What is 2x^2+2y^2+4x-8y-40=0 ?

The solution to 2x^2+2y^2+4x-8y-40=0 is Circle with (a,b)=(-1,2),r=5

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2x^2+2y^2+4x-8y-40=0

2 x^{2} + 2 y^{2} + 4 x - 8 y - 40 = 0

(a, b) = (- 1, 2), r = 5

Solution steps

2 x^{2} + 2 y^{2} + 4 x - 8 y - 40 = 0

Rewrite

2 x^{2} + 2 y^{2} + 4 x - 8 y - 40 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 2)^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (- 1, 2), r = 5

x^{2} + (y + 7)^{2} = 36

x^{2} + y^{2} - 8 x - 20 y + 67 = 0

x^{2} + y^{2} - 6 x = - 5

(x - 2)^{2} + (y + 2)^{2} = 20

x^{2} + y^{2} + 6 x + 16 y + 37 = 0

What is 2x^2+2y^2+4x-8y-40=0 ?
The solution to 2x^2+2y^2+4x-8y-40=0 is Circle with (a,b)=(-1,2),r=5