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 - 4)^{2} + (y + 7)^{2} = 8

4 x^{2} - 24 x + 4 y^{2} + 72 y = 76

(x - 3)^{2} + (y - 5)^{2} - 54 = 0

(x - (- 3))^{2} + (y - 0)^{2} = 6^{2}

(x - 6)^{(2)} + (y + 2)^{(2)} = 5

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