What is the center (x+4)^2+(y-2)^2=121 ?

The center (x+4)^2+(y-2)^2=121 is (-4,2)

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center (x+4)^2+(y-2)^2=121

center

(x + 4)^{2} + (y - 2)^{2} = 121

(- 4, 2)

Solution steps

(x + 4)^{2} + (y - 2)^{2} = 121

Rewrite

(x + 4)^{2} + (y - 2)^{2} = 121

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 4, 2)

3 (x + \frac{3}{4})^{2} + 3 (y - \frac{1}{2})^{2} = 36

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

(x + 4)^{2} + (y - 1)^{2} = 25

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

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

What is the center (x+4)^2+(y-2)^2=121 ?
The center (x+4)^2+(y-2)^2=121 is (-4,2)