What is the radius x^2+y^2-4x-2y+1=0 ?

The radius x^2+y^2-4x-2y+1=0 is r=2

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radius x^2+y^2-4x-2y+1=0

radius

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

r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 2

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

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

x^{2} - 6 x + y^{2} + 8 y = 0

x^{2} + y^{2} = 3 x + 4 y

x^{2} + y^{2} - 10 x + 10 y + 49 = 0