Difference between revisions of "2013 AMC 12B Problems/Problem 6"

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<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8</math>
 
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8</math>
  
it is 3 because it is a degenerate circle when graphThe solution was deleted.
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The answer is 2.  This problem can be solved by noticing that this is the equation of a circle with radius of 0When put into vertex form, the one point that satisfies this equation is the center of the circle.

Revision as of 10:12, 15 June 2019

The following problem is from both the 2013 AMC 12B #6 and 2013 AMC 10B #11, so both problems redirect to this page.

Problem

Real numbers $x$ and $y$ satisfy the equation $x^2 + y^2 = 10x - 6y - 34$. What is $x + y$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

The answer is 2. This problem can be solved by noticing that this is the equation of a circle with radius of 0. When put into vertex form, the one point that satisfies this equation is the center of the circle.