Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 25"

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Let the circle we are looking for be <math>(x-h)^{2}+(y-k)^{2}=r^{2}</math> where <math>(h,k)</math> is obviously the center. Plugging in points <math>(6,0)</math> and <math>(0,2)</math> gives us that <math>3k-h=8</math>. Seeing our answer choices, none of the points work, thus our answer is E.
 
Let the circle we are looking for be <math>(x-h)^{2}+(y-k)^{2}=r^{2}</math> where <math>(h,k)</math> is obviously the center. Plugging in points <math>(6,0)</math> and <math>(0,2)</math> gives us that <math>3k-h=8</math>. Seeing our answer choices, none of the points work, thus our answer is E.
  
== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 24|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 26|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 10:47, 23 July 2006

Problem

What is the center of the circle passing through the point $(6,0)$ and tangent to the circle $x^2 + y^2 = 4$ at $(0,2)$? (Two circles are tangent at a point $P$ if they intersect at $P$ and at no other point.)

$\mathrm{(A) \ }(0,-6) \qquad \mathrm{(B) \ } (1,-9) \qquad \mathrm{(C) \ } (-1,-9) \qquad \mathrm{(D) \ } (0,-9) \qquad \mathrm{(E) \ } \rm{none \ } \rm{of \ } \rm{these}$

Solution

Let the circle we are looking for be $(x-h)^{2}+(y-k)^{2}=r^{2}$ where $(h,k)$ is obviously the center. Plugging in points $(6,0)$ and $(0,2)$ gives us that $3k-h=8$. Seeing our answer choices, none of the points work, thus our answer is E.