What is the center 2x^2+2y^2+4x-8y-22=0 ?

The center 2x^2+2y^2+4x-8y-22=0 is (-1,2)

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center 2x^2+2y^2+4x-8y-22=0

center

(- 1, 2)

Solution steps

2 x^{2} + 2 y^{2} + 4 x - 8 y - 22 = 0

Rewrite

2 x^{2} + 2 y^{2} + 4 x - 8 y - 22 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 2)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 1, 2), r = 4

And the center is:

(- 1, 2)

What is the center 2x^2+2y^2+4x-8y-22=0 ?
The center 2x^2+2y^2+4x-8y-22=0 is (-1,2)