What is the center x^2+y^2+4x-5=0 ?

The center x^2+y^2+4x-5=0 is (-2,0)

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center x^2+y^2+4x-5=0

center

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

(- 2, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 0)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- 2, 0), r = 3

And the center is:

(- 2, 0)

4 x^{2} + 4 y^{2} + 12 x - 4 y - 90 = 0

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

x^{2} + y^{2} + 4 x - 6 y = 12

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

x^{2} + y^{2} = 3