What is x^2+y^2+4x-12y-3=0 ?

The solution to x^2+y^2+4x-12y-3=0 is Circle with (a,b)=(-2,6),r=sqrt(43)

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x^2+y^2+4x-12y-3=0

x^{2} + y^{2} + 4 x - 12 y - 3 = 0

(a, b) = (- 2, 6), r = 43

Solution steps

x^{2} + y^{2} + 4 x - 12 y - 3 = 0

Rewrite

x^{2} + y^{2} + 4 x - 12 y - 3 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 6)^{2} = (43)^{2}

Therefore the circle properties are:

(a, b) = (- 2, 6), r = 43

2 x^{2} - 8 x + 8 + 2 y^{2} + 4 x + 2 = 25

100 = x^{2} + y^{2} + 4 x + 60

(x + 3)^{2} + (y + 3)^{2} = 25

(x - 8)^{2} + (y - 3)^{2} = 9

(x - 4.5)^{2} + (y - 1.5)^{2} = 5.52

What is x^2+y^2+4x-12y-3=0 ?
The solution to x^2+y^2+4x-12y-3=0 is Circle with (a,b)=(-2,6),r=sqrt(43)