What is the center 2x^2+2y^2+16x-32y+134=0 ?

The center 2x^2+2y^2+16x-32y+134=0 is (-4,8)

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center 2x^2+2y^2+16x-32y+134=0

center

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

(- 4, 8)

Solution steps

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

Rewrite

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 8)^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (- 4, 8), r = 13

And the center is:

(- 4, 8)

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

(x - 0)^{2} + (y - 0)^{2} = 4

x^{2} + y^{2} + 10 x - 16 y + 75 = 0

- 20 - x^{2} = - 16 x - 20 y + y^{2}

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

What is the center 2x^2+2y^2+16x-32y+134=0 ?
The center 2x^2+2y^2+16x-32y+134=0 is (-4,8)