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设\(A\)为圆\({\left( {x - 1} \right)^2} + {y^2} = 1\)上的动点，\(PA\)是圆的切线且\(\left| {PA} \right| = 1\)，则点\(P\)的轨迹方程是___．
参考答案：\({\left( {x - 1} \right)^2} + {y^2} = 2\)
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