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

The solution to 3(x+1)^2+3(y-1)^2=6 is Circle with (a,b)=(-1,1),r=sqrt(2)

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

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

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

x^{2} + y^{2} + 32 x + 14 y + 296 = 0

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

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

x^{2} + y^{2} + 10 x - 6 y + 33 = 0

3 (x - 4)^{2} + 3 y^{2} = 54

What is 3(x+1)^2+3(y-1)^2=6 ?
The solution to 3(x+1)^2+3(y-1)^2=6 is Circle with (a,b)=(-1,1),r=sqrt(2)