What is the center x^2+y^2-12x-12y+36=0 ?

The center x^2+y^2-12x-12y+36=0 is (6,6)

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center x^2+y^2-12x-12y+36=0

center

x^{2} + y^{2} - 12 x - 12 y + 36 = 0

(6, 6)

Solution steps

x^{2} + y^{2} - 12 x - 12 y + 36 = 0

Rewrite

x^{2} + y^{2} - 12 x - 12 y + 36 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(6, 6)

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

x^{2} + y^{2} = 289

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

x^{2} + y^{2} - 20 x + 64 y - 371 = 0

4 x^{2} + 4 y^{2} - 8 y - 5 = 0

What is the center x^2+y^2-12x-12y+36=0 ?
The center x^2+y^2-12x-12y+36=0 is (6,6)