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

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

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

center

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

(11, - 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (11, - 2), r = 4

And the center is:

(11, - 2)

(x - \frac{3}{2})^{2} + (y - \frac{3}{2})^{2} = \frac{9}{4}

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

2 x^{2} + 2 y^{2} + 6 x - 8 y + 12 = 0

4 x^{2} + 12 x + 4 y^{2} + 16 y - 11 = 0

(x - 3)^{2} + (y - 5)^{2} = 4

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