What is the center x^2-4x+y^2+6y+12=0 ?

The center x^2-4x+y^2+6y+12=0 is (2,-3)

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center x^2-4x+y^2+6y+12=0

center

x^{2} - 4 x + y^{2} + 6 y + 12 = 0

(2, - 3)

Solution steps

x^{2} - 4 x + y^{2} + 6 y + 12 = 0

Rewrite

x^{2} - 4 x + y^{2} + 6 y + 12 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 3))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (2, - 3), r = 1

And the center is:

(2, - 3)

x^{2} + y^{2} - 10 x + 20 y + 61 = 0

(x + 5)^{2} + y^{2} = 25

(X - 5)^{2} + (Y - 6)^{2} = 36

(x + 3)^{2} + y^{2} = 48

x^{2} + y^{2} + 2 x - \frac{27}{2} y + 20 = 0