What is the center (x+1)^2+y^2=4 ?

The center (x+1)^2+y^2=4 is (-1,0)

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center (x+1)^2+y^2=4

center

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

(- 1, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, 0), r = 2

And the center is:

(- 1, 0)

(x + 3)^{2} + (y - 1)^{2} = 9

2 x^{2} + 2 y^{2} - 17 x = 0

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

(x - 2)^{2} + (y + 3)^{2} = 36

(x - 3)^{2} + (y + 1)^{2} = 25