如图,AB是⊙O的直径,P为AB延长线上一点,PC切⊙O于点C,PC=4,PB=2,则⊙O的半径等于(  ) A.1

2025-02-28 04:28:32
推荐回答(1个)
回答1:

∵PC切⊙O于点C,PC=4,PB=2,
∴PC 2 =PB×PA,即4 2 =2PA,
解得PA=8,
∴OA=OB=
1
2
(PA-PB)=3,
故⊙O的半径为3.
故选C.