Art of Problem Solving

2013 AMC 12B Problems/Problem 6: Difference between revisions

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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>


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.
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. - Honestly

Revision as of 09:12, 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. - Honestly