What is 2x^2+2y^2+8x+4y-3=0 ?

The solution to 2x^2+2y^2+8x+4y-3=0 is Circle with (a,b)=(-2,-1),r=sqrt(13/2)

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2x^2+2y^2+8x+4y-3=0

2 x^{2} + 2 y^{2} + 8 x + 4 y - 3 = 0

(a, b) = (- 2, - 1), r = \frac{13}{2}

Solution steps

2 x^{2} + 2 y^{2} + 8 x + 4 y - 3 = 0

Rewrite

2 x^{2} + 2 y^{2} + 8 x + 4 y - 3 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 1))^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 1), r = \frac{13}{2}

16 x^{2} + 16 y^{2} + 8 x + 32 y + 1 = 0

x^{2} + y^{2} + 4 x + 12 y - 104 = 0

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

x^{2} + (y + 3)^{2} = 64

x^{2} + y^{2} - 6 x - 10 y + 30 = 0

What is 2x^2+2y^2+8x+4y-3=0 ?
The solution to 2x^2+2y^2+8x+4y-3=0 is Circle with (a,b)=(-2,-1),r=sqrt(13/2)