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

The radius x^2+y^2-4x-2y-4=0 is r=3

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

radius

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

r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 1), r = 3

And the radius is:

r = 3

x^{2} + y^{2} - 4 x + 6 y + 4 = 0

4 x^{2} + 4 y^{2} - 16 x + 4 y - 47 = 0

3 (x + 5)^{2} + 3 (y + 2)^{2} = 27

x^{2} + y^{2} + 2 x + 10 y + 18 = 0

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