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

The solution to x^2+y^2-4x+2y+1=0 is Circle with (a,b)=(2,-1),r=2

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

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

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

(x - 6)^{2} + (y + 36)^{2} = 25

x^{2} + y^{2} - 7 y + 8 = 0

x^{2} + 6 x + y^{2} - 10 y = 135

z^{2} = 4 - y^{2}

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

What is x^2+y^2-4x+2y+1=0 ?
The solution to x^2+y^2-4x+2y+1=0 is Circle with (a,b)=(2,-1),r=2