Question

What is the minimum value of the expression $x^2+y^2-6x+4y+18$ for real $x$ and $y$?

Answer 1

x^2+y^2-6x+4y+18=0
x^2-6x+y^2+4y+18=0
x^2-6x+9+y^2+4y+4 =-18+9+4
(x-3)^2+(y+2)^2=-5
-5 is the minimum value of the given circle