专题九 解析几何第二十五讲 直线与圆
2019年
1.(2019北京理3)已知直线l的参数方程为word/media/image1_1.png (t为参数),则点(1,0)
到直线l的距离是
(A)word/media/image2_1.png (B)word/media/image3.emf (C)word/media/image4_1.png (D)word/media/image5_1.png
2.(2019江苏10)在平面直角坐标系中,P是曲线word/media/image7_1.png上的一个动点,
则点P到直线x+y=0的距离的最小值是 .
3(2019江苏18)如图,一个湖的边界是圆心为O的圆,湖的一侧有一条直线型公路l,湖上有桥AB(AB是圆O的直径).规划在公路l上选两个点P、Q,并修建两段直线型道路PB、QA.规划要求:线段PB、QA上的所有点到点O的距离均不小于圆O的半径.已知点A、B到直线l的距离分别为AC和BD(C、D为垂足),测得AB=10,AC=6,BD=12(单位:百米).
(1)若道路PB与桥AB垂直,求道路PB的长;
(2)在规划要求下,P和Q中能否有一个点选在D处?并说明理由;
(3)在规划要求下,若道路PB和QA的长度均为d(单位:百米).求当d最小时,P、Q两点间的距离.
4.(2019浙江12)已知圆word/media/image9_1.png的圆心坐标是,半径长是word/media/image11_1.png.若直线与圆word/media/image9_1.png相切于点word/media/image13_1.png,则word/media/image14_1.png=_____,word/media/image15_1.png=______.
2010-2018年
2010-2018年
一、选择题
1.(2018全国卷Ⅲ)直线d0131af79155ff003a5dd811c71194d2.png分别与9dd4e461268c8034f5c8564e155c67a6.png轴,415290769594460e2e485922904f345d.png轴交于7fc56270e7a70fa81a5935b72eacbe29.png,9d5ed678fe57bcca610140957afab571.png两点,点44c29edb103a2872f519ad0c9a0fdaaa.png在圆b87ba37273613b5ea611bc14800516f9.png上,则c33b3116eb585080aa1fa75e5ea79ca2.png面积的取值范围是
A.0115a0c32df0c9878381ac50ee93eefd.png B.c38dadf5912dd1148ff1726bb17081a2.png C.c67fcc6dac1c61d23e3c770e3a7df8ac.png D.88a1fbe7a683548a4b38854683f4cde2.png
2.(2018天津)已知圆a2af4505908dd5987ec22aa44d2a10ed.png的圆心为C,直线44ed3097c232416504971ed920d3afc1.png(e358efa489f58062f10dd7316b65649e.png为参数)与该圆相交于A,B两点,则533c003242f935720a3ff6d1bc2c631e.png的面积为 .
3.(2018北京)在平面直角坐标系中,记8277e0910d750195b448797616e091ad.png为点d5dd7429585145922afe9f88016dc7de.png到直线3da80ccce4d9f37c35fe5f0fcf3db902.png的距离,当7943b5fdf911af3ffcf9d8f738478e8a.png,6f8f57715090da2632453988d9a1501b.png变化时,8277e0910d750195b448797616e091ad.png的最大值为
A.1 B.2 C.3 D.4
4.(2017新课标Ⅲ)已知椭圆0d61f8370cad1d412f80b84d143e1257.png:3edead31d5bc3b76f260246dc82b727d.png的左、右顶点分别为e283f48f6f3d4077546b2b697c3eebad.png,17b99e166258f650036939b57689bdec.png,且以线段43029b769764a1f5cda92a3ae74927c2.png为直径的圆与直线2c978fed89aeff8bda4c2d786496ae83.png相切,则0d61f8370cad1d412f80b84d143e1257.png的离心率为
A.7eeb585c11985cd3d38bae4074cc419b.png B.227e9e6ea96659f752771b4ec095b788.png C.cd62d0e8ddbf98caac736cfe095bf266.png D.7964c6a339acf2ddea25a5ef0552b97e.png
5.(2017新课标Ⅲ)在矩形cb08ca4a7bb5f9683c19133a84872ca7.png中,5985309ccee9b7f6ce883983d55aad5e.png,399cbf60c74dc97bcb6fbe063ef9ec23.png,动点44c29edb103a2872f519ad0c9a0fdaaa.png在以点0d61f8370cad1d412f80b84d143e1257.png为圆心且与87a47565be4714701a8bc2354cbaea36.png相切的圆上.若3969918ec1f4d57c0539a76a47ecea86.png,则0b70fef760c3ae083b655845f05edb02.png的最大值为
A.3 B.44dd454ebaaa58a80fe43f4a19faf2df.png C.aa4e3cfb024c7ff30a8846913966dfb1.png D.2
6.(2015山东)一条光线从点fab77df574be82b033b484bf1a56d3c2.png射出,经415290769594460e2e485922904f345d.png轴反射后与圆7f3a3c4c8f29ca9854b8f2f9f6589116.png相切,则反射光线所在直线的斜率为
A.7bc58e4fbbaf06ca3c8588f9b4eb8d30.png或032f3b3130a138d1fadcdee5bfecab55.png B.aabae3a6729958d6b3334df7b566110c.png或fc18bf9360e6895b9b01a774b18e3780.png C.94bc85b7678c72c91cec63f0433b4440.png或e7fd6c5eb9668700cb8a5274ac925fb1.png D.3165b4cb8edddba985ea1791c02195a0.png或e5ebd788dbb2252020101c55da638a66.png
7.(2015广东)平行于直线158f87d2c7138cbd7639a54582fd5b22.png且与圆23670698b07c2735de26671c4834df9a.png相切的直线的方程是
A.d13acca7140c190547ed1685666fc71d.png或a3e02bf43e83d61e5e40295c4b8b1a90.png
B.234edfbe97f0e4a6c3af749cc9916988.png或cd3df5becfa0a4b430282044ec1d047e.png
C.c53ec1e9471366330e4cfa843566e252.png或1d4144eb601c7da83cd29eaf3b20853e.png
D.3d7e1d5b4c9d0e3c4809145f1c6afa86.png或477a5089ff898efe375cdd28123aed67.png
8.(2015新课标2)过三点6ecbb976424c1c99ee8da766e08c3bd2.png,f60a435622379d3acddfc67968545048.png,5f8715e3fafc74b2337b2bb83df4ca4c.png的圆交于415290769594460e2e485922904f345d.png轴于69691c7bdcc3ce6d5d8a1361f22d04ac.png、8d9c307cb7f3c4a32822a51922d1ceaa.png两点,则f0168aee1f144c5f40751161b1e3bb43.png=
A.265ebe73c520528b6825b8ff4002086d7.png B.8 C.465ebe73c520528b6825b8ff4002086d7.png D.10
9.(2015重庆)已知直线l:d3d27db7640cb5f281b1521b5a241f09.png是圆0d61f8370cad1d412f80b84d143e1257.png:669f609c20dfe3c17c6a63ceeb44d6cf.png的对称轴,过点161c7d3fb285bbb45bea939fb83468e2.png作圆0d61f8370cad1d412f80b84d143e1257.png的一条切线,切点为9d5ed678fe57bcca610140957afab571.png,则74ad26aaeeaa9e7d3038372cf8af55b9.png=
A.2 B.39b4572b4316b034d8778c77bd53db35.png C.6 D.4b92b22b68102253b5b7fa3dcdda6b6b.png
10.(2014新课标2)设点76d89ed27c1aebdee274dfa5aa960623.png,若在圆6f01700fbd354149d7a2d44fb27cccb1.png上存在点N,使得1bef99f83355e7e522b1a477640a7743.png,则0b21a666a81629962ade8afd967826ed.png的取值范围是
A.05132f14fb0a837981cc8b0715c9165f.png B.f2502e3c6e575966a6964cefa4d31c62.png C.3277956fdea971343a42429c9fba9ad6.png D.732a9e5d7bb253c1dc754b4c107dbc03.png
11.(2014福建)已知直线word/media/image104_1.png过圆word/media/image105_1.png的圆心,且与直线word/media/image106_1.png垂直,则word/media/image104_1.png的方程是
A.2fa0e44325e81d59a9e101da22cda56d.png B.df0209a61865201b18000fa63c40719e.png C.90bb1363e5b7d680c57da577677bf508.png D.df3e05b3fea0b3d2cc8d30b88cd4cf3c.png
12.(2014北京)已知圆2d4c8aea51ec67059c871a343b30a83d.png和两点89f80b4f0ca66123f0d421cb51b8b767.png,9f64078d60679cabae60dd70c09d8eba.png,若圆0d61f8370cad1d412f80b84d143e1257.png上存在点44c29edb103a2872f519ad0c9a0fdaaa.png,使得5deb7fab19e97ff3be4f808abfd97408.png,则6f8f57715090da2632453988d9a1501b.png的最大值为
A.8f14e45fceea167a5a36dedd4bea2543.png B.1679091c5a880faf6fb5e6087eb1b2dc.png C.e4da3b7fbbce2345d7772b0674a318d5.png D.a87ff679a2f3e71d9181a67b7542122c.png
13.(2014湖南)若圆9d4508e172c4e3aee8b8ec40466f24f9.png与圆e65667823c5dfb888357106dda9bd4d5.png外切,则73f54e93c7697bd0266977c4487ac527.png
A.3c59dc048e8850243be8079a5c74d079.png B.1f0e3dad99908345f7439f8ffabdffc4.png C.45c48cce2e2d7fbdea1afc51c7c6ad26.png D.27b54d477c9e59bb797e3d306beeeef3.png
14.(2014安徽)过点Pword/media/image129_1.png的直线word/media/image130_1.png与圆word/media/image131_1.png有公共点,则直线word/media/image130_1.png的倾斜角的取值范围是
