2013 AMC 10B Problem 11

This can be rewritten as $$x^2-10x+y^2+6y+34=0.$$ Completing the square yields $$x^2-10x+25+y^2+6y+9$$ $$=(x-5)^2+(y+3)^2$$ $$=0.$$ Since both squared terms are nonnegative and their sum is $0,$ both must be $0$. Thus $$(x-5)^2 = (y+3)^2=0.$$ Therefore, $x=5,y=-3,$ so $x+y=2.$ Thus, the correct answer is B.