What is 2x^2+2y^2+8x+20y+1=0 ?

The solution to 2x^2+2y^2+8x+20y+1=0 is Circle with (a,b)=(-2,-5),r=sqrt(57/2)

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2x^2+2y^2+8x+20y+1=0

2 x^{2} + 2 y^{2} + 8 x + 20 y + 1 = 0

(a, b) = (- 2, - 5), r = \frac{57}{2}

Solution steps

2 x^{2} + 2 y^{2} + 8 x + 20 y + 1 = 0

Rewrite

2 x^{2} + 2 y^{2} + 8 x + 20 y + 1 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 5))^{2} = (\frac{57}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 5), r = \frac{57}{2}

(x + 2)^{2} + (y + 5)^{2} = 64

(x - 12)^{2} + (y - 3)^{2} = 49

4 x^{2} + 4 y^{2} + 16 x + 24 y + 31 = 0

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

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

What is 2x^2+2y^2+8x+20y+1=0 ?
The solution to 2x^2+2y^2+8x+20y+1=0 is Circle with (a,b)=(-2,-5),r=sqrt(57/2)