A.word/media/image132_1.png B.word/media/image133_1.png C.word/media/image134_1.png D.word/media/image135_1.png
15.(2014浙江)已知圆a56b96f08bca8f661afe5440a0891457.png截直线d0131af79155ff003a5dd811c71194d2.png所得弦的长度为4,则实数0cc175b9c0f1b6a831c399e269772661.png的值是
A.-2 B.-4 C.-6 D.-8
16.(2014四川)设0f537d3dadfcf395fd7c33fdf62800ae.png,过定点7fc56270e7a70fa81a5935b72eacbe29.png的动直线3f91bf5d4760578eed138ad292e61b3f.png和过定点9d5ed678fe57bcca610140957afab571.png的动直线75359eab6bfe587fd6e9e510f63ba013.png交于点4b1ba35f3a26c92043b659cb00da4721.png,则036c7833b7359a00d1d8f0a563983870.png的取值范围是
A.3c7d2b6017152f301218ef9f3acd03d2.png B.d05fcb653cc03b2c853f665034b35e67.png C.9aac8cc69a3910197b33aac97c1990af.png D.12cc24b783377c8c510b19359e50edee.png
17.(2014江西)在平面直角坐标系中,6c30b42101939c7bdf95f4c1052d615c.png分别是9dd4e461268c8034f5c8564e155c67a6.png轴和415290769594460e2e485922904f345d.png轴上的动点,若以b86fc6b051f63d73de262d4c34e3a0a9.png为直径的圆0d61f8370cad1d412f80b84d143e1257.png与直线2ad7747769238e9ddbbce1349b68b1e5.png相切,则圆0d61f8370cad1d412f80b84d143e1257.png面积的最小值为
A.97aff147c020b5307ec9056842fa32fc.png B.36ea7bb6a150c1218f9c1a35fe59614a.png C.b0515152bec3a3dab40e26c7a5d40d8a.png D.18ce63ba9e29e460b5027ffb43080066.png
18.(2013山东)过点(3,1)作圆word/media/image161_1.png的两条切线,切点分别为A,B,则直线AB的方程为
A. B.word/media/image163_1.png
C. D.word/media/image165_1.png
19.(2013重庆)已知圆,圆word/media/image167_1.png,word/media/image168_1.png分别是圆word/media/image169_1.png上的动点,word/media/image170_1.png为轴上的动点,则word/media/image172_1.png的最小值为
A.word/media/image173_1.png B.word/media/image174_1.png C. D.word/media/image176_1.png
20.(2013安徽)直线被圆word/media/image178_1.png截得的弦长为
A.1 B.2 C.4 D.word/media/image179_1.png
21.(2013新课标2)已知点word/media/image180_1.png;word/media/image181_1.png;,直线word/media/image183_1.png将△word/media/image185_1.png分割为面积相等的两部分,则word/media/image186_1.png的取值范围是
A.word/media/image187_1.png B. C.word/media/image189_1.png D.word/media/image190_1.png
22.(2013陕西)已知点db7e83b803c05fdb00aead61b16d96f4.png在圆外, 则直线e9994df7545f18ac84dea5b4988605fe.png与圆O的位置关系是
A.相切 B.相交 C.相离 D.不确定
23.(2013天津)已知过点P(2,2) 的直线与圆word/media/image194_1.png相切, 且与直线垂直, 则word/media/image196_1.png
A. B.1 C.2 D.word/media/image198_1.png
24.(2013广东)垂直于直线word/media/image199_1.png且与圆word/media/image200_1.png相切于第一象限的直线方程是
A.word/media/image201_1.png B.
C.word/media/image203_1.png D.
25.(2013新课标2)设抛物线word/media/image205_1.png的焦点为word/media/image206_1.png,直线word/media/image207_1.png过且与word/media/image209_1.png交于word/media/image210_1.png,两点.若word/media/image212_1.png,则word/media/image207_1.png的方程为
A.word/media/image213_1.png或45ce3926b05ebb26d70d35d66c6b1511.png B.word/media/image215_1.png或
C.word/media/image217_1.png或 D.word/media/image219_1.png或word/media/image220_1.png
26.(2012浙江)设ea07f419f0bb38cd5639126242d758db.png,则“3872c9ae3f427af0be0ead09d07ae2cf.png”是“直线e6c5419e04a1206d2b1ba0ec48009362.png:f52602c647044c746a6f6c69f406719f.png与直线c7b5cb501695b127a4a5203ecdf63d70.png:f45a0a95722f423a79932d064d16d66d.png平行”的
A.充分不必要条件 B.必要不充分条件
C.充分必要条件 D.既不充分也不必要条件
27.(2012天津)设6f8f57715090da2632453988d9a1501b.png,2b2cdced699a4d98ecc359c3f0274ab2.png,若直线d056f6e645d2e66d96ed89d9f4233790.png与圆1cab4f4335c9c9904040396ca9fd3059.png相切,则cf1ade4c34301c43d30481ed5d6c52c1.png的取值范围是
A.895d4bcec5a31e1a9790ca54b1733e0a.png B.2098f8b8358834f97e9bd109bd90dbad.png
C.f5ad3f7f30906ca22a1383fe07e25623.png D.0e67955dd2f890e320632a51e2c31851.png
28.(2012湖北)过点8915d2698837b855a2407073fec34040.png的直线,将圆形区域3077bef346963dd57c2694eda2bf5cd2.png分为两部分,使得这两部分的面积之差最大,则该直线的方程为
A.2fa0e44325e81d59a9e101da22cda56d.png B.dc7db2d312a0c70592569e15b9436a80.png C.8d31af925ff32996a72a0ce3feb4fc4a.png D.9c5fcaa2e2f4318ae2127a5eb059a51f.png
29.(2012天津)在平面直角坐标系中,直线word/media/image243_1.png与圆word/media/image244_1.png相交于word/media/image245_1.png两点,则弦word/media/image246_1.png的长等于
A. B.word/media/image248_1.png C. D.word/media/image250_1.png
30.(2011北京)已知点A(0,2),B(2,0).若点C在函数y = x的图像上,则使得ΔABC的面积为2的点C的个数为
A.4 B.3 C.2 D.1
31.(2011江西)若曲线9824b26a51714309aa4afd370035ce53.png:a2af4505908dd5987ec22aa44d2a10ed.png与曲线932d0ec79260e01afd1dd960c7bc69bb.png:11268dcdcb12375129755075bd5d7ab3.png有四个不同的交点,则实数m的取值范围是
A.(437a1141f49c4ddc987721639026154c.png,227e9e6ea96659f752771b4ec095b788.png) B.(437a1141f49c4ddc987721639026154c.png,0)f3cef465dc643930d97faf93729b90dc.png(0,227e9e6ea96659f752771b4ec095b788.png)
C.[437a1141f49c4ddc987721639026154c.png,227e9e6ea96659f752771b4ec095b788.png] D.(196e4ef52826478d4c903f3f9b627aca.png,437a1141f49c4ddc987721639026154c.png)f3cef465dc643930d97faf93729b90dc.png (227e9e6ea96659f752771b4ec095b788.png,+b63148f54c8e3d3cfd0603baae351208.png)
32.(2010福建)以抛物线7cdd10dd857d8d8d5071ff43625ebbad.png的焦点为圆心,且过坐标原点的圆的方程为
A.fbdbb67ed4e53ec033d18170f4a83374.png B.43696a783e4808afdc377efaca4a7aa6.png C.6bf1ff254c08becd354237ffeca29fc2.png D.a2af4505908dd5987ec22aa44d2a10ed.png
33.(2010广东)若圆心在9dd4e461268c8034f5c8564e155c67a6.png轴上、半径为aa4e3cfb024c7ff30a8846913966dfb1.png的圆f186217753c37b9b9f958d906208506e.png位于415290769594460e2e485922904f345d.png轴左侧,且与直线53edf3d6981150675b1dedfbfa860e69.png 相切,则圆f186217753c37b9b9f958d906208506e.png的方程是
A.f8e37e3f158c79adc20dfd5e008a1b86.png B.9f28525cc39f216a1651a71e1d9e7145.png
C.4eb252f4f78fc1695fab8ca31a2db230.png D.5b8397681cc04626700dd799c4d6945c.png
二、填空题
34.(2018江苏)在平面直角坐标系4fdf4e11d662a2fa39a87dcb39945bb6.png中,A为直线a8bfcd100949dac286176ae47d191d2d.png上在第一象限内的点,4e4123f6ca8915469429f3d255985da1.png,以b86fc6b051f63d73de262d4c34e3a0a9.png为直径的圆C与直线l交于另一点D.若828d95174b1d106a1d1f13596db58c9d.png,则点A的横坐标为 .
35.(2017江苏)在平面直角坐标系4fdf4e11d662a2fa39a87dcb39945bb6.png中,807277c81f5f94d00e6cba4f023b4cdb.png,65a1ee5833df8730a1f64ef4d9f5891d.png,点44c29edb103a2872f519ad0c9a0fdaaa.png在圆f186217753c37b9b9f958d906208506e.png:d3aaec74bca557c97b87efc4284390e6.png上,若5c9e8c9c56cb2333120237da670fbff9.png,则点44c29edb103a2872f519ad0c9a0fdaaa.png的横坐标的取值范围是 .
