What is the center (x+14)^2+(y-11)^2=16 ?

The center (x+14)^2+(y-11)^2=16 is (-14,11)

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center (x+14)^2+(y-11)^2=16

center

(x + 14)^{2} + (y - 11)^{2} = 16

(- 14, 11)

Solution steps

(x + 14)^{2} + (y - 11)^{2} = 16

Rewrite

(x + 14)^{2} + (y - 11)^{2} = 16

in the form of the standard circle equation

(x - (- 14))^{2} + (y - 11)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 14, 11), r = 4

And the center is:

(- 14, 11)

x^{2} + y^{2} - 6 x + 10 y + 9 = 0

x^{2} + y^{2} + 6 x + 8 y + 24 = 0

x^{2} + 8 x + y^{2} - 8 y = - 16

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

(x - 3)^{2} + (y - 5)^{2} = 121

What is the center (x+14)^2+(y-11)^2=16 ?
The center (x+14)^2+(y-11)^2=16 is (-14,11)