What is x^2+y^2+(-6)+4y-3=0 ?

The solution to x^2+y^2+(-6)+4y-3=0 is Circle with (a,b)=(0,-2),r=sqrt(13)

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x^2+y^2+(-6)+4y-3=0

x^{2} + y^{2} + (- 6) + 4 y - 3 = 0

(a, b) = (0, - 2), r = 13

Solution steps

x^{2} + y^{2} - 6 + 4 y - 3 = 0

Rewrite

x^{2} + y^{2} - 6 + 4 y - 3 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 2))^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (0, - 2), r = 13

x^{2} - 14 x + y^{2} + 10 y + 58 = 0

(x - (- 3))^{2} + (y - 4)^{2} = 5^{2}

4 x^{2} + 8 x + 4 y^{2} - 12 y = 23

x^{2} + y^{2} + 8 x + 4 y - 5 = 0

x^{2} + y^{2} - 2 x - 15 + 14 y = 0

What is x^2+y^2+(-6)+4y-3=0 ?
The solution to x^2+y^2+(-6)+4y-3=0 is Circle with (a,b)=(0,-2),r=sqrt(13)