36.(2015湖北)如图,圆0d61f8370cad1d412f80b84d143e1257.png与9dd4e461268c8034f5c8564e155c67a6.png轴相切于点948ffaadd256621e93ad852dd73843bb.png,与415290769594460e2e485922904f345d.png轴正半轴交于两点6c30b42101939c7bdf95f4c1052d615c.png(B在A的上方),且3b0cacedaaf3e0b62baf51349c5fa266.png.
(Ⅰ)圆0d61f8370cad1d412f80b84d143e1257.png的标准方程为 ;
(Ⅱ)过点7fc56270e7a70fa81a5935b72eacbe29.png任作一条直线与圆6f01700fbd354149d7a2d44fb27cccb1.png相交于c307df251313e30dbffc899da627b0d6.png两点,下列三个结论:
①351295a4321e0a658fd8ed5ed6893546.png; ②df9a57748efb8eed38e8d24c8b21c2e0.png; ③7e02a68534fc1b4f8f5929e98db7a47c.png.
其中正确结论的序号是 . (写出所有正确结论的序号)
37.(2014江苏)在平面直角坐标系word/media/image305_1.png中,直线word/media/image306_1.png被圆word/media/image307_1.png截得的弦长为 .
38.(2014重庆)已知直线与圆心为word/media/image309_1.png的圆相交于word/media/image311_1.png两点,且为等边三角形,则实数word/media/image313_1.png_________.
39.(2014湖北)直线e6c5419e04a1206d2b1ba0ec48009362.png:bf29c8e990bf2695276097d1b9124c7d.png和c7b5cb501695b127a4a5203ecdf63d70.png:c1344a9edd512a7f9a9b1a436c4a6dc3.png将单位圆word/media/image318_1.png分成长度相等的四段弧,则word/media/image319_1.png________.
40.(2014山东)圆心在直线72e097dce32234e370f566f9d4c87b88.png上的圆0d61f8370cad1d412f80b84d143e1257.png与415290769594460e2e485922904f345d.png轴的正半轴相切,圆0d61f8370cad1d412f80b84d143e1257.png截9dd4e461268c8034f5c8564e155c67a6.png轴所得弦的长为68d9c09d99cc222af7e825a07a0f3065.png,则圆0d61f8370cad1d412f80b84d143e1257.png的标准方程为 .
41.(2014陕西)若圆word/media/image327_1.png的半径为1,其圆心与点word/media/image328_1.png关于直线word/media/image329_1.png对称,则圆word/media/image327_1.png的标准方程为____.
42.(2014重庆)已知直线word/media/image330.wmf与圆心为word/media/image331_1.png的圆word/media/image332_1.png相交于word/media/image333_1.png两点,且word/media/image334_1.png,则实数word/media/image335_1.png的值为_________.
43.(2014湖北)已知圆6f01700fbd354149d7a2d44fb27cccb1.png和点,若定点bbe2500d604132f3c70cba1a6b423a6e.pngf05adc45a8f1abd1b43b33277af2df94.png和常数6af8e2f02f674b41b6ccf43debc252d2.png满足:对圆f186217753c37b9b9f958d906208506e.png上任意一点,都有62606159a07b692928975a62af7025ec.png,则
(Ⅰ)word/media/image344_1.png ;
(Ⅱ)word/media/image345_1.png .
44.(2013浙江)直线a8f02a635cee8d7bf4d90f4c33dcdf70.png被圆8316df36a9e8a301b8f429422b098fe0.png所截得的弦长等于__________.
45.(2013湖北)已知圆:word/media/image349_1.png,直线word/media/image350_1.png:word/media/image351_1.png(word/media/image352_1.png).设圆word/media/image353_1.png上到直线word/media/image354_1.png的距离等于1的点的个数为,则word/media/image356_1.png .
46.(2012北京)直线5dbad057040ec6eb5aa5841786e25d33.png被圆8f58134edaed7f8986a3c8b89c7f4d15.png截得的弦长为 .
47.(2011浙江)若直线c3f18cab68e13730919683ab26ef2db9.png与直线f0a49c4362dff597bfc8155fe3aeca87.png互相垂直,则实数6f8f57715090da2632453988d9a1501b.png=__.
48.(2011辽宁)已知圆C经过A(5,1),B(1,3)两点,圆心在x轴上,则C的方程为__.
49.(2010新课标)圆心在原点上与直线2fa0e44325e81d59a9e101da22cda56d.png相切的圆的方程为 .
50.(2010新课标)过点A(4,1)的圆C与直线8d31af925ff32996a72a0ce3feb4fc4a.png相切于点bb90a8ced9967a586665ae435024d494.png,则圆C的方程
为 .
三、解答题
51.(2016年全国I)设圆d84c3506be3e744f601db6de5b2b1b24.png的圆心为7fc56270e7a70fa81a5935b72eacbe29.png,直线2db95e8e1a9267b7a1188556b2013b33.png过点e618053a4fbef2d9e22729df4f7dca35.png且与9dd4e461268c8034f5c8564e155c67a6.png轴不重合,2db95e8e1a9267b7a1188556b2013b33.png交圆7fc56270e7a70fa81a5935b72eacbe29.png于0d61f8370cad1d412f80b84d143e1257.png,f623e75af30e62bbd73d6df5b50bb7b5.png两点,过9d5ed678fe57bcca610140957afab571.png作4144e097d2fa7a491cec2a7a4322f2bc.png的平行线交e182ebbc166d73366e7986813a7fc5f1.png于点3a3ea00cfc35332cedf6e5e9a32e94da.png.
(I)证明71a2383fd24bb1dae472a784b685b944.png为定值,并写出点3a3ea00cfc35332cedf6e5e9a32e94da.png的轨迹方程;
(word/media/image377.gif)设点3a3ea00cfc35332cedf6e5e9a32e94da.png的轨迹为曲线9824b26a51714309aa4afd370035ce53.png,直线2db95e8e1a9267b7a1188556b2013b33.png交9824b26a51714309aa4afd370035ce53.png于69691c7bdcc3ce6d5d8a1361f22d04ac.png,8d9c307cb7f3c4a32822a51922d1ceaa.png两点,过9d5ed678fe57bcca610140957afab571.png且与2db95e8e1a9267b7a1188556b2013b33.png垂直的直线与圆7fc56270e7a70fa81a5935b72eacbe29.png交于44c29edb103a2872f519ad0c9a0fdaaa.png,f09564c9ca56850d4cd6b3319e541aee.png两点,求四边形9dc0b6c3f36e7bb4b0dd24f3d7cd193b.png面积的取值范围.
52.(2014江苏)如图,为了保护河上古桥,规划建一座新桥BC,同时设立一个圆形保护区.规划要求:新桥BC与河岸AB垂直;保护区的边界为圆心M在线段OA上并与BC相切的圆.且古桥两端O和A到该圆上任意一点的距离均不少于80m. 经测量,点A位于点O正北方向60m处, 点C位于点O正东方向170m处(OC为河岸),word/media/image385_1.png.
(I)求新桥BC的长;
(II)当OM多长时,圆形保护区的面积最大?
53.(2013江苏)如图,在平面直角坐标系word/media/image387_1.png中,点,直线word/media/image389_1.png.设圆0d61f8370cad1d412f80b84d143e1257.png 的半径为1,圆心在word/media/image391_1.png上.
word/media/image392.emf
(I)若圆心word/media/image393_1.png也在直线word/media/image394_1.png上,过点word/media/image395_1.png作圆的切线,求切线的方程;
(II)若圆word/media/image393_1.png上存在点,使word/media/image398_1.png,求圆心的横坐标的取值范围.
54.(2013新课标2)在平面直角坐标系word/media/image400_1.png中,已知圆在word/media/image402_1.png轴上截得线段长为word/media/image403_1.png,在word/media/image404_1.png轴上截得线段长为word/media/image405_1.png.
(I)求圆心的轨迹方程;
(II)若word/media/image407_1.png点到直线的距离为word/media/image409_1.png,求圆word/media/image410_1.png的方程.
55.(2011新课标)在平面直角坐标系5b26bfdd93caf0c383de7f512d198cb5.png中,曲线410e1475dd0dc59e4d4ce3367df8323b.png与坐标轴的交点都在圆C上.
(I)求圆C的方程;
(II)若圆C与直线3caf230b0b0e32bd495161753c39d724.png交于A,B两点,且f60099a27c198d27be935f9a8b4c189e.png求0cc175b9c0f1b6a831c399e269772661.png的值.
56.(2010北京)已知椭圆C的左、右焦点坐标分别是757e056403c0d00581e72d7f80d885de.png,3afd15f9e4e58bdcea53a6bebaa83c14.png,离心率是7eeb585c11985cd3d38bae4074cc419b.png,直线72549d20fb89e8be1a06280d121f1b66.png椭圆C交与不同的两点69691c7bdcc3ce6d5d8a1361f22d04ac.png,8d9c307cb7f3c4a32822a51922d1ceaa.png,以线段943afaf25ac17fe7bc39fdaae916e3a4.png为直径作圆44c29edb103a2872f519ad0c9a0fdaaa.png,圆心为44c29edb103a2872f519ad0c9a0fdaaa.png.
(I)求椭圆C的方程;
(II)若圆44c29edb103a2872f519ad0c9a0fdaaa.png与9dd4e461268c8034f5c8564e155c67a6.png轴相切,求圆心44c29edb103a2872f519ad0c9a0fdaaa.png的坐标;
(Ⅲ)设31e18a4993b5ad2b0d5a567c815a2f2e.png是圆44c29edb103a2872f519ad0c9a0fdaaa.png上的动点,当e358efa489f58062f10dd7316b65649e.png变化时,求415290769594460e2e485922904f345d.png的最大值.
