What is x^2+y^2+2x+4y+1/3 >0 ?

The solution to x^2+y^2+2x+4y+1/3 >0 is Circle with (a,b)=(-1,-2),r=sqrt(14/3)

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x^2+y^2+2x+4y+1/3 >0

x^{2} + y^{2} + 2 x + 4 y + \frac{1}{3} > 0

(a, b) = (- 1, - 2), r = \frac{14}{3}

Solution steps

x^{2} + y^{2} + 2 x + 4 y + \frac{1}{3} > 0

Rewrite

x^{2} + y^{2} + 2 x + 4 y + \frac{1}{3} > 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 2))^{2} = (\frac{14}{3})^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 2), r = \frac{14}{3}

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

(x^{2} - 4 x) + (y^{2} + 2 y) = 31

(x + 9)^{2} + (y + 2)^{2} = 4

x^{2} + y^{2} + 6 x - 4 y - 36 = 0

(x + 2)^{2} + (y - 3)^{2} = 4^{2}

What is x^2+y^2+2x+4y+1/3 >0 ?
The solution to x^2+y^2+2x+4y+1/3 >0 is Circle with (a,b)=(-1,-2),r=sqrt(14/3)