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

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

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

center

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

(- 3, 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 3, 2), r = 4

And the center is:

(- 3, 2)

x^{2} + y^{2} + 4 x - 6 y - 23 = 0

2 x^{2} + 2 y^{2} - 8 x + 7 = 0

4 x^{2} + 4 y^{2} - 16 x + 40 y + 76 = 0

x^{2} + y^{2} \leq 9

x^{2} + 10 x + y^{2} - 16 - 11 = 0