专题九 解析几何
第二十五讲 直线与圆
答案部分
2019年
1.解析 由直线l的参数方程消去t,可得其普通方程为.则点(1,0)到直线l的距离是word/media/image430_1.png.故选D.
2. 解析 解法一:由,得word/media/image432_1.png,设斜率为word/media/image433_1.png的直线与曲线word/media/image434_1.png切于word/media/image435_1.png,由,解得word/media/image437_1.png.
所以曲线上,点word/media/image439_1.png到直线word/media/image440_1.png的距离最小,最小值为word/media/image441_1.png.
解法二:由题意可设点的坐标为word/media/image443_1.pngword/media/image444_1.png,则点到直线word/media/image445_1.png的距离
word/media/image446_1.png,当且仅当word/media/image447_1.png等号成立,
所以点到直线word/media/image445_1.png的距离的最小值为4.
3.解析 解法一:
(1)过A作word/media/image448_1.png,垂足为E.
由已知条件得,四边形ACDE为矩形,.'
因为PB⊥AB,
所以word/media/image450_1.png.
所以.
因此道路PB的长为15(百米).
(2)①若P在D处,由(1)可得E在圆上,则线段BE上的点(除B,E)到点O的距离均小于圆O的半径,所以P选在D处不满足规划要求.
②若Q在D处,联结AD,由(1)知,
从而word/media/image454_1.png,所以∠BAD为锐角.
所以线段AD上存在点到点O的距离小于圆O的半径.
因此,Q选在D处也不满足规划要求.
综上,P和Q均不能选在D处.
(3)先讨论点P的位置.
当∠OBP<90°时,线段PB上存在点到点O的距离小于圆O的半径,点P不符合规划要求;
当∠OBP≥90°时,对线段PB上任意一点F,OF≥OB,即线段PB上所有点到点O的距离均不小于圆O的半径,点P符合规划要求.
设word/media/image455_1.png为l上一点,且word/media/image456_1.png,由(1)知,word/media/image455_1.pngB=15,
此时word/media/image457_1.png;
当∠OBP>90°时,在中,word/media/image459_1.png.
由上可知,d≥15.
再讨论点Q的位置.
由(2)知,要使得QA≥15,点Q只有位于点C的右侧,才能符合规划要求.当QA=15时,.此时,线段QA上所有点到点O的距离均不小于圆O的半径.
综上,当PB⊥AB,点Q位于点C右侧,且CQ=word/media/image461_1.png时,d最小,此时P,Q两点间的距离PQ=PD+CD+CQ=17+word/media/image461_1.png.
因此,d最小时,P,Q两点间的距离为17+word/media/image461_1.png(百米).
解法二:(1)如图,过O作OH⊥l,垂足为H.
以O为坐标原点,直线OH为y轴,建立平面直角坐标系.
因为BD=12,AC=6,所以OH=9,直线l的方程为y=9,点A,B的纵坐标分别为3,−3.
因为AB为圆O的直径,AB=10,所以圆O的方程为x2+y2=25.
从而A(4,3),B(−4,−3),直线AB的斜率为word/media/image463_1.png.
因为PB⊥AB,所以直线PB的斜率为,
直线PB的方程为word/media/image465_1.png.
所以P(−13,9),.
因此道路PB的长为15(百米).
(2)①若P在D处,取线段BD上一点E(−4,0),则EO=4<5,所以P选在D处不满足规划要求.
②若Q在D处,联结AD,由(1)知D(−4,9),又A(4,3),
所以线段AD:word/media/image467_1.png.
在线段AD上取点M(3,word/media/image468_1.png),因为word/media/image469_1.png,
所以线段AD上存在点到点O的距离小于圆O的半径.
因此Q选在D处也不满足规划要求.
综上,P和Q均不能选在D处.
(3)先讨论点P的位置.
当∠OBP<90°时,线段PB上存在点到点O的距离小于圆O的半径,点P不符合规划要求;
当∠OBP≥90°时,对线段PB上任意一点F,OF≥OB,即线段PB上所有点到点O的距离均不小于圆O的半径,点P符合规划要求.
设word/media/image455_1.png为l上一点,且word/media/image456_1.png,由(1)知,word/media/image455_1.pngB=15,此时word/media/image455_1.png(−13,9);
当∠OBP>90°时,在中,word/media/image459_1.png.
由上可知,d≥15.
再讨论点Q的位置.
由(2)知,要使得QA≥15,点Q只有位于点C的右侧,才能符合规划要求.当QA=15时,设Q(a,9),由word/media/image470_1.png,得a=,所以Q(word/media/image472_1.png,9),此时,线段QA上所有点到点O的距离均不小于圆O的半径.
综上,当P(−13,9),Q(,9)时,d最小,此时P,Q两点间的距离
word/media/image474_1.png.
因此,d最小时,P,Q两点间的距离为(百米).
4.解析:解法一:如图,由圆心与切点的连线与切线垂直,得word/media/image477_1.png,解得word/media/image478_1.png.所以圆心为(0,-2),则半径word/media/image479_1.png.
解法二:由,得word/media/image478_1.png,所以word/media/image481_1.png.
2010-2018年
1.A【解析】圆心d8888c9eb26f75e00c7f92f18903d91d.png到直线的距离e42dfb7e448a0e6f7992f5f9d8547924.png,
所以点44c29edb103a2872f519ad0c9a0fdaaa.png到直线的距离5ecd83499cc61721ab39718e964ebc83.png.根据直线的方程可知7fc56270e7a70fa81a5935b72eacbe29.png,9d5ed678fe57bcca610140957afab571.png两点的坐标分别为feff727935034b6b0e721169137a7e2b.png,e6ac77364f663f305a7667622ec2216f.png,所以70a3768038dd60b719c0bb6dcb21b017.png,
所以c33b3116eb585080aa1fa75e5ea79ca2.png的面积56172554a70abbfb0824a471ff5d3745.png.
因为5ecd83499cc61721ab39718e964ebc83.png,所以9884fcd0cc5abca9bbecbb588dcba123.png,即c33b3116eb585080aa1fa75e5ea79ca2.png面积的取值范围是0115a0c32df0c9878381ac50ee93eefd.png.故选A.
2.93b05c90d14a117ba52da1d743a43ab1.png【解析】直线的普通方程为2fa0e44325e81d59a9e101da22cda56d.png,圆的标准方程为1f2fd99c453bb316a10a33d694756460.png,
圆心为a731de22239b7792f78c8c95c4a7b7c0.png,半径为1,点0d61f8370cad1d412f80b84d143e1257.png到直线2fa0e44325e81d59a9e101da22cda56d.png的距离b83f510d39e313404df9f5ec214b4c7f.png,所以8512ebbe87031640677475d3d08b4b60.png,所以dafd4f72fc191b1d7f13937c870423f8.png.
3.C【解析】由题意可得105876ca8b28982cbc61b083fc405186.png
50a73039ae39bf29c5187b1335da7b2d.png
(其中e2ed4e3dc40eab85946095662e5c6d1e.png,6dff9a1b75b32e5d6e49d6927be330b8.png),∵57372c1e766fc3415f208225ba8d23e9.png,
∴8a5f130a2c815507dabae04119ba6c37.png,6cb5f52cd4d05f3f3542e32709e5b961.png,
∴当e6753e61990bc639ae1869683cb421b7.png时,8277e0910d750195b448797616e091ad.png取得最大值3,故选C.
4.A【解析】以线段43029b769764a1f5cda92a3ae74927c2.png为直径的圆是5e7ded6b3a54ab3a99b931185210e0fc.png,直线2c978fed89aeff8bda4c2d786496ae83.png与圆相切,所以圆心到直线的距离c7322ab07f408794f9f56104b3cf1880.png,整理为b26acbc306319b794db0c5fc53c568b6.png,
即446fabba1a0a4bd4ea1690b207a5d579.png,即b2b97548e92f562e90e3c040a48794dc.png ,22dae513502bbc4f7ae60e7e36c49d14.png,故选A.
5.A【解析】如图建立直角坐标系,
word/media/image519.emf
则8a78c427aecb40dc2eb14dddc091f9ba.png,ade0d24e6b7f8643b80aa52f34cd5b94.png,074019d8d06db524fbad2357aeee990e.png,4b1ba35f3a26c92043b659cb00da4721.png,由等面积法可得圆的半径为335b365c715b256b402c181b42cf3d72.png,
所以圆的方程为fde115166e0c427ac7d24410d0a7e110.png,
所以2ffc88041419fc8e97d267b71e1e1b6a.png,b22cc8796b9b6cb167aec93f05416f1d.png,5c0c6c0435d95bac4e337cb4401ba948.png,
由3969918ec1f4d57c0539a76a47ecea86.png,得0d44eece08449eabebc8a7aaa4d173cb.png,所以0b70fef760c3ae083b655845f05edb02.png=cde8e1dee42a55c15d49cb4b8f0d75ba.png,
设bbf680051e245d73dd6944554e98e3d5.png,即d7d9a8328cae6708fa90bb5337bac787.png,
点4b1ba35f3a26c92043b659cb00da4721.png在圆上,所以圆心到直线d7d9a8328cae6708fa90bb5337bac787.png的距离小于半径,
所以e2fef30ff1864ce97972b4518d3ed785.png,解得5df2661ac8536191a0263b0be62d1e58.png,所以fbade9e36a3f36d3d676c1b808451dd7.png的最大值为3,
即0b70fef760c3ae083b655845f05edb02.png的最大值为3,选A.
