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

The solution to x^2+y^2+4x-12y=0 is Circle with (a,b)=(-2,6),r=2sqrt(10)

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

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

(a, b) = (- 2, 6), r = 210

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 6)^{2} = (210)^{2}

Therefore the circle properties are:

(a, b) = (- 2, 6), r = 210

x^{2} - 14 x + y^{2} + 8 y - 40 = 0

x^{2} + y^{2} - 6 x - 10 y - 15 = 0

x^{2} + y^{2} + 6 x - 12 y + 20 = 0

x^{2} + y^{2} + 6 x - 8 y + 9 = 0

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

What is x^2+y^2+4x-12y=0 ?
The solution to x^2+y^2+4x-12y=0 is Circle with (a,b)=(-2,6),r=2sqrt(10)