What is the center 2x^2+2y^2-4x+12y-24=0 ?

The center 2x^2+2y^2-4x+12y-24=0 is (1,-3)

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center 2x^2+2y^2-4x+12y-24=0

center

2 x^{2} + 2 y^{2} - 4 x + 12 y - 24 = 0

(1, - 3)

Solution steps

2 x^{2} + 2 y^{2} - 4 x + 12 y - 24 = 0

Rewrite

2 x^{2} + 2 y^{2} - 4 x + 12 y - 24 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 3))^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (1, - 3), r = 22

And the center is:

(1, - 3)

x^{2} + 12 x + y^{2} + 12 y + 71 = 0

x^{2} + 6 x + 1 + y^{2} = 4

(x + 4)^{2} + (y + 2)^{2} = 25

(x - 12)^{2} + (y - 7)^{2} = 196

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

What is the center 2x^2+2y^2-4x+12y-24=0 ?
The center 2x^2+2y^2-4x+12y-24=0 is (1,-3)