6.D【解析】fab77df574be82b033b484bf1a56d3c2.png关于415290769594460e2e485922904f345d.png轴对称点的坐标为39eef66b9e465ea3f9c680e9494fb357.png,设反射光线所在直线为
e3058bd6903839d0c18765ae5b5122d2.png,即689f244b7fd4c827e27c71cb55162f4d.png,则1d6f9488761371acf61319c30234d26a.png,
99fe28972d688597387f32e766dae073.png,解得8291d9f0e28185e57b9f1aafc1dd047f.png或e5ebd788dbb2252020101c55da638a66.png.
7.A 【解析】 设所求直线的方程为55e7813f3319f943a6bd045e0c3e79c6.png9cc34e0d7a4d4107eb772ebdd88397d9.png,则66537606a13e2f56436ec6aca82e52d9.png,所以66e6709fccd42910c753956b6ee321fd.png,故所求直线的方程为d13acca7140c190547ed1685666fc71d.png或a3e02bf43e83d61e5e40295c4b8b1a90.png.
8.C【解析】设过ce04be1226e56f48da55b6c130d45b94.png三点的圆的方程为6855aa4a76a27e711587b788b428b1cd.png,
则5925b5d802dfb3a043d643b0dead1b01.png,解得171e0c12abaa5dc35b13d8d6dbff51a8.png,
所求圆的方程为3f31fe1d6c1a8551b697d58397669715.png,令e11729b0b65ecade3fc272548a3883fc.png,得3ae49fe305ac187b0af0b42a388b8690.png,
设f61013f9118579f81cc45a3929f20dc1.png,a9ad8ab6b42a6cb06661c7d5d31425aa.png,则98e64ad0fe01b0afa90ea6aa07bfcb51.png,16f972ff24d945a49bef28b4406b151c.png,
所以8452504903e5c866be663344b3730209.png.
9.C【解析】圆标准方程为word/media/image566_1.png,圆心为,半径为word/media/image568_1.png,
因此,word/media/image570_1.png,即,
word/media/image572_1.png.选C.
10.A【解析】当点69691c7bdcc3ce6d5d8a1361f22d04ac.png的坐标为fb0ce7c2864d45cd277575f863f6af1c.png时,圆上存在点2e94dd8cf87b7e48bbedc7382795ecec.png,使得797c3b85cd774f860fe06abde3095943.png,所以73547acb4a3fc3002fd596aa027dbdc1.png符合题意,排除B、D;当点69691c7bdcc3ce6d5d8a1361f22d04ac.png的坐标为6bac7945dbe12bc5b826168e2cbbca8b.png时,5af7cf4e023d52bb7e6d4e755953ebe3.png,过点69691c7bdcc3ce6d5d8a1361f22d04ac.png作圆f186217753c37b9b9f958d906208506e.png的一条切线409a83d0a49f06cdb614d3dcdfd10f2e.png,连接7fbe4d00defc3d618a6904950a97a91b.png,则在c7de568c83f059b3ef60573afeb7569f.png中,bbed4e74bfa31d4e88aa0a3fbe7cd02e.png,则5ef7b8bbd33eedda401a49a957bb0010.png,故此时在圆f186217753c37b9b9f958d906208506e.png上不存在点8d9c307cb7f3c4a32822a51922d1ceaa.png,使得1bef99f83355e7e522b1a477640a7743.png,即6da1538d0619854ffc5ee10c16214508.png不符合题意,排除C,故选A.
11.D【解析】直线2db95e8e1a9267b7a1188556b2013b33.png过点3a65da281bf256a8a600d46e0c99625f.png,斜率为c4ca4238a0b923820dcc509a6f75849b.png,所以直线2db95e8e1a9267b7a1188556b2013b33.png的方程为df3e05b3fea0b3d2cc8d30b88cd4cf3c.png.
12.B【解析】因为圆0d61f8370cad1d412f80b84d143e1257.png的圆心为8666756b9458fe068d3d3769b68bcdef.png,半径为1,f2449f71e31d099860d6c76043fe5022.png,所以以原点为圆心、以6f8f57715090da2632453988d9a1501b.png为半径与圆0d61f8370cad1d412f80b84d143e1257.png有公共点的最大圆的半径为6,所以6f8f57715090da2632453988d9a1501b.png的最大值为6,故选B.
13.C【解析】由题意得dceba6afec84752abc2f9b4713ede5cb.png,f0e77f4e193be5625e245a1aa98bbd51.png,
20127e6db9bfb1a4228bbc7c52b40523.png,所以3b8aca087620bf651b2c1c8f077c0236.png.
14.D【解析】设直线2db95e8e1a9267b7a1188556b2013b33.png的倾斜角为7943b5fdf911af3ffcf9d8f738478e8a.png,由题意可知b89506eea0555fb5fe00df349b191113.png.
15.B【解析】圆的标准方程为1d1734e64334e4ca6f19c5af3c865ee4.png,则圆心471b902a6815b4b29f3856b7d98702d2.png,半径4b43b0aee35624cd95b910189b3dc231.png满足0d9159456466fd8121af7a81ee266d57.png,则圆心0d61f8370cad1d412f80b84d143e1257.png到直线d0131af79155ff003a5dd811c71194d2.png的距离1339b4d35b97c4e63ae6db3ac5aafe5d.png,
所以64daba5e0665cf6a96fa7c1498fd0dce.png,故d16e273bbc81c8757bfeb5ad272f0e40.png
16.B【解析】易知直线3f91bf5d4760578eed138ad292e61b3f.png过定点8e42fb75e061fda32fc147bd1200b6d5.png,直线75359eab6bfe587fd6e9e510f63ba013.png过定点f89af4ce8092ddb5519a4d6c5b6c52ea.png,且两条直线相互垂直,故点44c29edb103a2872f519ad0c9a0fdaaa.png在以b86fc6b051f63d73de262d4c34e3a0a9.png为直径的圆上运动,
故f5980d3e3add307decb36b2ebbf8b2c0.pngc7ed468cdad466f5685e3941d706f54e.png
0a26ec7018c8025abada9a8f7cd36269.png.故选B.
17.A【解析】由题意可知以线段b86fc6b051f63d73de262d4c34e3a0a9.png为直径的圆C过原点f186217753c37b9b9f958d906208506e.png,要使圆0d61f8370cad1d412f80b84d143e1257.png的面积最小,只需圆0d61f8370cad1d412f80b84d143e1257.png的半径或直径最小.又圆0d61f8370cad1d412f80b84d143e1257.png与直线2ad7747769238e9ddbbce1349b68b1e5.png相切,所以由平面几何知识,知圆的直径的最小值为点f186217753c37b9b9f958d906208506e.png到直线2ad7747769238e9ddbbce1349b68b1e5.png的距离,此时a754cfd2e36f6f49cc108a491e97972f.png,得00c0ff68ccf1523622c017eb303ec1da.png,圆0d61f8370cad1d412f80b84d143e1257.png的面积的最小值为69434a7c941f373a7cdd30aa7ce416e6.png.
18.A【解析】根据平面几何知识,直线b86fc6b051f63d73de262d4c34e3a0a9.png一定与点(3,1),(1,0)的连线垂直,这两点连线的斜率为93b05c90d14a117ba52da1d743a43ab1.png,故直线b86fc6b051f63d73de262d4c34e3a0a9.png的斜率一定是02e3c1cfd03c87011d3b438dee3b08c4.png,只有选项A中直线的斜率为02e3c1cfd03c87011d3b438dee3b08c4.png.
19.A【解析】圆C1,C2的圆心分别为C1,C2,由题意知|PM|≥|PC1|-1,|PN|≥|PC2|-3,
∴|PM|+|PN|≥|PC1|+|PC2|-4,故所求值为|PC1|+|PC2|-4的最小值.
又C1关于x轴对称的点为C3(2,-3),
所以|PC1|+|PC2|-4的最小值为|C3C2|-4=505eee89f8a145ac57c9e1634769c720.png,
故选A.
20.C【解析】圆心word/media/image628_1.png,圆心到直线的距离word/media/image629_1.png,半径,所以最后弦长为word/media/image631_1.png.
21.B【解析】(1)当word/media/image183_1.png过word/media/image180_1.png与word/media/image632_1.png的中点word/media/image633_1.png时,符合要求,此word/media/image634_1.png,
(2)当word/media/image183_1.png位于②位置时word/media/image635_1.png,word/media/image636_1.png,
令得word/media/image638_1.png,∵,∴word/media/image640_1.png
(3) 当word/media/image183_1.png位于③位置时,word/media/image642_1.png,
令word/media/image643_1.png,即word/media/image644_1.png,
化简得word/media/image645_1.png,∵,
∴,解得word/media/image647_1.png
word/media/image648.emf
综上:word/media/image649_1.png,选B
22.B【解析】点M(a, b)在圆bac03f0ffc7d608da9f4129763cb49e7.png外,∴a49acc194ff582e08d87cefe78087ef7.png.圆18fc4c8c1ba5b3f94dbd306cf916f9e3.png到直线e9994df7545f18ac84dea5b4988605fe.png距离17a7beb1370067deca9bc7a5c7b19fbc.png=圆的半径,故直线与圆相交.所以选B.
