What is the center 3x^2+3y^2-18x+6y+2=0 ?

The center 3x^2+3y^2-18x+6y+2=0 is (3,-1)

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center 3x^2+3y^2-18x+6y+2=0

center

3 x^{2} + 3 y^{2} - 18 x + 6 y + 2 = 0

(3, - 1)

Solution steps

3 x^{2} + 3 y^{2} - 18 x + 6 y + 2 = 0

Rewrite

3 x^{2} + 3 y^{2} - 18 x + 6 y + 2 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 1))^{2} = (\frac{2 7}{3})^{2}

Therefore the circle properties are:

(a, b) = (3, - 1), r = \frac{2 7}{3}

And the center is:

(3, - 1)

(x - 13)^{2} + (y + 12)^{2} = 64

x^{2} + y^{2} - \frac{2}{3} x - \frac{5}{3} y - \frac{5}{9} = 0

\frac{( x + 2 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

9 x^{2} - 42 x + 9 y^{2} + 84 y + 65 = 0

x^{2} + y^{2} - 8 y + 7 = 0

What is the center 3x^2+3y^2-18x+6y+2=0 ?
The center 3x^2+3y^2-18x+6y+2=0 is (3,-1)