What is 2x^2+2y^2+20x+8y+40=0 ?

The solution to 2x^2+2y^2+20x+8y+40=0 is Circle with (a,b)=(-5,-2),r=3

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2x^2+2y^2+20x+8y+40=0

2 x^{2} + 2 y^{2} + 20 x + 8 y + 40 = 0

(a, b) = (- 5, - 2), r = 3

Solution steps

2 x^{2} + 2 y^{2} + 20 x + 8 y + 40 = 0

Rewrite

2 x^{2} + 2 y^{2} + 20 x + 8 y + 40 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 5, - 2), r = 3

(x - (- 1))^{2} + (y - 4)^{2} = 2^{2}

x, x^{2} - 6 x + Y (x)^{2} + 4 Y = 36

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

1 = 16 - x^{2} - y^{2}

x^{2} + y^{2} - 8 x + 18 y - 3 = 0

What is 2x^2+2y^2+20x+8y+40=0 ?
The solution to 2x^2+2y^2+20x+8y+40=0 is Circle with (a,b)=(-5,-2),r=3