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

The solution to 2(x-3)^2+2y^2=2 is Circle with (a,b)=(3,0),r=1

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

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

(a, b) = (3, 0), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 3)^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (3, 0), r = 1

x^{2} + y^{2} + 6 x - 1 = 0

(x - \frac{2}{5})^{2} + (y - \frac{1}{7})^{2} = \frac{4}{9}

(x - 4)^{2} + (y + 5)^{2} = 4

4 x^{2} + 4 y^{2} = 1156

(x - 3)^{2} + (y - 0)^{2} = 9

What is 2(x-3)^2+2y^2=2 ?
The solution to 2(x-3)^2+2y^2=2 is Circle with (a,b)=(3,0),r=1