What is the center (x-1)^2+(y-3)^2=16 ?

The center (x-1)^2+(y-3)^2=16 is (1,3)

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center (x-1)^2+(y-3)^2=16

center

(x - 1)^{2} + (y - 3)^{2} = 16

(1, 3)

Solution steps

(x - 1)^{2} + (y - 3)^{2} = 16

Rewrite

(x - 1)^{2} + (y - 3)^{2} = 16

in the form of the standard circle equation

(x - 1)^{2} + (y - 3)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (1, 3), r = 4

And the center is:

(1, 3)

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

x^{2} + y^{2} \leq 1

x^{2} + y^{2} - 4 x - 8 y - 16 = 0

y^{2} + x^{2} \leq 4

x^{2} - 9 = - y^{2}