What is the center (x+3)^2+(y-4)^2=16 ?

The center (x+3)^2+(y-4)^2=16 is (-3,4)

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center (x+3)^2+(y-4)^2=16

center

(x + 3)^{2} + (y - 4)^{2} = 16

(- 3, 4)

Solution steps

(x + 3)^{2} + (y - 4)^{2} = 16

Rewrite

(x + 3)^{2} + (y - 4)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 3, 4)

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

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

(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{9}{4}

(x - (- 5))^{2} + (y - 7)^{2} = 11

x^{2} + y^{2} - 6 x - 10 y + 2 = 0