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

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

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

radius

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

r = 8

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 8

And the radius is:

r = 8

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

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

(x + 4)^{2} + (y - 3)^{2} = 25

(x - 1)^{2} + (y - 4)^{2} = 25

(x + 2)^{2} + (y - 1)^{2} = 9