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

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

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

radius

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

r = 2

Solution steps

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

Rewrite

x^{2} + 4 x + y^{2} - 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 - 3)^{2} = 36

x^{2} + y^{2} = \frac{1}{4}

x^{2} + y^{2} - 8 x + 12 y = - 43

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

x^{2} + y^{2} = 12