Question

已知圆O的半径为1，PA、PB为该圆的两条切线，A、B为两切点，那么

PA

•

PB

的最小值为-3+2

2

-3+2

2

．

Answer 1

设PA与PO的夹角为a，则|PA|=|PB|=

1

tanα

y=

PA

•

PB

=|

PA

||

PB

|cos2α
=

1

(tanα)2

•cos2α=

cos2α

sin2α

•cos2α
=

1+cos2α

1-cos2α

•cos2α
记cos2a=u．则y=

u(u+1)

1-u

=(-u-2)+

2

1-u

=-3+(1-u)+

2

1-u

≥-3+2

2

即

PA

•

PB

的最小值为-3+2

2

故答案为：-3+2

2