23.C【解析】设直线斜率为word/media/image655_1.png,则直线方程为,即word/media/image657_1.png,圆心到直线的距离word/media/image659_1.png,即word/media/image660_1.png,解得.因为直线与直线垂直,所以word/media/image662_1.png, 即,选C.
24.A【解析】∵圆心到直线的距离等于word/media/image664_1.png,排除B、C;相切于第一象限排除D,选A.直接法可设所求的直线方程为:,再利用圆心到直线的距离等于word/media/image664_1.png,求得word/media/image666_1.png.
25.C【解析】抛物线7cdd10dd857d8d8d5071ff43625ebbad.png的焦点坐标为ec6014ffc3d92709aa6a4fea11bb3788.png,准线方程为e28b7964a01e56385d1d9fa4da54388c.png,设f99a5e7a20e963b917af6f0a746b0589.png,
32a60b125f2f670bebc6a7a1ef8a9f5c.png,则因为|AF|=3|BF|,所以5777dab5e4bec6c4a3d05f2fff17c54a.png,所以6bf748ac3351ba82402d73365c3877e5.png,
因为8f9e9bbe65ab9490c2f1266ff396713b.png=338e1a52a95796ddbb1a9bf2610683771.png,f9a3b8e9e501458e8face47cae8826de.png=98f43fce8dbdf3c4f8d0ac91f0de1d43d.png,所以f9a3b8e9e501458e8face47cae8826de.png=3,8f43fce8dbdf3c4f8d0ac91f0de1d43d.png=word/media/image678_1.png,当f9a3b8e9e501458e8face47cae8826de.png=3时,word/media/image679_1.png,
所以此时,若word/media/image681_1.png,则word/media/image682_1.png,
此时word/media/image683_1.png,此时直线方程为。若word/media/image685_1.png,
则,此时word/media/image687_1.png,此时直线方程为word/media/image688_1.png.
所以word/media/image689_1.png的方程是或word/media/image688_1.png,选C.
26.A【解析】“直线e6c5419e04a1206d2b1ba0ec48009362.png:f52602c647044c746a6f6c69f406719f.png与直线c7b5cb501695b127a4a5203ecdf63d70.png:f45a0a95722f423a79932d064d16d66d.png平行”的充要条件是0109272297d6cfc350a9b302f12323ee.png,解得,3872c9ae3f427af0be0ead09d07ae2cf.png或efb2883242a9cf0a3be9f72d5c18f644.png,所以是充分不必要条件。
27.D【解析】∵直线d056f6e645d2e66d96ed89d9f4233790.png与圆1cab4f4335c9c9904040396ca9fd3059.png相切,∴圆心fb0ce7c2864d45cd277575f863f6af1c.png到直线的距离为1c7c5c66c41a453c502f0bea580d405e.png,所以cd7293b24a8e5c107013a816af52d2d8.png,
设b203ba3307850fc3d8576478bfac897b.png,则09d9642f2d772daedea43a05182b17d2.png,解得723874fa498d5647d48b4bf2e9010821.png.
28.A【解析】要使直线将圆形区域分成两部分的面积之差最大,必须使过点44c29edb103a2872f519ad0c9a0fdaaa.png的圆的弦长达到最小,所以需该直线与直线7457cdd15d09bfc6c4dbb5d2b6f87390.png垂直即可.又已知点8915d2698837b855a2407073fec34040.png,则c9ba52e3eee0de9bedc2d0f92b8f2915.png,故所求直线的斜率为5499493768447afb7e8fc8f1af32eb51.png1.又所求直线过点8915d2698837b855a2407073fec34040.png,故由点斜式得,所求直线的方程为
8138872c2b4b1dfeabbdd45187485f03.png,即2fa0e44325e81d59a9e101da22cda56d.png.故选A.
29.B【解析】圆word/media/image244_1.png的圆心到直线word/media/image243_1.png的距离word/media/image707_1.png
弦word/media/image246_1.png的长.
30.A【解析】设点2807071352ec386f356a7e0fcc03cb8f.png,直线b86fc6b051f63d73de262d4c34e3a0a9.png的方程是2fa0e44325e81d59a9e101da22cda56d.png,70a3768038dd60b719c0bb6dcb21b017.png,由于75b781a7c7441078ffd5053329c34092.png的面积为2,则这个三角形中b86fc6b051f63d73de262d4c34e3a0a9.png边上的高2510c39011c5be704182423e3a695e91.png满足方程e975c08b52f703d3d852cd7c87738849.png,即ae1d30c4a5ad8cce6c34b0dc76f1c6ad.png,
由点到直线的距离公式得e86ebe1aa51b2902fcd7b83852d89f3e.png,即118a5598730bd105e7dc67828b3ff1e2.png,解得有4个实根,
故这样的点C有4个.
31.B【解析】c1886a9694e10ba05b61d7c5102747c2.png,932d0ec79260e01afd1dd960c7bc69bb.png表示两条直线即9dd4e461268c8034f5c8564e155c67a6.png轴和直线2db95e8e1a9267b7a1188556b2013b33.png:8b1b9828810c86d3030da3c0a702465b.png,
显然9dd4e461268c8034f5c8564e155c67a6.png 轴与9824b26a51714309aa4afd370035ce53.png有两个交点,由题意2db95e8e1a9267b7a1188556b2013b33.png与932d0ec79260e01afd1dd960c7bc69bb.png相交,所以9824b26a51714309aa4afd370035ce53.png的圆心到2db95e8e1a9267b7a1188556b2013b33.png的距离
14928e45628a1bd90ca9806a3f13775f.png,解得2da719c9a2eab9fc1f8e04d4b895fe8a.png,又当e6753e61990bc639ae1869683cb421b7.png时,
直线2db95e8e1a9267b7a1188556b2013b33.png与9dd4e461268c8034f5c8564e155c67a6.png轴重合,此时只有两个交点,不符合题意.故选B.
32.D【解析】因为已知抛物线的焦点坐标为ec6014ffc3d92709aa6a4fea11bb3788.png,即所求圆的圆心,又圆过原点,所以圆的半径为7c6270537cb2cf1c86fd46bbc6975dd3.png,故所求圆的方程为1f2fd99c453bb316a10a33d694756460.png,即268101c1c52a733001dc73894809928e.png,选D.
33.D【解析】设圆心54aa5dada59776d7edb7eb268fac2b22.png,则59f4751259cce0c795faeb47699adbc2.png,即51ac9177bba41b6a59590321283b8de7.png,解得e402cc044782247fdcac3cec67609bad.png,所以圆f186217753c37b9b9f958d906208506e.png的方程为5b8397681cc04626700dd799c4d6945c.png.
34.3【解析】因为828d95174b1d106a1d1f13596db58c9d.png,所以248b33f82999360c311305e5385e903b.png,又点0d61f8370cad1d412f80b84d143e1257.png为b86fc6b051f63d73de262d4c34e3a0a9.png的中点,所以6def38b2ee9758afb4d18ae597fb9508.png,设直线2db95e8e1a9267b7a1188556b2013b33.png的倾斜角为7943b5fdf911af3ffcf9d8f738478e8a.png,直线b86fc6b051f63d73de262d4c34e3a0a9.png的斜率为8ce4b16b22b58894aa86c421e8759df3.png,则8ee317d4cd87d95c4c6cb4ebbcfa798e.png,52d00dd57712f852e001315a0ad288d6.png.又4e4123f6ca8915469429f3d255985da1.png,所以直线b86fc6b051f63d73de262d4c34e3a0a9.png的方程为cdc8108073fc867f7687c5b49241d7db.png,又7fc56270e7a70fa81a5935b72eacbe29.png为直线2db95e8e1a9267b7a1188556b2013b33.png:c4533a5f878d2fe1864fb9d4d5046e05.png上在第一象限内的点,联立直线b86fc6b051f63d73de262d4c34e3a0a9.png与直线2db95e8e1a9267b7a1188556b2013b33.png的方程,得5d967b7738dcdf894987df4d6fd668dc.png,解得70d8798e4aeb4c7d7270a5d1c004c87a.png,所以点7fc56270e7a70fa81a5935b72eacbe29.png的横坐标为3.
35.403aba68d858e7cd68208d54a593b446.png【解析】设4b1ba35f3a26c92043b659cb00da4721.png,由5c9e8c9c56cb2333120237da670fbff9.png,得cd207f1d91fb0c5983641c5ee88df4a2.png,
word/media/image768_1.png
如图由cd207f1d91fb0c5983641c5ee88df4a2.png可知,44c29edb103a2872f519ad0c9a0fdaaa.png在d0d785b720c6bc3084597e32c49b8743.png上,
由2f5e55e1832a03d14ac5138c5c4fb80e.png,解得f1f907bde9c283d5824975aebc650414.png,865c654302d413ead569ff73c6fdb549.png,
所以44c29edb103a2872f519ad0c9a0fdaaa.png点横坐标的取值范围为403aba68d858e7cd68208d54a593b446.png.
36.(Ⅰ)d54890cba9180b25eaf09418fc811430.png;(Ⅱ)①②③
【解析】(Ⅰ)由题意,设194e55f0209673fe0f3c228d48b978c7.png(4b43b0aee35624cd95b910189b3dc231.png为圆0d61f8370cad1d412f80b84d143e1257.png的半径),因为6cd5cfa70b4c8ab67cc227c035fc99bf.png,
所以e53161725b514a63a9b0c6a7e60d8b35.png,所以圆心2abd02ba71bd7913332500ddf88b69e9.png,
故圆0d61f8370cad1d412f80b84d143e1257.png的标准方程为d54890cba9180b25eaf09418fc811430.png.
