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

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

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

center

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

(3, - 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(3, - 1)

x^{2} + (y - 4)^{2} = 49

x^{2} + y^{2} + 2 x - 4 y - 31 = 0

x^{2} + y^{2} = 8

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

x^{2} + (y - 1)^{2} = 25