2422直线和圆的位置关系同步测控优化训练含答案1112118.pdf-得力文库

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1、（51.RtAABCAB=6 cm,AC=3 cm.（1）_ C2 cmAB（2）_ C4 cmAB（3）_ CAB 2._ 3.00r=5 cm,P1OP=5 cm,1OO A.B.C.D.4.003,01d,10dA.d=3 B.d3 C.d3（10124-2-2-1,ZAOB=30,MOAM2 cm）M.MOA 0M=_ cm（DM0B 2.00R,I001d,dRA.dR B.dR C.feR D.dR 3.RtAABCZC=90,AB=10,AC=6,CCAB0C（A.8 B.4 C.9.6 D.4.8 4.00m,1m C.d D.d 2 2 5.24.2.2A.B.C.D.624

2、-2-2-2,PAPB00AB.0P=4,PA=23,ZAOB7RtAABCZABC=90,DAC0OADBQB00E24223(1).ZCAB24-2-2-3(2),24-2-2-3 242 2 3(2)CE=CE824-2-2-4,900AB,0A=AB.DEEBDEC.iiE:ZACB=-ZOAC.3 A.900 B.100 C.110 24-2-2-4（30124225,O,AB=CD,ABE.CD 24-2-2-5 224-2 26,PA.PBAB,O,ZOAB=25%ZAPB 24226 324-2-2-7ABCDADBC,ZD=90,AD+BC=AB,AB 0O,OOCDD 15

3、 A 24-2-2-7424-2-2-8AB0CDABCD=BD,D DEACE,DE0O 24-2-2-8 524-2-2-9,ABCD2,MBCPMC,PMC,AB0O,POOADF,E.CDFP 24229 624-2-2-10ABOMNC,ADMND,BEMNE,BEF,AD=3 cm,BE=7 cm,(1)0O(2)DE 24-2-2-10 724 22 11,OA hjOBRBCOAC.0A*jOBPDACD,BCM.(1)CD=PB;(2)DNBC,DN0B 24-2-2-11 0O|O,xyAB.(1)24-2-2-12,AGOiyC,OAB AC5 BC 5(2)001M(2

4、,2),ABOAd,d+AB 24-2-2-12(51.RtAABCAB=6 cm,AC=3 cm.(1)C2 cmAB(2)C4 cmAB(3)CAB3 R cm,AB33 -cm.2(1)(2)(3)cm 2 2._ 3.00r=5 cm,P1OP=5 cm,1OO A.B.C.D.PD4.003,01d,100dA.d=3 B.d3 C.d3 1B(101.24-2-2-1,ZAOB=30,MOAM2 cm)M.MOA 0M=cmQM0B24-2-2-1 OM=2x2=442.00R,101d,dR）A.dR B.dR D.dm C.d D.d-2 2 C5.A.B.C.D.B624-2

5、-2-2,PAPB）0AB.0P=4,PA=23,rZAOB（）A.90 B.100C.11OD.12O0 TPAPBOAB.PAOA,PBOB.ZAPO=ZBPO.VOP=4,PA=2 V3,AOA=2.A ZAPO=ZBPO=30,ZAPB=60./.ZAOB=120.D7RtAABCZABC=90,DAC0OADBQB00E24223(1).ZCAB24-2-2-3(2),24-2-2-3 24-2-2-3(2)CE=CEAE AE OD.V ZABC=90,CB0OE,ZABE=90 .AEOODACOAEOD#CE.AE,AE=CE.DAC AD=CD=BD Z1=Z2.AEBD（D

6、O.Z1=ZDAE.Z2=ZDAE.AAE=CE.DE,AEOO ZADE=90.YDACDEACAE=CE.24-2-2-4,000AB,OA=AB.DEEBDEC.ACBZOAC.24-2-2-4 OEAE,AAFDEF,DEE0EDC.V BC DE,OEAFBC.Z1=ZACB,Z2=Z3.VOA=OE,/.Z4=Zr3./.Z4=Z2.AOBFECAE=AC.AZ1=Z2.Z4=Z2=Z1,ZACB=l ZOAC 3（30124 22 5,0,AB=CD,ABE.CD24-2-2-5 OE,OFXCDF.VABE,AOEAB.VOFCD,AB=CD,AOE=OF./.CD O 2.2

