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

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

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center 2x^2+2y^2+2x-y=0

center

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

(- \frac{1}{2}, \frac{1}{4})

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - \frac{1}{4})^{2} = (\frac{5}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, \frac{1}{4}), r = \frac{5}{4}

And the center is:

(- \frac{1}{2}, \frac{1}{4})

36 x^{2} + 36 y^{2} - 12 x - 27 y - 8 = 0

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

x^{2} + (y - 4)^{2} = 49

x^{2} + y^{2} + 2 x - 4 y - 31 = 0

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

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