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解:(1)過點(diǎn)O作OG⊥DE,垂足G,連接OB,OD,
∵DE=8,
∴DG=GE=
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DE=4,
∵AB切⊙O于點(diǎn)B,
∴OB⊥AB,
∵DE⊥AB,
∴四邊形BCGO是矩形,
∴OG=CB=3,CG=OB,
∴在Rt△ODG中,r=OD=
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=5,
∴CG=OB=5,
∴CD=CG-DG=5-4=1;
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(2)∵△ACM∽△NEM,
∴∠NEM=90°,
∴NE∥AB,
連接OB并反向延長交NE于H,
∴∠OHE=180°-∠ABO=90°,
∴NH=EH=
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EN,
∵∠OHE=∠NEM=∠ACM=∠BCM=90°,
∴四邊形BCHE為矩形,
∴BC=EH,
又∵BC=CA=
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AB,
∴EN=AB;
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(3)解:存在.
如圖,設(shè)AM切⊙O于點(diǎn)R,RM=x,EM=y,
∵CB=3,DE=8.
∴DM=DE+EM=y+8,
∴RM
2=EM•DM,
即 x
2=y(y+8)①,
∵CF是AB的垂直平分線,
∴AC=BC=3,
∴AB=6,
∵AB與AR都是⊙O的切線,
∴AR=AB=6,
∴AM=6+x,CM=CD+DM=9+y,
∵在Rt△ACM中,AM
2=AC
2+CM
2,
∴(6+x)
2=9+(9+y)
2②,
聯(lián)立①②:
②-①得:12x-10y-54=0,
∴x=
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③,
③代入①,整理得:11y
2+18y-27=0,
即(11y-81)(y+9)=0,
解得:y=
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或y=-9(舍去),
∴CM=9+y=
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<17.
∴當(dāng)a=
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時,使得AM與⊙O相切.
分析:(1)首先過點(diǎn)O作OG⊥DE,垂足G,連接OB,OD,由垂徑定理可得DG的長,又由AB切⊙O于點(diǎn)B,易得四邊形BCGO是矩形,然后在Rt△ODG中,利用勾股定理,即可求得⊙O的半徑長,繼而求得CD的長;
(2)由△ACM∽△NEM,易得NE∥AB,然后連接OB并反向延長交NE于H,易證得NH=EH=
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EN,四邊形BCHE為矩形,繼而可得BC=CA=
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AB,則可證得:EN=AB;
(3)首先設(shè)AM切⊙O于點(diǎn)R,RM=x,EM=y,由切割線定理,可得RM
2=EM•DM,即 x
2=y(y+8)①,由勾股定理可得在Rt△ACM中,AM
2=AC
2+CM
2,即(6+x)
2=9+(9+y)
2②,聯(lián)立①②,即可求得EM的長,繼而求得答案.
點(diǎn)評:此題考查了切線的性質(zhì)、切線長定理、切割線定理、垂徑定理、矩形的判定與性質(zhì)以及勾股定理等知識.此題綜合性很強(qiáng),難度較大,注意掌握輔助線的作法,注意數(shù)形結(jié)合思想與方程思想的應(yīng)用.