What is the center 2x^2+2y^2+28x-8y+4=0 ?

The center 2x^2+2y^2+28x-8y+4=0 is (-7,2)

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center 2x^2+2y^2+28x-8y+4=0

center

2 x^{2} + 2 y^{2} + 28 x - 8 y + 4 = 0

(- 7, 2)

Solution steps

2 x^{2} + 2 y^{2} + 28 x - 8 y + 4 = 0

Rewrite

2 x^{2} + 2 y^{2} + 28 x - 8 y + 4 = 0

in the form of the standard circle equation

(x - (- 7))^{2} + (y - 2)^{2} = (51)^{2}

Therefore the circle properties are:

(a, b) = (- 7, 2), r = 51

And the center is:

(- 7, 2)

x^{2} + y^{2} + 4 x - 2 y + 1 = 0

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

(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

What is the center 2x^2+2y^2+28x-8y+4=0 ?
The center 2x^2+2y^2+28x-8y+4=0 is (-7,2)