(Ⅱ)由02d4d187c9779fbcb733577afd54822e.png,解得932fcbca2521636c9afeb443bbfa09fa.png或3d6ea56e0fa977caf843b325ce72f01d.png,
因为9d5ed678fe57bcca610140957afab571.png在7fc56270e7a70fa81a5935b72eacbe29.png的上方,所以a7f536af281c4f3562047b844c803b1d.png,33dfb08cfc89375ffc2dc513297a1903.png.
不妨令直线943afaf25ac17fe7bc39fdaae916e3a4.png的方程为e11729b0b65ecade3fc272548a3883fc.png,c8ac519bb7463c165423fb5dbcae813d.pnga263792f24431eedaec005a80696abb0.png,
所以5a9134fa9541507d9cc4e01151fcf515.png,9539d67a74684cf82ea425c2ca9d0ac8.png,9c85f89dfb89c1ec105a667378e4cc4b.png,b4f2d8a3c3f0b7d0098bdc2580778f36.png,
所以8f76c9c28efa8da2217509b5d901764b.png,0914f07e0d14fdf0b836291ae5fc81b3.png,
所以11d9f892657be9f96f15dd27c0a8f9e7.png,所以20519e98a05f1bc09dfd3c467f539d84.png.
ef637dc75abf1c33193babfdc089465e.png.
正确结论的序号①②③.
37.5a5957c0da4987ffbb2e1128d033e042.png【解析】圆心f316c9be1045212dcf370348af09c3d4.png到直线word/media/image306_1.png的距离d99f93a4e1885e47fe234ea08980bab4.png.
直线word/media/image306_1.png被圆word/media/image307_1.png截得的弦长为f24fda6f08c75cd85223ca3dd0eb536e.png5a5957c0da4987ffbb2e1128d033e042.png.
38.2d5405d8117ccf850b5778f0d451d939.png【解析】由题意知圆心b205acf41cc2d8cef96e812d4a3187f7.png到直线的距离等于91a24814efa2661939c57367281c819c.png,
即c93310dede81f555e9ffcf72b136b855.png,解得c180349fce3ecfd00b45ab555003a3f3.png.
39.2【解析】由题意得,直线e6c5419e04a1206d2b1ba0ec48009362.png截圆所得的劣弧长为6d1a6127d3610e7b68659478ed0c2ae2.png,则圆心到直线e6c5419e04a1206d2b1ba0ec48009362.png的距离为a00b629a6429aaa56a0373d8de9efd68.png,即86db195567cd6ba3cab92d6ae470301d.png,得b214dad391380b4216ea6f4f7a3aabd0.png,同理可得a78f360850e79bc1c156969b13769928.png,则ae47224248c297645be05a70a727f183.png.
40.ac350e40acf89d07afe734521398d995.png【解析】设圆心为0f403edc44e3ff5ade179c1f1a660676.png,则圆的半径为f8bc2fbe2c937ea5b5e8839cbea69491.png,圆心到9dd4e461268c8034f5c8564e155c67a6.png轴的距离为92eb5ffee6ae2fec3ad71c777531578f.png,所以67f297e0c33adb543872078decba8a7c.png,解得3c94d884933477acdc14fc70da4b987a.png,所以圆0d61f8370cad1d412f80b84d143e1257.png的标准方程为
ac350e40acf89d07afe734521398d995.png.
41.8229a5b648d5734db39f83d67a1a4008.png【解析】因为点word/media/image328_1.png关于直线word/media/image329_1.png对称的点的坐标为b6dbc33006b907f2db1855810abfce98.png,所以所求圆的圆心为b6dbc33006b907f2db1855810abfce98.png,半径为1,于是圆C的标准方程为8229a5b648d5734db39f83d67a1a4008.png.
42.0或6【解析】圆d7aa32b0c582222819b91c8e0ccca765.png的标准方程为82f06b5ebc5b1efdba8ff4e8a9e3406f.png,所以圆心为2eb52c0c36b4338ffa4e480ef26d1a13.png,半径为3.因为0c906b9808314d82b6de7ef5a3bd29c7.png,所以圆心word/media/image331_1.png到曲线word/media/image330.wmf的距离为b8fecf2a25ddd5226c7451ea3287c0a9.png,
即13c0077f305c9a5d468838549a1373ff.png,所以ded681eaa02d11064c9a469dd1b3e04c.png或6.
43.80f3224bfd06da9a14287ce934225fb6.png【解析】设5b4499a2f2d85c2d050d1c86b2a2c826.png,则90afcea7c55a26129172a3779ac19493.png,
8242ef4374b37e84c6cb9d8226131399.png,
∵6af8e2f02f674b41b6ccf43debc252d2.png为常数,∴d68d06b6a845776b2bd5c2a05359cce0.png,解得345bb96155ce2279e6859123e8c5dfce.png或881bf6c7c58829980716611162525f32.png(舍去),∴8e01ac32295bb8045da6d0624ac22a8a.png.
解得f80bb9c5ecfb4dbe6a7c2133a37ea671.png或77bdbb914746244175ea2a27dbfe4ee0.png(舍去).
44.【解析】已知圆心为444619b1235019694b99286985e6bf28.png,半径为5,圆心到直线a8f02a635cee8d7bf4d90f4c33dcdf70.png的距离为5f1b67acb1c8c1ae0cf1a47497c6c073.png,所以弦长ae5449fd801c306971a2c8e2b9c009e3.png.
45.4【解析】由题意圆心到该直线的距离为1,而圆半径为aa4e3cfb024c7ff30a8846913966dfb1.png>2,故圆上有4个点到该直线的距离为1.
46.44dd454ebaaa58a80fe43f4a19faf2df.png【解析】圆心(0,2)到直线5dbad057040ec6eb5aa5841786e25d33.png的距离为8277e0910d750195b448797616e091ad.png=a8f2abf2309443b3e1dc4c7418ebada2.png,圆的半径为2,所以所求弦长为200aba789557d70ab1dd65acb0f6814d2.png
47.1【解析】当e6753e61990bc639ae1869683cb421b7.png时,两直线不垂直,故78e4335685327737023df15f5f5d53b8.png.因为直线c3f18cab68e13730919683ab26ef2db9.png与直线f0a49c4362dff597bfc8155fe3aeca87.png的斜率分别为93b05c90d14a117ba52da1d743a43ab1.png和7496340ee5c4b4aeb0059a0d00d30828.png,由513fc9a9cdaa6b64e86e4ff97a07bfb4.png,故40aa2227f8ab9f9737e2ce467090bb9c.png.
48.834b7a1088958c4493c0477393b59138.png【解析】以题意设圆0d61f8370cad1d412f80b84d143e1257.png的方程为ca760080894bb66611ddc015c8b357bb.png,把所给的两点坐标代入方程得187a9231c9cc7bea629419ce5d58011e.png,解得0809ee9a5e12d09d359eee370fa770d2.png,所以圆C:834b7a1088958c4493c0477393b59138.png.
49.964271f0c7226f06dd1446efa82f5b1e.png【解析】由题意可知原点到直线2fa0e44325e81d59a9e101da22cda56d.png的距离为圆的半径,
即982ba66b0d8ea667202fd31dd8ac7da3.png,所求圆的方程为964271f0c7226f06dd1446efa82f5b1e.png.
50.daef519518be796a5acae6da80ac1ebe.png【解析】设圆0d61f8370cad1d412f80b84d143e1257.png的方程为a092d3e88c87f2b51324b933d0c1c290.png,由题意得
5302b3cce02f4030e863a8111f22d12c.png,解得22ddc73aeaeb68bfbec06624ebf1c387.png,所以圆C的方程为daef519518be796a5acae6da80ac1ebe.png.
51.【解析】(Ⅰ)因为word/media/image874_1.png,,故word/media/image876_1.png,
所以,故word/media/image878_1.png.
又圆word/media/image879_1.png的标准方程为word/media/image880_1.png,从而word/media/image881_1.png,所以.
由题设得word/media/image883_1.png,,,由椭圆定义可得点word/media/image886_1.png的轨迹方程为:
(word/media/image888_1.png).
(Ⅱ)当word/media/image889_1.png与word/media/image890_1.png轴不垂直时,设的方程为word/media/image892_1.png,,word/media/image894_1.png.
由word/media/image895_1.png得word/media/image896_1.png.
则word/media/image897_1.png,.
所以word/media/image899_1.png.
过点且与word/media/image901_1.png垂直的直线:word/media/image903_1.png,word/media/image904_1.png到word/media/image905_1.png的距离为word/media/image906_1.png,所以
.故四边形word/media/image908_1.png的面积
.
可得当word/media/image910_1.png与word/media/image911_1.png轴不垂直时,四边形word/media/image908_1.png面积的取值范围为word/media/image912_1.png.
当与word/media/image914_1.png轴垂直时,其方程为,word/media/image916_1.png,word/media/image917_1.png,四边形word/media/image908_1.png的面积为12.
综上,四边形word/media/image908_1.png面积的取值范围为word/media/image912_1.png.
52.【解析】(I)如图,以O为坐标原点,OC所在直线为x轴,建立平面直角坐标系xOy.
由条件知A(0, 60),C(170, 0),
直线BC的斜率k BC=-tan∠BCO=-fa02b68ab3ebb2cf37dabd34cdfc6b97.png.
又因为AB⊥BC,
所以直线AB的斜率k AB=9df743fb4a026d67e85ab08111c4aedd.png.
