What is the center x^2+y^2-6x+4y+4=0 ?

The center x^2+y^2-6x+4y+4=0 is (3,-2)

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center x^2+y^2-6x+4y+4=0

center

x^{2} + y^{2} - 6 x + 4 y + 4 = 0

(3, - 2)

Solution steps

x^{2} + y^{2} - 6 x + 4 y + 4 = 0

Rewrite

x^{2} + y^{2} - 6 x + 4 y + 4 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 2))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (3, - 2), r = 3

And the center is:

(3, - 2)

x^{2} + y^{2} + 2 x - 6 y = 0

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

x^{2} + y^{2} - 4 x + 6 y + 4 = 0

2 x^{2} + 2 y^{2} - 3 x = 0

x^{2} + y^{2} - 12 y + 32 = 0