What is the center 2x^2+2y^2+8x+16y+14=0 ?

The center 2x^2+2y^2+8x+16y+14=0 is (-2,-4)

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center 2x^2+2y^2+8x+16y+14=0

center

2 x^{2} + 2 y^{2} + 8 x + 16 y + 14 = 0

(- 2, - 4)

Solution steps

2 x^{2} + 2 y^{2} + 8 x + 16 y + 14 = 0

Rewrite

2 x^{2} + 2 y^{2} + 8 x + 16 y + 14 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 4), r = 13

And the center is:

(- 2, - 4)

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

(x + 4)^{2} + (y - 10)^{2} = 274

x^{2} + y^{2} - 10 x + 12 y + 25 = 0

y^{2} + x^{2} - 12 x + 18 y + 116 = 0

16 x^{2} + 16 y^{2} + 8 x - 3 y = - 1

What is the center 2x^2+2y^2+8x+16y+14=0 ?
The center 2x^2+2y^2+8x+16y+14=0 is (-2,-4)