.中考数学圆压轴题

.中考数学圆压轴题1/9CBOAD1推理运算如图，AB为O直径，CD为弦，且CDAB，垂足为H．（1）OCD的平分线CE交O于E，连结OE．求证：E为ADB的中点；（2）如果O的半径为1，3CD，①求O到弦AC的距离；②填空：此时圆周上存在个点到直线AC的距离为12．2如图6，在Rt△ABC中，∠ABC=90°，D是AC的中点，⊙O经过A、B、D三点，CB的延长线交⊙O于点E.(1)求证AE=CE；(2)EF与⊙O相切于点E，交AC的延长线于点F，若CD=CF=2cm，求⊙O的直径；（3）若nCDCF（n0），求sin∠CAB.3已知：如图，在半径为4的⊙O中，AB，CD是两条直径，M为OB的中点，CM的延长线交⊙O于点E，且EM＞MC．连结DE，DE=15.(1)求证：AMMBEMMC；(2)求EM的长；（3）求sin∠EOB的值.4如图，已知⊙O的直径AB＝2，直线m与⊙O相切于点A，P为⊙O上一动点（与点A、点B不重合），PO的延长线与⊙O相交于点C，过点C的切线与直线m相交于点D．（1）求证：△APC∽△COD．（2）设AP＝x，OD＝y，试用含x的代数式表示y．（3）试探索x为何值时，△ACD是一个等边三角形．5如图，在以O为圆心的两个同心圆中，AB经过圆心O，且与小圆相交于点A、与大圆相交于点B．小圆的切线AC与大圆相交于点D，且CO平分∠ACB．（1）试判断BC所在直线与小圆的位置关系，并说明理由；（2）试判断线段AC、AD、BC之间的数量关系，并说明理由；（3）若8cm10cmABBC，，求大圆与小圆围成的圆环的面积．（结果保留π）6在Rt△ABC中，BC=9，CA=12，∠ABC的平分线BD交AC与点D，DE⊥DB交AB于点E．（1）设⊙O是△BDE的外接圆，求证:AC是⊙O的切线;（2）设⊙O交BC于点F，连结EF，求EFAC的值．ABDEOCHABCEDOM.中考数学圆压轴题2/97如图，点A，B在直线MN上，AB＝11厘米，⊙A，⊙B的半径均为1厘米．⊙A以每秒2厘米的速度自左向右运动，与此同时，⊙B的半径也不断增大，其半径r（厘米）与时间t（秒）之间的关系式为r＝1+t（t≥0）．（1）试写出点A，B之间的距离d（厘米）与时间t（秒）之间的函数表达式；（2）问点A出发后多少秒两圆相切？8如图，在△ABC中，∠BAC＝90°，BM平分∠ABC交AC于M，以A为圆心，AM为半径作⊙A交BM于N，AN的延长线交BC于D，直线AB交⊙A于P、K两点，作MT⊥BC于T．(1)求证：AK＝MT；(2)求证：AD⊥BC；(3)当AK＝BD时，求证：BMACBPBN．9如图，AB为O的直径，CDAB于点E，交O于点D，OFAC于点F．（1）请写出三条与BC有关的正确结论；（2）当30D，1BC时，求圆中阴影部分的面积．10如图，已知O的直径AB垂直于弦CD于点E，过C点作CGAD∥交AB的延长线于点G，连接CO并延长交AD于点F，且CFAD．（1）试问：CG是O的切线吗？说明理由；（2）请证明：E是OB的中点；（3）若8AB，求CD的长．11如图11，⊙P与⊙O相交于A、B两点，⊙P经过圆心O，点C是⊙P的优弧上任意一点（不与点A、B重合），连结AB、AC、BC、OC。（1）指出图中与∠ACO相等的一个角；（2）当点C在⊙P上什么位置时，直线CA与⊙O相切？请说明理由；（3）当∠ACB=60°时，两圆半径有怎样的大小关系？请说明你的理由。12如图，⊙O是△ABC的外接圆，且AB=AC，点D在弧BC上运动，过点D作DE∥BC，DE交AB的延长线于点E，连结AD、BD．（1）求证：∠ADB=∠E；（3分）（2）当点D运动到什么位置时，DE是⊙O的切线？请说明理由．（3分）（3）当AB=5，BC=6时，求⊙O的半径．（4分）ABNMPBCDTNMAK(第27题图)ADFEOCBG（第10题图）CBAOFDEOEDCBA（第12题图）.中考数学圆压轴题3/91（1）OCOE，EOCE······················································（1分）又OCEDCE，EDCE．OECD∥．························································································（2分）又CDAB，90AOEBOE．E为ADB的中点．···············································································（3分）（2）①CDAB，AB为O的直径，3CD，1322CHCD．·············································································（4分）又1OC，332sin12CHCOBOC．60COB，····················································································（5分）30BAC．作OPAC于P，则1122OPOA．······················································（6分）②3········································································································（7分）2证明：（1）连接DE，∵∠ABC=90°∴∠ABE=90°，∴AE是⊙O直径．