What is the center (x+2)^2+y^2=16 ?

The center (x+2)^2+y^2=16 is (-2,0)

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center (x+2)^2+y^2=16

center

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

(- 2, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 0)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 2, 0), r = 4

And the center is:

(- 2, 0)

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

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

x^{2} + z^{2} + 4 x - 4 z = - 4

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

(x - 1)^{2} + (y - 5)^{2} = 17