What is 2x^2+2y^2+4x+12y-2=0 ?

The solution to 2x^2+2y^2+4x+12y-2=0 is Circle with (a,b)=(-1,-3),r=sqrt(11)

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2x^2+2y^2+4x+12y-2=0

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

(a, b) = (- 1, - 3), r = 11

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 3))^{2} = (11)^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 3), r = 11

(x - 12)^{2} + (y - 4)^{2} = 4

x^{2} + y^{2} + 12 x - 6 y + 36 = 0

(x - 8)^{2} + (y - 1)^{2} = 1

(x - 3)^{2} + (y - 2)^{2} = 36

x^{2} + y^{2} - 4 x - 2 y - 3 = 0

What is 2x^2+2y^2+4x+12y-2=0 ?
The solution to 2x^2+2y^2+4x+12y-2=0 is Circle with (a,b)=(-1,-3),r=sqrt(11)