What is the center (x+4)^2+(y-(-1))^2=12 ?

The center (x+4)^2+(y-(-1))^2=12 is (-4,-1)

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center (x+4)^2+(y-(-1))^2=12

center

(x + 4)^{2} + (y - (- 1))^{2} = 12

(- 4, - 1)

Solution steps

(x + 4)^{2} + (y - (- 1))^{2} = 12

Rewrite

(x + 4)^{2} + (y - (- 1))^{2} = 12

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 1))^{2} = (12)^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 1), r = 12

And the center is:

(- 4, - 1)

3 x^{2} + 3 y^{2} - 27 x - 9 y - 24 = 0

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

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

9 x^{2} + 12 x + 9 y^{2} - 18 y - 23 = 0

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

What is the center (x+4)^2+(y-(-1))^2=12 ?
The center (x+4)^2+(y-(-1))^2=12 is (-4,-1)