Difference between revisions of "1994 AHSME Problems/Problem 4"
(Created page with "==Problem== In the <math>xy</math>-plane, the segment with endpoints <math>(-5,0)</math> and <math>(25,0)</math> is the diameter of a circle. If the point <math>(x,15)</math> is ...") |
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<math> \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20 </math> | <math> \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20 </math> | ||
==Solution== | ==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>. | ||
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+ | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] |
Revision as of 15:11, 28 June 2014
Problem
In the -plane, the segment with endpoints
and
is the diameter of a circle. If the point
is on the circle, then
Solution
We see that the center of this circle is at . The radius is
. So the equation of this circle is
Substituting
yields
so
.
--Solution by TheMaskedMagician