题目

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2-5-9A.9                B.8                C.6                D.5 答案：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

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