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

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

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

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

(a, b) = (2, 0), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 0), r = 1

x^{2} + y^{2} - 6 x - 2 y - 15 = 0

x^{2} + y^{2} - \frac{23}{2} x + \frac{11}{2} - \frac{13}{2} = 0

(x + 5)^{2} + (y - 8)^{2} = 25

x^{2} + y^{2} - 6 x - 8 y + 21 = 0

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

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