What is the center (x+4)^2+(y-6)^2=18 ?

The center (x+4)^2+(y-6)^2=18 is (-4,6)

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center (x+4)^2+(y-6)^2=18

center

(x + 4)^{2} + (y - 6)^{2} = 18

(- 4, 6)

Solution steps

(x + 4)^{2} + (y - 6)^{2} = 18

Rewrite

(x + 4)^{2} + (y - 6)^{2} = 18

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 6)^{2} = (18)^{2}

Therefore the circle properties are:

(a, b) = (- 4, 6), r = 18

And the center is:

(- 4, 6)

x^{2} + y^{2} = 20

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

x^{2} + (y + 3)^{2} = 81

x^{2} + y^{2} - 4 x - 2 y - 31 = 0

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