What is the center 3x^2+3y^2-6x+12y=0 ?

The center 3x^2+3y^2-6x+12y=0 is (1,-2)

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center 3x^2+3y^2-6x+12y=0

center

3 x^{2} + 3 y^{2} - 6 x + 12 y = 0

(1, - 2)

Solution steps

3 x^{2} + 3 y^{2} - 6 x + 12 y = 0

Rewrite

3 x^{2} + 3 y^{2} - 6 x + 12 y = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 2))^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (1, - 2), r = 5

And the center is:

(1, - 2)

x^{2} + y^{2} + 10 x + 6 y + 18 = 0

(x - 4)^{2} + (y - 15)^{2} = 169

(x - 2)^{2} + (y + 5)^{2} = 16

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

(x + 5)^{2} + y^{2} = 4