What is the center 3x^2+3y^2-18x-12y=108 ?

The center 3x^2+3y^2-18x-12y=108 is (3,2)

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center 3x^2+3y^2-18x-12y=108

center

3 x^{2} + 3 y^{2} - 18 x - 12 y = 108

(3, 2)

Solution steps

3 x^{2} + 3 y^{2} - 18 x - 12 y = 108

Rewrite

3 x^{2} + 3 y^{2} - 18 x - 12 y = 108

in the form of the standard circle equation

(x - 3)^{2} + (y - 2)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (3, 2), r = 7

And the center is:

(3, 2)

[x - 7]^{2} + [y - 3]^{2} = 49

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

2 x^{2} + 2 y^{2} + 8 x - 4 y - 15 = 0

x^{2} + y^{2} + 4 x + 14 y + 52 = 0

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

What is the center 3x^2+3y^2-18x-12y=108 ?
The center 3x^2+3y^2-18x-12y=108 is (3,2)