What is x^2+y^2+12x-4y+31=0 ?

The solution to x^2+y^2+12x-4y+31=0 is Circle with (a,b)=(-6,2),r=3

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x^2+y^2+12x-4y+31=0

x^{2} + y^{2} + 12 x - 4 y + 31 = 0

(a, b) = (- 6, 2), r = 3

Solution steps

x^{2} + y^{2} + 12 x - 4 y + 31 = 0

Rewrite

x^{2} + y^{2} + 12 x - 4 y + 31 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 6, 2), r = 3

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

x^{2} + y^{2} + 6 x - 8 y - 15 = 0

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

x^{2} + y^{2} + 12 x + 4 y + 8 = 0

(x - 0.04)^{2} + (y - 0.069)^{2} = 0.2 1^{2}

What is x^2+y^2+12x-4y+31=0 ?
The solution to x^2+y^2+12x-4y+31=0 is Circle with (a,b)=(-6,2),r=3