Let the circle (X-2) ^ 2y-3) ^ 2 = 1, a point P (x0, Y0) and a point P (x0, Y0). The tangent to the circle is m, O is the origin of the coordinates, and | PM | = | Po |

Let the circle (X-2) ^ 2y-3) ^ 2 = 1, a point P (x0, Y0) and a point P (x0, Y0). The tangent to the circle is m, O is the origin of the coordinates, and | PM | = | Po |

I found the original title, and now the excerpt is as follows:
1. Draw a tangent from a point P (a, b) outside the circle (X-2) ^ 2 + (Y-3) ^ 2 = 1 to the circle, the tangent point is Q, and O is the origin; (1) if Po = PQ, find the relationship between a and B; (2) under the condition of (1), find the coordinates of the point P that makes | PQ | the smallest
(1) From Po = PQ, we can get: (A-2) ^ 2 + (B-3) ^ 2-1 = a ^ 2 + B ^ 2, and simplify to 2A + 3B = 6, that is, a and B satisfy the relationship;
(2) Let | PQ | be the minimum, that is, let | Po | be the minimum, and let the vertical line 3a-2b passing through the origin 2A + 3B = 6 = 0. The vertical foot is the point P (12 / 13, 18 / 13)
I hope I can help you!