What is 2x^2+2y^2-8x+12y-8=0 ?

The solution to 2x^2+2y^2-8x+12y-8=0 is Circle with (a,b)=(2,-3),r=sqrt(17)

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2x^2+2y^2-8x+12y-8=0

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

(a, b) = (2, - 3), r = 17

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 3))^{2} = (17)^{2}

Therefore the circle properties are:

(a, b) = (2, - 3), r = 17

(x + 1)^{2} + (y + 6)^{2} = 100

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

x^{(2)} + y^{(2)} - 6 x + 4 y - 12 = 0

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

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

What is 2x^2+2y^2-8x+12y-8=0 ?
The solution to 2x^2+2y^2-8x+12y-8=0 is Circle with (a,b)=(2,-3),r=sqrt(17)