What is 2(x-3)^2+2y^2=8 ?

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

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2(x-3)^2+2y^2=8

2 (x - 3)^{2} + 2 y^{2} = 8

(a, b) = (3, 0), r = 2

Solution steps

2 (x - 3)^{2} + 2 y^{2} = 8

Rewrite

2 (x - 3)^{2} + 2 y^{2} = 8

in the form of the standard circle equation

(x - 3)^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (3, 0), r = 2

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

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

(x + 2)^{2} + (y - 1)^{2} = 25

9 x^{2} + 9 y^{2} + 42 x - 54 y + 49 = 0

(x - \frac{7}{4})^{2} + (y + \frac{3}{4})^{2} = \frac{169}{16}

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