Make a circle (x + 1) ^ 2 + (Y-2) ^ 2 = 1 tangent through point P, and the tangent point is m. if the length of PM = the length of Po, then the minimum value of PM

Make a circle (x + 1) ^ 2 + (Y-2) ^ 2 = 1 tangent through point P, and the tangent point is m. if the length of PM = the length of Po, then the minimum value of PM

Let the center of the circle be point Q, then the coordinate of point q is (- 1,2), and QM = 1 is the radius of the circle;
It is known that PM tangent circle q is in M, PM ⊥ QM,
There are: PM & # 178; + QM & # 178; = PQ & # 178;
Let the coordinates of point p be (a, b),
PM²+QM² = PO²+QM² = a²+b²+1 ,
PQ² = (a+1)²+(b-2)² ,
There are: a-178; + b-178; + 1 = (a + 1) & 178; + (b-2) & 178,
A = 2b-2,
Because PM & # 178; = Po & # 178; = A & # 178; + B & # 178; = (2b-2) & # 178; + B & # 178; = 5B & # 178; - 8b + 4 = 5 (B-4 / 5) & # 178; + 4 / 5 ≥ 4 / 5,
Therefore, PM ≥ (2 / 5) √ 5, that is, the minimum value of PM is (2 / 5) √ 5