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

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

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

center

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

(- 1, \frac{1}{8})

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{1}{8})^{2} = (\frac{65}{8})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{1}{8}), r = \frac{65}{8}

And the center is:

(- 1, \frac{1}{8})

(x + 2)^{2} + (y + 5)^{2} = 16

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

4 x^{2} + 4 y^{2} - 16 = 0

x^{2} + y^{2} + 12 y + 32 = 0

x^{2} - 6 x + y^{2} = 7