Question

如图2-5-9,AB、CD是⊙O的两条平行切线,B、D为切点,AC为⊙O的切线,切点为E点,若AB=4,CD=9,则⊙O的半径为(    )

图2-5-9

A.9                B.8                C.6                D.5

Answer 1

解析:连结OB,并作BO的延长线,过A作AF⊥CD,F为垂足.

∵AB切⊙O于B,∴OB⊥AB.

∵AB∥CD,∴BO⊥CD.

∴BO经过D点.∴BD为⊙O直径.

又∵AF⊥CD,

∴四边形ABDF是矩形.

在Rt△ACF中,AF=.

由切线长定理得AB=AE,CE=CD.

∴AC=AE+CE=AB+CD=13,CF=CD-DF=CD-AB=5.

∴AF==12,OB=6.

答案:C