What is the center x^2+y^2-12x-12y=4 ?

The center x^2+y^2-12x-12y=4 is (6,6)

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center x^2+y^2-12x-12y=4

center

x^{2} + y^{2} - 12 x - 12 y = 4

(6, 6)

Solution steps

x^{2} + y^{2} - 12 x - 12 y = 4

Rewrite

x^{2} + y^{2} - 12 x - 12 y = 4

in the form of the standard circle equation

(x - 6)^{2} + (y - 6)^{2} = (219)^{2}

Therefore the circle properties are:

(a, b) = (6, 6), r = 219

And the center is:

(6, 6)

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

x^{2} + z^{2} + 4 x - 4 z = - 4

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

(x - 1)^{2} + (y - 5)^{2} = 17

(x - 2)^{2} + (y + 3)^{2} = \frac{16}{9}