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} - 8 x - 12 y + 43 = 0

x^{2} - 2 x + y^{2} + 4 y = 11

x^{2} + y^{2} - 9 x + 4 y - 181 = 0

x^{2} + y^{2} - 32 x - 60 y + 1122 = 0

x^{2} + y^{2} < 9