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

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

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

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

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

Solution steps

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

Rewrite

x^{2} - 2 x + y^{2} + 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

x^{2} + y^{2} - 20 x - 2 y + 52 = 0

4 x^{(2)} + 4 y^{(2)} + 24 x - 8 y - 9 = 0

x^{2} + 4 x + y^{2} - 6 y + 9 = 0

(x - 5)^{2} + (y - 6)^{2} = 36

x^{2} + y^{2} + 8 x - 18 y - 47 = 0

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