Draw a point P (x0, Y0) from the circle C: x ^ 2 + y ^ 2-2x-2y-2 = 0 to the tangent line of the circle, the tangent point is m, O is the coordinate origin, and | PM | = | Po | Draw a point P (x0, Y0) from the circle C: x ^ 2 + y ^ 2-2x-2y-2 = 0 to the tangent line of the circle, the tangent point is m, O is the coordinate origin, and | PM | = | Po |, find the minimum P point coordinate of | PM |

Draw a point P (x0, Y0) from the circle C: x ^ 2 + y ^ 2-2x-2y-2 = 0 to the tangent line of the circle, the tangent point is m, O is the coordinate origin, and | PM | = | Po | Draw a point P (x0, Y0) from the circle C: x ^ 2 + y ^ 2-2x-2y-2 = 0 to the tangent line of the circle, the tangent point is m, O is the coordinate origin, and | PM | = | Po |, find the minimum P point coordinate of | PM |

It is easy to know that the center of the circle C is (1,1) and the circle passes through the origin o. ∵ PM | = | Po |. It can be seen that △ POC ≡ Δ PMC | Po is also tangent, then there is Pythagorean theorem in △ POC. We can know that Po ^ 2 + OC ^ 2 = PC ^ 2 |, and simplify it to x0 + Y0 = 0, so when p is on the fixed line: x + y = 0, the shortest PM is Po, and when p is the origin, the shortest value = 0P (0