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

The solution to x^2+y^2+4x+2y=1 is Circle with (a,b)=(-2,-1),r=sqrt(6)

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

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

(a, b) = (- 2, - 1), r = 6

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 1))^{2} = (6)^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 1), r = 6

x^{2} + y^{2} - 22 x + 20 y + 196 = 0

x^{2} + y^{2} + 3 x - 5 y + \frac{25}{4} = 0

4 x^{2} + 4 y^{2} - 4 x + 12 y + 9 = 0

x^{2} + y^{2} + \frac{1}{2} x - \frac{3}{2} y = \frac{71}{16}

(X + 1)^{2} + (Y - 4)^{2} = 25

What is x^2+y^2+4x+2y=1 ?
The solution to x^2+y^2+4x+2y=1 is Circle with (a,b)=(-2,-1),r=sqrt(6)