If x.y satisfies x square + y square-4x + 1 = 0, find the maximum value of 3 / x, Y-X minimum value, x square + y square maximum value

x² + y² -4x + 1 =0;
(x-2)² + y² = 3;
So radius r = √ 3, center of circle: (2,0)
1)
Let k = Y / X; y = KX;
Put it into the original equation
（1+k²）x² -4x +1 =0;
△= 16-4(1+k²)>=0;
k²

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