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

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

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

radius

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

r = 22

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 2), r = 22

And the radius is:

r = 22

x^{2} + y^{2} + 10 x - 4 y - 7 = 0

x^{2} + y^{2} = 20

x^{2} + y^{2} - 2 x + 6 y + 9 = 0

y^{2} + x^{2} - 6 y + 5 = 0

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