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

The solution to 2x^2+2y^2-12x+4y-12=0 is Circle with (a,b)=(3,-1),r=4

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

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

(a, b) = (3, - 1), r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 1))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (3, - 1), r = 4

(x - 3)^{2} + (y - 5)^{2} = 144

x^{2} + y^{2} - 16 x + 40 y + 67 = 0

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

(x + 3)^{2} + y^{2} = 1

(x - 0)^{2} + (y - 0)^{2} = 7^{2}

What is 2x^2+2y^2-12x+4y-12=0 ?
The solution to 2x^2+2y^2-12x+4y-12=0 is Circle with (a,b)=(3,-1),r=4