What is the center 2x^2+2y^2+8x-12y-72=0 ?

The center 2x^2+2y^2+8x-12y-72=0 is (-2,3)

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center 2x^2+2y^2+8x-12y-72=0

center

2 x^{2} + 2 y^{2} + 8 x - 12 y - 72 = 0

(- 2, 3)

Solution steps

2 x^{2} + 2 y^{2} + 8 x - 12 y - 72 = 0

Rewrite

2 x^{2} + 2 y^{2} + 8 x - 12 y - 72 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 3)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 7

And the center is:

(- 2, 3)

(x + 3)^{2} + (y - 1)^{2} = 5

x^{2} - 10 x + y^{2} + 8 y = - 20

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

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

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

What is the center 2x^2+2y^2+8x-12y-72=0 ?
The center 2x^2+2y^2+8x-12y-72=0 is (-2,3)