What is 2x^2+2y^2+4x-12y+11=0 ?

The solution to 2x^2+2y^2+4x-12y+11=0 is Circle with (a,b)=(-1,3),r= 3/(sqrt(2))

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2x^2+2y^2+4x-12y+11=0

2 x^{2} + 2 y^{2} + 4 x - 12 y + 11 = 0

(a, b) = (- 1, 3), r = \frac{3}{2}

Solution steps

2 x^{2} + 2 y^{2} + 4 x - 12 y + 11 = 0

Rewrite

2 x^{2} + 2 y^{2} + 4 x - 12 y + 11 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, 3), r = \frac{3}{2}

a^{2} + b^{2} = 2009

x^{2} + (y - 4)^{2} = 9

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

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

x^{2} + (y - 1)^{2} = 25

What is 2x^2+2y^2+4x-12y+11=0 ?
The solution to 2x^2+2y^2+4x-12y+11=0 is Circle with (a,b)=(-1,3),r= 3/(sqrt(2))