What is the center (x-3)^2+y^2=16 ?

The center (x-3)^2+y^2=16 is (3,0)

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center (x-3)^2+y^2=16

center

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

(3, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 3)^{2} + (y - 0)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (3, 0), r = 4

And the center is:

(3, 0)

x^{2} - 4 x + y^{2} = 21

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

x^{2} + y^{2} + 14 x + 16 y + 14 = 0

5 x^{2} + 5 y^{2} + 10 x - 5 y + 3 = 0

(x - 126)^{2} + (y - 102.5)^{2} \geq 149. 5^{2}