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

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

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

center

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

(- 2, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 2, 0)

x^{2} + y^{2} = 125

x^{2} + y^{2} + 4 x + 4 y = 1

(x + 11)^{2} + (y + 8)^{2} = 147

(x - 5)^{2} + (y - 6)^{2} = 36

(x - 5)^{2} + (y + 7)^{2} = 9