What is the radius 2x^2+2y^2-4x+12y-24=0 ?

The radius 2x^2+2y^2-4x+12y-24=0 is r=sqrt(22)

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radius 2x^2+2y^2-4x+12y-24=0

radius

2 x^{2} + 2 y^{2} - 4 x + 12 y - 24 = 0

r = 22

Solution steps

2 x^{2} + 2 y^{2} - 4 x + 12 y - 24 = 0

Rewrite

2 x^{2} + 2 y^{2} - 4 x + 12 y - 24 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 3))^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (1, - 3), r = 22

And the radius is:

r = 22

x^{2} + y^{2} - 6 x - 8 y - 3 = 0

(x + 2)^{2} + y^{2} = 9

(x + 3)^{2} + y^{2} = 100

4 x^{2} + 4 y^{2} = 128

(x - 2)^{2} + (y - 1)^{2} = 49

What is the radius 2x^2+2y^2-4x+12y-24=0 ?
The radius 2x^2+2y^2-4x+12y-24=0 is r=sqrt(22)