# 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 $x+y$? $\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

## 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 $x^2 - 10x + y^2 + 6y + 34 = 0$. We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$. Notice this is a circle with radius $0$, which only contains one point. 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

~ pi_is_3.14

## Video Solution

–no one

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