What is x^2+y^2+4y-117=0 ?

The solution to x^2+y^2+4y-117=0 is Circle with (a,b)=(0,-2),r=11

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x^2+y^2+4y-117=0

x^{2} + y^{2} + 4 y - 117 = 0

(a, b) = (0, - 2), r = 11

Solution steps

x^{2} + y^{2} + 4 y - 117 = 0

Rewrite

x^{2} + y^{2} + 4 y - 117 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 2))^{2} = 1 1^{2}

Therefore the circle properties are:

(a, b) = (0, - 2), r = 11

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} + \frac{5}{2} x + 10 y + 5 = 0

(x - 7)^{2} + (y + 2)^{2} = 100

x^{2} + y^{2} - 14 x - 4 y + 17 = 0

(x - 2)^{2} + (y - 4)^{2} = 17

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

What is x^2+y^2+4y-117=0 ?
The solution to x^2+y^2+4y-117=0 is Circle with (a,b)=(0,-2),r=11