What is the standard form of the equation of the circle x2 -4x + y2 + 6y + 12 = 0?

We take the equation
                            x^2 -4x + y^2 + 6y + 12 = 0
and complete the square for x and y.
                             x^2 -4x + 4 - 4 + y^2 + 6y + 9 - 9 + 12= 0
                                            (x-2)^2 - 4 + (y+3)^2 - 9 = -12
                                                       (x-2)^2+ (y+3)^2  = 1
Therefore the answer is (x-2)^2+ (y+3)^2  = 1

In this problem it decreased 20%