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

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