What is the center x^2+(y-1)^2=16 ?

The center x^2+(y-1)^2=16 is (0,1)

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center x^2+(y-1)^2=16

center

x^{2} + (y - 1)^{2} = 16

(0, 1)

Solution steps

x^{2} + (y - 1)^{2} = 16

Rewrite

x^{2} + (y - 1)^{2} = 16

in the form of the standard circle equation

(x - 0)^{2} + (y - 1)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (0, 1), r = 4

And the center is:

(0, 1)

x^{2} + y^{2} + 2 x - 8 y - 19 = 0

2 x^{2} + 2 y^{2} + 28 x - 8 y + 4 = 0

x^{2} + y^{2} + 4 x - 2 y + 1 = 0

2 x^{2} + 32 x + 2 y^{2} - 12 = - 128

(x - 3)^{2} + (y + 3)^{2} = 8