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

The radius x^2+y^2-2y-8=0 is r=3

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

radius

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

r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - 1)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (0, 1), r = 3

And the radius is:

r = 3

(x + 3)^{2} + (y - 3)^{2} = 8

(x - \frac{3}{2})^{2} + y^{2} = \frac{9}{4}

x^{2} + y^{2} + 14 x + 12 y + 21 = 0

3.5 (x + 2.2)^{2} + 3.5 (y - 11.1)^{2} - 21 = 0

x^{2} + y^{2} + 6 x - 2 y + 1 = 0