What is the center x^2+y^2-6x-12y-55=0 ?

The center x^2+y^2-6x-12y-55=0 is (3,6)

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center x^2+y^2-6x-12y-55=0

center

x^{2} + y^{2} - 6 x - 12 y - 55 = 0

(3, 6)

Solution steps

x^{2} + y^{2} - 6 x - 12 y - 55 = 0

Rewrite

x^{2} + y^{2} - 6 x - 12 y - 55 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - 6)^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (3, 6), r = 10

And the center is:

(3, 6)

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

(x - 2)^{2} + (y - 1)^{2} = 49

x^{2} - 8 x + y^{2} + 6 y + 21 = 0

(x - 2)^{2} + (y + 9)^{2} = 1

16 x^{2} + 16 y^{2} + 8 x - 32 y = - 1

What is the center x^2+y^2-6x-12y-55=0 ?
The center x^2+y^2-6x-12y-55=0 is (3,6)