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

Revision as of 12:23, 27 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 $x + y$ ?

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

If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . Notice this is a circle with radius $0$ , which only contains one point. We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$ . So, the only point is $(5, -3)$ , so the sum is $5 + (-3) = 2 \implies \boxed{\textbf{(B)}}$

@@ Line 6: / Line 6: @@
 <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.  - Honestly
+If we move every term dependent on <math>x</math> or <math>y</math> to the LHS, we get <math>x^2 - 10x + y^2 + 6y = -34</math>. Adding <math>34</math> to both sides, we have <math>x^2 - 10x + y^2 + 6y + 34 = 0</math>. Notice this is a circle with radius <math>0</math>, which only contains one point. We can split the <math>34</math> into <math>25</math> and <math>9</math> to get <math>(x - 5)^2 + (y + 3)^2 = 0</math>. So, the only point is <math>(5, -3)</math>, so the sum is <math>5 + (-3) = 2 \implies \boxed{\textbf{(B)}}</math>

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

Revision as of 12:23, 27 June 2019

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