························································（1分）∴∠ADE=90°，∴DE⊥AC．···········································（2分）又∵D是AC的中点，∴DE是AC的垂直平分线．∴AE=CE．··································································（3分）（2）在△ADE和△EFA中，∵∠ADE=∠AEF=90°，∠DAE=∠FAE，∴△ADE∽△EFA．·······················································（4分）∴AEADAFAE，∴AEAE26．·························································（5分）∴AE=23cm．·····················································································（6分）（3）∵AE是⊙O直径，EF是⊙O的切线，∴∠ADE=∠AEF=90°，.中考数学圆压轴题4/9∴Rt△ADE∽Rt△EDF．∴DFDEEDAD．·············································（7分）∵nCDCF，AD=CD，∴CF=nCD，∴DF=（1+n）CD，∴DE=n1CD．······（8分）在Rt△CDE中，CE2=CD2+DE2=CD2+（n1CD）2=（n+2）CD2．∴CE=2nCD．·················································································（9分）∵∠CAB=∠DEC，∴sin∠CAB=sin∠DEC=CECD=21n=22nn．（103解：⑴连接AC，EB，则∠CAM=∠BEM.……………1分又∠AMC=∠EMB,∴△AMC∽△EMB．∴EMMBAMMC，即AMMBEMMC．………3分(2)∵DC为⊙O的直径，∴∠DEC=90°，EC=22228(15)7.DCDE………………………4分∵OA=OB=4，M为OB的中点，∴AM=6，BM=2．…………………………………5分设EM=x，则CM=7－x．代入(1),得62(7)xx.解得x1=3，x2=4．但EM＞MC，∴EM=4.…………………………………………7分(3)由(2)知，OE=EM=4．作EF⊥OB于F，则OF=MF=41OB=1．………………8分在Rt△EOF中，EF=,15142222OFOE…………………………9分∴sin∠EOB=415OEEF.……………………………………………………………10分4（1）∵PC是⊙O的直径，CD是⊙O的切线∠PAC＝∠OCD＝90°，显然△DOA≌△DOC···················1分∴∠DOA＝∠DOC······························2分∴∠APC＝∠COD······························3分APCCOD△∽△····························4分（2）由APCCOD△∽△，得APOCPCOD··················6分12xy，2yx····························7分（3）若ACD△是一个等边三角形，则6030ADCODC，·······8分于是2ODOC，可得2y，1x故，当1x时，ACD△是一个等边三角形··················10分5解：（1）BC所在直线与小圆相切，理由如下：过圆心O作OEBC，垂足为E，ABCEDOMF.中考数学圆压轴题5/9AC是小圆的切线，AB经过圆心O，OAAC，又CO平分ACBOEBC，．OEOA．BC所在直线是小圆的切线．（2）ACBDBC理由如下：连接OD．AC切小圆O于点A，BC切小圆O于点E，CECA．在RtOAD△与RtOEB△中，90OAOEODOBOADOEB，，，RtRtOADOEB△≌△（HL）EBAD．BCCEEB，BCACAD．（3）90BAC，8106ABBCAC，，．BCACAD，4ADBCAC．圆环的面积2222πππ()SODOAODOA又222ODOAAD，224π16πcmS．说明：若第（1）、（2）题中结论已证出，但在证明前未作判断的不扣分．6（1）证明：由已知DE⊥DB，⊙O是Rt△BDE的外接圆，∴BE是⊙O的直径，点O是BE的中点，连结OD，··················································································································1分∵90C，∴90DBCBDC．又∵BD为∠ABC的平分线，∴ABDDBC．∵OBOD，∴ABDODB．∴90ODBBDC，即∴90ODC····················································4分又∵OD是⊙O的半径，∴AC是⊙O的切线．·············································5分（2）解：设⊙O的半径为r，在Rt△ABC中，22222912225ABBCCA，∴15AB····························································7分∵AA，9