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

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

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

center

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

(2, - 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(2, - 1)

+ x^{2} + y^{2} - 14 x + 2 y + 41 = 0

x^{2} + 4 x + y^{2} - 2 y + 1 = 0

x^{2} + y^{2} - 8 x + 18 y + 16 = 0

(x - 3)^{2} + (y + 5)^{2} = 25

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

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