28 June 2014

Problem

Solution