2013 AMC 12B Problems/Problem 6

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 $x + y$ ?

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

Solution 1

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 $5 + (-3) = 2 \implies \boxed{\textbf{(B)}}$ . ~ asdf334

Solution 2

If we move every term including $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34.$ We can complete the square to find that this equation becomes $(x - 5)^2 + (y + 3)^2 = 0.$ Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$ . Equality holds when the value inside the parhentheses is equal to $0$ . We find that $(x,y) = (5,-3)$ and the sum we are looking for is $5+(-3)=2 \implies \boxed{\textbf{(B)}}.$ - Honestly

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 12 Problems and Solutions

{{AMC10 box|year=2013|ab=B|num-b=6|num-a=8} The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2013 AMC 12B Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See Also