What is 2x^2+2y^2+16x-32y+134=0 ?

The solution to 2x^2+2y^2+16x-32y+134=0 is Circle with (a,b)=(-4,8),r=sqrt(13)

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2x^2+2y^2+16x-32y+134=0

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

(a, b) = (- 4, 8), r = 13

Solution steps

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

Rewrite

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 8)^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (- 4, 8), r = 13

x^{2} + y^{2} = \frac{22}{3}

x^{2} + y^{2} - 8 x - 6 y = - 21

x^{2} - 14 x + y^{2} - 10 y + 7 = 0

(x - 3)^{2} + (y - 5)^{2} = 9^{2}

x^{2} + (y + \frac{1}{5})^{2} = \frac{9}{25}

What is 2x^2+2y^2+16x-32y+134=0 ?
The solution to 2x^2+2y^2+16x-32y+134=0 is Circle with (a,b)=(-4,8),r=sqrt(13)