What is the center (x+2)^2+(y+5)^2=16 ?

The center (x+2)^2+(y+5)^2=16 is (-2,-5)

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center (x+2)^2+(y+5)^2=16

center

(x + 2)^{2} + (y + 5)^{2} = 16

(- 2, - 5)

Solution steps

(x + 2)^{2} + (y + 5)^{2} = 16

Rewrite

(x + 2)^{2} + (y + 5)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 5), r = 4

And the center is:

(- 2, - 5)

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

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

x^{2} + y^{2} + 12 y + 32 = 0

x^{2} - 6 x + y^{2} = 7

(x - 3)^{2} + y^{2} = 36

What is the center (x+2)^2+(y+5)^2=16 ?
The center (x+2)^2+(y+5)^2=16 is (-2,-5)