What is the radius x^2+y^2+6x-2y-8=0 ?

The radius x^2+y^2+6x-2y-8=0 is r=3sqrt(2)

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radius x^2+y^2+6x-2y-8=0

radius

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

r = 32

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 1)^{2} = (32)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 1), r = 32

And the radius is:

r = 32

x^{2} + y^{2} + 2 x + 4 y - 20 = 0

(x - 2)^{2} + (y - 5)^{2} = 9

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

(x - 2)^{2} + (y + 3)^{2} = 9

3 y^{2} + 3 x^{2} + 24 x + 32 = 0

What is the radius x^2+y^2+6x-2y-8=0 ?
The radius x^2+y^2+6x-2y-8=0 is r=3sqrt(2)