Đường thẳng - đường tròn nguyễn tuấn lâm

GIUPBANQNTAp • ~Sl1Ji'lf(jNC CIAO GIllA DUONG THANG vA DUONG TRON • B ai toan vi su' tuong giao giiia duong thdng va duong trim thuang xudt hien trong cdc di thi tuyen sinh vao Dai h9C, Cao acing va co thi tcng dung di giai mot s6 bai todn dai s6. I. LiTHUYET Cho dirong trim (C) co tam 1, ban kinh R va dirong thang L'l. D~t h = d (I, L'l). TrU'Ong hQ1> h = R. L'l la tiep tuyen cua dirong trim (C) M~ T-~~4 I Hinh 1 N~u tir di~m M ngoai (C) ke hai tiep tuyen MA va ME d~n duong trim (C) (h. 1) thi • L'lMAl = L'lMEl, 1M2=R2 + AM- =R2 +BM2. • AB 1 1M tai trung di~m H cua AB. • SIAMB = 2S IAM = Al.AM = AH.IM. ·:·TrU'Ong hQ1>h < R. L'l dt (C) tai hai diem phan biet P va Q (h. 2 . Hinh 2 GQi H la trung diem cua doan PQ thi • Tam giac IPQ can tai 1, SIPQ = ~ R2 sinPIQ. NGUYEN TUAN LAM (GV THPT Thanh Nhan, TP H6 Chi Minh) • IH l PQ, R2 = IH2 + HP2 = IH2 + PQ2 . 4 • DQ dai doan PQ IOn nhelt khi L'l di qua tam 1 cua dirong tron (C). .:. Truong hQ'P h > R. L'l va (C) khong co di~m chung. • Tir mot diem belt ki tren L'l luon ke duoc hai tiep tuy€n d~n (C). . • Lely diem K n~m tren duong tron (C) thi h - R ~ d(K,L'l) ~ h + R . II. cAc THi DV MINH HQA .:. Truong hQ1>L'l ti~p xuc VOl (C) * Thi du 1. Cho duong tron (C): .xl + Y = 2. n« phuong trinh tii p tuyen L'l cua (C) sao cho L'l ctit cac tia Ox, Oy fdn luot tai A, B va dien tich tam giac OAB nho nhat. Lai giiii. Duong tron (C) co tam trung g6c toa dQ 0 va ban kinh R = fi. Ti~p tuyen L'l qua A(a;O) ,B(O;b)(a>O, b>O) co P'T x +y =l<=>bx+ay-ab=O . a b Ta co d( O,L'l) = R d, J labl = fi a 2 +b 2 <=> ab = ~2(a2 +b 2 ) ~ 2M => ab ~ 4, nen SOAB = !OA.GB = !ab ~ 2. Dang thirc xay ra 2 2 khi a=b=2. V~yPT L'l: x+y-2=O.D *Thi dl}2. (Ciiu VI.a.! Khbi B 2009) Cho duong tron (C): (x - 2)2 + y2 =.± va cac 5 diarng thang L'l1:X-y=O,L'l2:X-7y=O. Xac dinh toa d(J tam K va tinh ban kinh cua duong iron (C l ), biit rang duang iron (C l ) tii p xuc vai cac duong thdng L'll ,L'l2 va tam K thuoc duang iron (C). lrC:>~~ tic:><: ,~ • : ~~~;; 'CJ"u6itte V www.VNMATH.com Lai giai. GQi K(a;b)E(C)~(a-2)2 +b 2 =~; (C]) ti~p XUC fl],fl2 ~ la ~bl = la ~bl . "\12 5 2 T 'd' , {5(a-2)2 +5b 2 =4 ir 0 ta co 5l a - b l = la- 7b l· Gifti M nay ta duoc (a; b) = ( ~; ~ ). Ban kinh dirong trim (C]) la R = la ~bl = 2J2 .0 "\12 5 *Thi d1}.3. (ellu V.a.l Kh8i B 2006) Cho duong trim (C): x 2 +y2 -2x-6y+6=0 va diim M( -3; 1). G9i TJ va T2 fa cac tiep diim cua cac tii p tuyen ke tit M din (C). Viit phuong trinh duong thdng TJT 2 . Liri giai. Duong tron (C) c6 tam 1(1;3) ban kinh R = 2 ,IM = 215 > R nen M nfun ngoai (C). Ta c6 MI; = MI;. =.J MJ2 - R2 = 4 nen T;,T2 thuoc dirong tron (c,) c6 tam M va ban kinh R'=4. PT dirong tron (C') la r+y+6x-2y-6=0. Tir d6 (C) ciit (C,) tai hai di~m T; ,T2 . Xet he