What is the center 2x^2+32x+2y^2-12=-128 ?

The center 2x^2+32x+2y^2-12=-128 is (-8,0)

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center 2x^2+32x+2y^2-12=-128

center

2 x^{2} + 32 x + 2 y^{2} - 12 = - 128

(- 8, 0)

Solution steps

2 x^{2} + 32 x + 2 y^{2} - 12 = - 128

Rewrite

2 x^{2} + 32 x + 2 y^{2} - 12 = - 128

in the form of the standard circle equation

(x - (- 8))^{2} + (y - 0)^{2} = (6)^{2}

Therefore the circle properties are:

(a, b) = (- 8, 0), r = 6

And the center is:

(- 8, 0)

(x - 3)^{2} + (y + 3)^{2} = 8

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

x^{2} - 4 x + y^{2} + 8 y = - 4

x^{2} + y^{2} - 10 x - 12 y + 45 = 0

x^{2} + y^{2} - 8 x + 2 y + 13 = 0

What is the center 2x^2+32x+2y^2-12=-128 ?
The center 2x^2+32x+2y^2-12=-128 is (-8,0)