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

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

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

radius

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

r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 2

And the radius is:

r = 2

(x + \frac{1}{2})^{2} + (y - \frac{4}{5})^{2} = \frac{9}{4}

(x + 1)^{2} + (y + 3)^{2} = 5

x^{2} + y^{2} - 4 x - 2 y + 1 = 0

x^{2} + y^{2} + 7 x + 5 y + 16 = 0

(x + \frac{8}{3})^{2} + y^{2} = 1