What is the center (x+11)^2+(y+5)^2=4 ?

The center (x+11)^2+(y+5)^2=4 is (-11,-5)

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center (x+11)^2+(y+5)^2=4

center

(x + 11)^{2} + (y + 5)^{2} = 4

(- 11, - 5)

Solution steps

(x + 11)^{2} + (y + 5)^{2} = 4

Rewrite

(x + 11)^{2} + (y + 5)^{2} = 4

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 11, - 5), r = 2

And the center is:

(- 11, - 5)

2 x^{2} + 2 y^{2} + 8 x - 12 y - 72 = 0

(x + 3)^{2} + (y - 1)^{2} = 5

x^{2} - 10 x + y^{2} + 8 y = - 20

2 x^{2} + 2 y^{2} - 4 x + 12 y - 24 = 0

x^{2} + 12 x + y^{2} + 12 y + 71 = 0

What is the center (x+11)^2+(y+5)^2=4 ?
The center (x+11)^2+(y+5)^2=4 is (-11,-5)