What is x^2+y^2+4x+12y-104=0 ?

The solution to x^2+y^2+4x+12y-104=0 is Circle with (a,b)=(-2,-6),r=12

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x^2+y^2+4x+12y-104=0

x^{2} + y^{2} + 4 x + 12 y - 104 = 0

(a, b) = (- 2, - 6), r = 12

Solution steps

x^{2} + y^{2} + 4 x + 12 y - 104 = 0

Rewrite

x^{2} + y^{2} + 4 x + 12 y - 104 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 6), r = 12

x^{2} + y^{2} + 2 x - 8 y = 8

x^{2} + (y + 3)^{2} = 64

x^{2} + y^{2} - 6 x - 10 y + 30 = 0

2 x^{2} + 2 y^{2} - 12 x + 4 y - 15 = 0

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

What is x^2+y^2+4x+12y-104=0 ?
The solution to x^2+y^2+4x+12y-104=0 is Circle with (a,b)=(-2,-6),r=12