What is 2x^2+2y^2=4 ?

The solution to 2x^2+2y^2=4 is Circle with (a,b)=(0,0),r=sqrt(2)

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2x^2+2y^2=4

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

(a, b) = (0, 0), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 2

x^{2} + y^{2} - 12 x - 10 y = 12

(x + 2)^{2} + (y - 6)^{2} = 2 2^{2}

4 x^{2} + 4 y^{2} + 40 - 32 y - 5 = 0

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

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

What is 2x^2+2y^2=4 ?
The solution to 2x^2+2y^2=4 is Circle with (a,b)=(0,0),r=sqrt(2)