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

The center 2x^2+2y^2-4x+12y-12=0 is (1,-3)

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

center

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

(1, - 3)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 3))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (1, - 3), r = 4

And the center is:

(1, - 3)

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

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

(x - 1)^{2} + (y + 2)^{2} = 9

x^{2} + z^{2} + 4 x - 4 z = - 4

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

What is the center 2x^2+2y^2-4x+12y-12=0 ?
The center 2x^2+2y^2-4x+12y-12=0 is (1,-3)