Solution:

This can be rewritten as $$x^2-10x+y^2+6y+34=0.$$

From this, completing the square yields $$x^2-10x+25+y^2+6y+9$$$$=(x-5)^2+(y+3)^2$$$$=0.$$ Since both of these squared terms must be greater than or equal to $0,$ and their sum equals $0,$ both values must both be $0$ yielding $$(x-5)^2 = (y+3)^2=0.$$

As such, $x=5,y=-3.$ This makes the sum $x+y=2.$

Thus, the correct answer is B .