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

The radius 3(x+1)^2+3(y-1)^2=6 is r=sqrt(2)

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

radius

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

r = 2

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 radius is:

r = 2

x^{2} + y^{2} - 10 x - 6 y = 30

(x - 1)^{2} + (y + 3)^{2} = 16

x^{2} + y^{2} - 16 x + 8 y + 79 = 0

x^{2} + 8 x + y^{2} - 2 y = 9

x^{2} + y^{2} + 12 y - 108 = 0

What is the radius 3(x+1)^2+3(y-1)^2=6 ?
The radius 3(x+1)^2+3(y-1)^2=6 is r=sqrt(2)