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

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

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

center

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

(- 1, - 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, - 2), r = 3

And the center is:

(- 1, - 2)

(x + 4)^{2} + (y - \frac{3}{4})^{2} = 1

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

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

x^{2} + y^{2} - 14 x - 12 y + 59 = 0

(x + 5)^{2} + (y + 1)^{2} = 9