设点B的坐标为(a,b),
则k BC=f106a270d08a930d7076852789c41cea.png
k AB=a2ad0e00dbf5aeed004be3d85d0ee55d.png
解得a=80,b=120.
所以BC=21cf62cae15da0847b7e99a0e55de276.png.
因此新桥BC的长是150 m.
(II)设保护区的边界圆M的半径为r m,OM=d m,(0≤d≤60).
由条件知,直线BC的方程为82d36045681200fca9c360488714b3e6.png,即b6cce71fd97dbccce883b91b5119477a.png
由于圆M与直线BC相切,故点M(0,d)到直线BC的距离是r,
即f23393ffeac9851f083d8ded6a76bb0c.png.
因为O和A到圆M上任意一点的距离均不少于80 m,
所以9148cb090e32f4f0e75b19b33cbb15ee.png即bb9ddfd5a37db8dcdb361fbbc9110f30.png解得5e9d6ccc54304f11ed24fc6fad4e54f9.png
故当d=10时,1c9bce8d97bdbf05637cc377d0a1949d.png最大,即圆面积最大.
所以当OM = 10 m时,圆形保护区的面积最大.
解法二: (I)如图,延长OA, CB交于点F.
因为tan∠BCO=fa02b68ab3ebb2cf37dabd34cdfc6b97.png.所以sin∠FCO=27abf3c3c0ceec6fce6416dc3fcf1951.png,cos∠FCO=463e10b4289d71d8f76004d317ee77b5.png.
因为OA=60,OC=170,所以OF=OC tan∠FCO=5663ce87f11811849fe52301104386f6.png.
CF=7083af84168f1e44918e573b468a6bf0.png,从而baae0f095954efd1e89874e8c54aac47.png.
因为OA⊥OC,所以cos∠AFB=sin∠FCO==27abf3c3c0ceec6fce6416dc3fcf1951.png,
又因为AB⊥BC,所以BF=AF cos∠AFB==74508e0f72dd0c05a811e42f67303f4e.png,从而BC=CF-BF=150.
因此新桥BC的长是150 m.
(II)设保护区的边界圆M与BC的切点为D,连接MD,则MD⊥BC,且MD是圆M的半
径,并设MD=r m,OM=d m(0≤d≤60).
因为OA⊥OC,所以sin∠CFO =cos∠FCO,
故由(1)知,sin∠CFO =69ae485236e9e93109517d6fb05bfc0f.png所以1c9bce8d97bdbf05637cc377d0a1949d.png.
因为O和A到圆M上任意一点的距离均不少于80 m,
所以9148cb090e32f4f0e75b19b33cbb15ee.png即bb9ddfd5a37db8dcdb361fbbc9110f30.png解得5e9d6ccc54304f11ed24fc6fad4e54f9.png
故当d=10时,1c9bce8d97bdbf05637cc377d0a1949d.png最大,即圆面积最大.
所以当OM = 10 m时,圆形保护区的面积最大.
53.【解析】(I)由题设点b146132bf6b5fa1dc8f4cf59a832739d.png,又0d61f8370cad1d412f80b84d143e1257.png也在直线de1edca1d87138fa1b61b2997e47ca84.png上,f8a4e3124f1d8c6236dcfe040f27aef4.png
54c1cb3a05b906cdc837a4ae65b75649.png,由题,过A点切线方程可设为9a6cad5b07077532a1ca331582f4d4e9.png,
即a562620c0e4f0633b8d7827ffda5239b.png,则cfd085537caee4aa215dc7415f71f13f.png,解得:59f8863c0f7e3dd16cd6336614e02537.png,
∴所求切线为2afffe4b624abdd27735b7626f7a810d.png或026df2b115c2efe19a3358c476094c70.png
(II)设点b146132bf6b5fa1dc8f4cf59a832739d.png,70d3efd00780cbdcf86fd855877c442f.png,e04e791657a0b330c0391a3f8f1bce51.png,f683dc267c1118ee43842ff903b0f207.png,18fc4c8c1ba5b3f94dbd306cf916f9e3.png,
219de64bd12d55f36655bf9a3bfcfe9a.png,即b14fd467e1f33b2c12926328569adf5e.png,又点69691c7bdcc3ce6d5d8a1361f22d04ac.png在圆0d61f8370cad1d412f80b84d143e1257.png上,
0054d566156f120afdf281f595e84a1c.png,两式相减得
cda31632e71f16a3b82911763e4460c9.png,由题以上两式有公共点,
b5fc22be95d2e668a1b967050138cdc0.png
整理得:28b8a71392a02065fb4feef73a0df82f.png,即e91741dc692ad4265c37565624de0688.png,
令166fae7b14199d2ff5e5208415381162.png,则
c3f6798a0e45294ae4bc991ed82cedc6.png,解得:e180af67f89964c3a71734c11f373c88.png,145c9932ccd8a81829b42a87d8c29edb.png,解得:36530210d108d325efee8845828b1f55.png.
54.【解析】(I)设4b0fff314fb6dbbe87d5fcb163ca03e8.png,圆44c29edb103a2872f519ad0c9a0fdaaa.png的半径为4b43b0aee35624cd95b910189b3dc231.png.
由题设147a22b73f13988a0c0bd8f5d3cb67b2.png,从而90ba86afd1f1149c7a617123ec186bab.png
故点的轨迹方程为a447c64b88d12198b4c221fc7c6c9c13.png.
(II)设3dd2b46fe0a6e95d1eed6b4e912a5229.png,由已知得9d3c14a69b3132e5595e1d8363f4b4aa.png.
又44c29edb103a2872f519ad0c9a0fdaaa.png点在双曲线a447c64b88d12198b4c221fc7c6c9c13.png上,从而得c91b6931092ac03e55d8e96dcf33768b.png
由ff3892877f825766f28be59e7d19c2dd.png得027679805d477a60f12234b050cf2f55.png此时,圆44c29edb103a2872f519ad0c9a0fdaaa.png的半径6c8be55a4326663116d40fc7139afd07.png.
故圆44c29edb103a2872f519ad0c9a0fdaaa.png的方程为4b39a121ab8b6a2e04b3745a1452b094.png或69943cce1bef5301f45b8f8a8aa77c0e.png
55.【解析】(I)曲线ca6f043e66bb517fd6f210739e783393.png与415290769594460e2e485922904f345d.png轴的交点为b6dbc33006b907f2db1855810abfce98.png,与9dd4e461268c8034f5c8564e155c67a6.png轴的交点为
(8e00eecfa245ffbf08e27813f34a7d09.png
故可设C的圆心为1dd33d0d26cf450b8c16b5db91d897ba.png,则有e0276639235f38b2a91bca55b1ef6ee3.png解得b73c3280b6f85a6ac520af103083f535.png.
则圆C的半径为ed46a34a5b52378345e0790af6a1a9ff.png
所以圆C的方程为9a719559639201181a44f7ca9988a02c.png
(II)设f99a5e7a20e963b917af6f0a746b0589.png,32a60b125f2f670bebc6a7a1ef8a9f5c.png,其坐标满足方程组:11b561c5b939f9dd39a3729bf8a26cfe.png
消去415290769594460e2e485922904f345d.png,得到方程44c8648c0913d1fd7f7efa2088446789.png
由已知可得,判别式d462af7dafb04eafe3ac04c0663a4fb4.png
因此,6452dcaa535dc0743c9e616d80e3a911.png从而
cef36c4c0eeb7f5c8f72ef8fc161f7b9.png ①
由于da859706a04bd0256e614efa5d550e02.png,可得3e4b6f74e880c71cf8d2312425943458.png
又89ee72c141408e2f6fca37abd893417e.png所以
22068abc885ac09f70adb62521471877.png ②
由①,②得ba77036f543f07a168320c280152a8ad.png,满足235204737f42c95192ba97f69d6f309a.png故e120f01632d4482c8fa5e0eb8d8731e1.png
56.【解析】(I)因为1ee72f6945454619077924d52d6d2bc6.png,且604591a06cf5526118adefdfc240c271.png,所以29c63c5d7c19a40abe4a827b2ba113b8.png
所以椭圆C的方程为d41128e4a7fbf6ec15ed023f56f41f6b.png
(II)由题意知9b64105632fd5be377a5dafb34e2af21.png
由7a935b01105ed1b9e84bd2faf0ee79af.png 得ab8261e0e5cadb9f316dc67e4a3caf82.png
所以圆44c29edb103a2872f519ad0c9a0fdaaa.png的半径为6856516c6555f6c7d9994c3a8402690a.png
解得f8e087bbf8a1d96f8e97d8eceb361cb5.png,所以点44c29edb103a2872f519ad0c9a0fdaaa.png的坐标是(0,4863c35c869fd19cbcec934a6b75f9e3.png)
(Ⅲ)由(Ⅱ)知,圆44c29edb103a2872f519ad0c9a0fdaaa.png的方程64ca2a5bcbb550f155cb365acbd97be9.png。因为点31e18a4993b5ad2b0d5a567c815a2f2e.png在圆44c29edb103a2872f519ad0c9a0fdaaa.png上.
所以6df9de37f97a09fccbe45e2625888a19.png
设c49bc3cf9cba87cddd7e1ba89c47ebd8.png,则21e7158f5bea04f2af5c7bda13b00613.png
当b070ef27980ad7f8364f3c770803de68.png,即fde488b9c0adec0c51f27da99210f7c0.png,且e11729b0b65ecade3fc272548a3883fc.png,415290769594460e2e485922904f345d.png取最大值2.
十年高考理科数学真题 专题九 解析几何 二十五 直线与圆及答案