What is the center 3x^2+3y^2-12x+18y+3=0 ?

The center 3x^2+3y^2-12x+18y+3=0 is (2,-3)

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center 3x^2+3y^2-12x+18y+3=0

center

3 x^{2} + 3 y^{2} - 12 x + 18 y + 3 = 0

(2, - 3)

Solution steps

3 x^{2} + 3 y^{2} - 12 x + 18 y + 3 = 0

Rewrite

3 x^{2} + 3 y^{2} - 12 x + 18 y + 3 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 3))^{2} = (23)^{2}

Therefore the circle properties are:

(a, b) = (2, - 3), r = 23

And the center is:

(2, - 3)

x^{2} + y^{2} - 8 x - 10 y + 40 = 0

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

3 x^{2} + 3 y^{2} - 48 = 0

4 x^{2} + 4 y^{2} - 12 x + 4 y - 26 = 0

2 x^{2} + 2 y^{2} + 12 x - 2 y - 3 = 0

What is the center 3x^2+3y^2-12x+18y+3=0 ?
The center 3x^2+3y^2-12x+18y+3=0 is (2,-3)