What is the center x^2+y^2-6x+12y+9=9 ?

The center x^2+y^2-6x+12y+9=9 is (3,-6)

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center x^2+y^2-6x+12y+9=9

center

x^{2} + y^{2} - 6 x + 12 y + 9 = 9

(3, - 6)

Solution steps

x^{2} + y^{2} - 6 x + 12 y + 9 = 9

Rewrite

x^{2} + y^{2} - 6 x + 12 y + 9 = 9

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 6))^{2} = (35)^{2}

Therefore the circle properties are:

(a, b) = (3, - 6), r = 35

And the center is:

(3, - 6)

x^{2} + 7 x + y^{2} - 10 y + 6 = 0

x^{2} + y^{2} + 2 x - 1 y - 4 = 0

(x - 5)^{2} + (y - 3)^{2} = 64

12 x^{2} + 12 y^{2} - 18 x + 4 y - 5 = 0

(x - 3)^{2} + (y - 4)^{2} = 49