2013 AMC 10B Problem 11

Bring all terms to the left and group by variable:
$$x^2 - 10x + y^2 + 6y + 34 = 0.$$
Complete the square in each variable. Since $x^2 - 10x = (x-5)^2 - 25$ and $y^2 + 6y = (y+3)^2 - 9$, the equation becomes
$$(x-5)^2 + (y+3)^2 - 25 - 9 + 34 = 0 \quad\Longrightarrow\quad (x-5)^2 + (y+3)^2 = 0.$$
A sum of two squares of real numbers is zero only when both squares are zero, so $x = 5$ and $y = -3$. Then $x + y = 2$.

The answer is $\boxed{\textbf{(B)}\ 2}$.

One equation in two unknowns normally has infinitely many solutions, but when it can be written as a sum of squares equal to zero, the real-number constraint pins down every variable. The constant $34 = 25 + 9$ is the signal: it is exactly what completing both squares needs. Geometrically, the "circle" has radius $0$.

• Treating the equation as having many solutions and trying to eliminate a variable, instead of noticing the sum of squares must vanish.
• Sign slip in completing the square, e.g. writing $(y-3)^2$ and getting $y = 3$, which gives $x + y = 8$ (choice (E)).