7、4226,PAPBAB,ZOAB=25%ZAPB TPAPB 30 AB,.PA=PB.A0A PA.V ZOAB=25,/.ZPAB=65.ZAPB=180-65x2=50.OB,(1)PAPB OO AB,OAPA,OBAB.ZOAP+ZOBP=180.ZAPB+ZAOB=I80.VOA=OB AZOAB=ZOBA=25.ZAOB=130.A ZAPB=50 OPABC,(2).PA、PB 0O AB,OAPA,OPAB.OP ZAPB,AZAPC=ZOAB=25.ZAPB=50 324-2-2-7ABCDAD/7BC,ZD=90,AD+BC=ABAB 00.00CDp 24-2-2-7

8、ilEOO CD0CDOAOBAB）2 OOECDE.VOECD,A ZOEC=90 V ZD=90,ZOEC=ZDADOE.V AD BC,ADBC 0E.Zxxk.Com VOA=OB,ACE=DE.AOE=-（AD+BC）.2 VAD+BC=AB,OE=AB.2 OOCD 4.24-2-2-8AB00CDABCD=BD,DDEACE,DE0OilEDE00OD,ODDE12/18 ODAD.CD=BD,AZ1=Z2.VOA=OD.Z2=Z3./Z1=Z3.AEOD.TAEDE,AODDE.ADEOO5.24-2-2-9,ABCD2,MBCPMC,PMC,ABOO,P0OADF,E.CDF

9、PCDFP3ABCDZA=ZB=90.AFBPOOTPF0OFE=EA,PE=PB CDFPAD+DC+CB=2x3=6 6.24-2-2-10ABOMNC,ADMND,BEMNE,BEF,AD=3cm,BE=7cm,(1)90(2)DE 24-2-2-10：(1)OC,OC(2)AF,ADEFAD=EF,DE=AF,RtAABFAF OC.VMNC,OC MN.VADMN,BEMN,ADOCBE VOA=OB,.ocADEBAOC=-(AD+BE)=5 cm.2(DO5 cm.AF.VABO ZAFB=90.ZAFE=90.ZADE=ZDEF=90,ADEF.rDE=AF AD=EF=3 c

10、m.Rt/kABF BF=BE-EF=4 cm,AB=2OC=10cm.AF=JAB-B宀 JlO?42=2Q(cm),14/18/.DE=2 V2?cm.72422 11,A0BBBC0ACQA0BPDAC D,BCM.(1)CD=PB;(2)DNBC,DN0B(l)TBC0AC,DP0AP,ZDCM=Z BPM=90,MC=MP.ZDMC=ZBME/.ADCMABPM.CDPB (2)BBHDN,H.V HD/BC,BC CD,A HD CD.ZBCD=ZCDH=Z BHD=90.BCDH.BH=CD VCD=PB,/.BH=PB.DNOBOOiO,xyAB.(1)24-2-2-12,AO

11、iyC,OABAC5 BC 5(2)0OM(2,2),ABOAd,d+AB ACAC OA=3k(k0),A(J 3 ZAOB=90.BC 5 AAB=5k,OB=4k.VOA OB=AB OG=2SAAOB,12 3kx4k=5x.5.k=l.AOA=3,OB=4.AB=5 AX3,0)V ZAOB=90,.AB00VAC G0A,ABA AC.ZBAC=90.,A,AB 4 Rt/kABC.J=-,BC 5 .OC=BC-OB=-.4 ACykx+b,3k+b=0.3 9 o.k=,b=b=-.4 4 4 3 9 ACy=-x-.(2)d+AB AAOBOAOB.ABPQT.:.BQ=BT.AP=AT.OQ=OP=2 AB=BT+AT=OB +OA-=OA+OB d.2 2 d+AB=df+OA+OB-d=OA+OB.xN,AN=OB,OMBM.AM.MN.VM(2,2),.OM ZAOB.OM=2V2.ZBOM=ZMON=45 .AM=BM I ZMAN=ZOBM.OB=AN,AABOMAANM ZBOM=ZANM=45,ZANM=ZMON.OA+OB=OA+AN=ON=yjoM1+MN2xOM=VI x22=4.d+AB4.

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