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} - 6 x + 6 y + 2 = 0

2 x^{2} + 2 y^{2} - 12 x + 4 y - 30 = 0

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

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

(x - 3)^{2} + (y + 2)^{2} = \frac{16}{9}