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

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

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center 3(x+1)^2+3(y-1)^2=6

center

3 (x + 1)^{2} + 3 (y - 1)^{2} = 6

(- 1, 1)

Solution steps

3 (x + 1)^{2} + 3 (y - 1)^{2} = 6

Rewrite

3 (x + 1)^{2} + 3 (y - 1)^{2} = 6

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, 1), r = 2

And the center is:

(- 1, 1)

x^{2} + y^{2} - 8 x + 12 = 0

x^{2} + y^{2} - 6 x - 8 y - 11 = 0

x^{2} - 2 x + 3 y + y^{2} - 10 = 0

x^{2} + y^{2} + 8 x - 14 y + 1 = 0

x^{2} - 14 x + y^{2} + 8 y + 40 = 0

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