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

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

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

center

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

(- 4, - 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 2))^{2} = (26)^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 2), r = 26

And the center is:

(- 4, - 2)

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

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

x^{2} + y^{2} = 21

x^{2} + y^{2} - 20 x + 36 y + 353 = 0