What is the center 2x^2+2y^2-4x+12y-18=0 ?

The center 2x^2+2y^2-4x+12y-18=0 is (1,-3)

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center 2x^2+2y^2-4x+12y-18=0

center

2 x^{2} + 2 y^{2} - 4 x + 12 y - 18 = 0

(1, - 3)

Solution steps

2 x^{2} + 2 y^{2} - 4 x + 12 y - 18 = 0

Rewrite

2 x^{2} + 2 y^{2} - 4 x + 12 y - 18 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 3))^{2} = (19)^{2}

Therefore the circle properties are:

(a, b) = (1, - 3), r = 19

And the center is:

(1, - 3)

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

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

x^{2} + 6 x + y^{2} = 7

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

x^{2} + 6 x + y^{2} - 2 y - 26 = 0

What is the center 2x^2+2y^2-4x+12y-18=0 ?
The center 2x^2+2y^2-4x+12y-18=0 is (1,-3)