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

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

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

center

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

(- 1, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 1, 0), r = 1

And the center is:

(- 1, 0)

2 x^{2} + 2 y^{2} + 8 x - 4 y - 15 = 0

x^{2} + y^{2} + 4 x + 14 y + 52 = 0

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

x^{2} + y^{2} + 14 x - 2 y + 25 = 0

(x - 1)^{2} + (y - 4)^{2} = 25