What is x^2+y^2+2x+14y-31=0 ?

The solution to x^2+y^2+2x+14y-31=0 is Circle with (a,b)=(-1,-7),r=9

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x^2+y^2+2x+14y-31=0

x^{2} + y^{2} + 2 x + 14 y - 31 = 0

(a, b) = (- 1, - 7), r = 9

Solution steps

x^{2} + y^{2} + 2 x + 14 y - 31 = 0

Rewrite

x^{2} + y^{2} + 2 x + 14 y - 31 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 7))^{2} = 9^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 7), r = 9

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

x^{2} + y^{2} + 1 x + 5 y - 56 = 0

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} - \frac{1}{8} = 0

x^{2} + y^{2} - 16 x - 2 y + 65 = 64

x^{2} - 8 x - 16 + y^{2} + 2 y + 1 = 25

What is x^2+y^2+2x+14y-31=0 ?
The solution to x^2+y^2+2x+14y-31=0 is Circle with (a,b)=(-1,-7),r=9