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

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

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

radius

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

r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 2

x^{2} + y^{2} + 2 x - 4 y = 5.7

4 x^{2} + 4 y^{2} - 20 x + 40 y = - 5

2 x^{2} + 2 y^{2} + 6 x - 8 y + 12 = 0

16 x^{2} + 16 y^{2} - 8 y - 15 = 0

x^{2} + y^{2} + 4 x + 6 y = 23