PT {X2 +y2 - 2x - 6y +6 = 0 . x 2 +y2 +6x - 2y - 6 = 0 ~ 2x+ y-3 = 0 (1). Do T;,T2 la giao diem cua (C) va (c,) nen toa dQ cua cac diem T; ,T2 thoa man d~ng thirc (1). Do d6 PT duong thang T;T 2 la 2x+y-3=0.0 *Thi d1} 4. Cho duirng trim (Q:(x_l)2 +(y_2)2=4 va diim A (2; 1). DU'Cmg thdng d thay dJi di qua A ciit (C) tai hai diim T] va h Hai tiep tuyen cua (C) tai hai diim T] va T2 cdt nhau tai diim M Tim quy tich cua diim M Liri giai: Duong tron (C) c6 tam 1(1; 2) ban kinh R = 2, lA = J2 < R nen di~m A (y trong duong tron (C), suy ra dirong thang d qua A luon ciit (C) tai hai di~m T] va T 2 . Gift su M (xo;Yo), PT dirong tron (C') tam M va ban kinh R' = MJj = .JMJ2 _R2 la x 2 +y2 - 2xox- 2yoY +2xo +4yo -1 = O. Tuong tir Till du 3, ta tim diroc PT T;T2: (2xo - 2)x+ (2yo -4)y - 2xo -4yo + 2 = O. Do duong thang T; T2di qua A (2; 1) nen (2xo -2).2+(2Yo-4).1-2xo -4Yo+2=0 ~ Xo - Yo - 3 = 0 (2) Di~m M c6 toa dQ thoa man (2), suy ra quy tich di~m Mia dirong thang x - y - 3 = 0 .0 *Thi d1}.5. Cho duong trim (C): x 2 +y2 = 4 va duong thiing fl : x +y - 6 = O. Gia su Mfa diim thuoc duong thang fl. a) Chung minh rang tit M luon ke duac hai tiip tuyen MJj , MT2 din (C) (T; va T2 fa hai tiip diim) va duong thdng T;T 2 luon di qua mot diim cd dinh. b) Tim toa d6 diim M di doan thdng T;T 2 co ~A dai b ~ 815 ao ai ang . 5 c) Tim vi tri diim M di dien tich tu giac OT;MT2 dat gia tri nho nhdt. Liri gidi: a) Duong tron (C) c6 tam 0(0; 0) va R = 2, ME fl, d(O,fl) = 3J2 > R nen Mn~m ngoai, duOn~ tr~n. Vi vay ill M luon ke diroc hai tiep tuyen den duong tron (C). Do Met: nen M(t,6-tj, Me =MI2 =.JMJ2 _R2 . Tuong tu Thi du 3, ta c6 T;, T2 I~ giao diem cua dirong tron (C) va dirong tron (C') (C' c6 tam M va ban kinh R' = MJj ). Tir d6 ta c6 PT cua duong thang T;T 2 la tx+(6-t)y-4=0. Tir day suy ra dirong thang T;T 2 luon di qua d ·;:. ;. dinh H(2 2) rem co. 3;3 . b) Tac6 SOTlM72 =l T;T 2 .GM=0T;.MJj 2 = R JM02 - R2. Suy ra 1r<:>~r. tic:><: ~ ~~~ji & CfuOitte · V www.VNMATH.com 4J5 O'M = 2.J M02 - 4 => OM = 2J5 5 ¢::> t 2 + (6- t)2 = 20 ¢::> t = 4 hoac t = 2 . V~y co hai diem M, (4;2) va M2 (2;4). c) Ta co S01\MI2 = OTj.MY; = R JM02 - R2 = 2.J M02 - 4. Tir do S01\MI2 nho nhat khi dQ dai MO nho nhat. Khi do diem M la hinh chien cua 0 len ~, suy ra M(3;3).0 *Thi dl} 6. (Ciiu V.a.2 Kh6i D 2007) Cho duong trim (C): (x_l)2 +(y+2)2 =9 va dutmg thdng t:.: 3x - 4 y + m = O. Tim m di tren t:. co duy nhdt mot diim P ma tir do co thi ke duac hai tiep tuyen PA, PB d~n duong trim (C) (A, B la cac tiep diim) sao cho tam giac PAB d~u. Loi gidi. Duong trim (C) co tam I (1;-2) va ban kinh R = 3. Tam giac PAB deu nen IP = 21A = 2R = 6 nen P thuoc duong trim (c,) tam I ban kinh R' = 6. Tren ~ co duy nhat di€m P thoa man yeu du bai toan ¢::> d(J,~) = 6 => m = 19;m = -41.0 ·:·TfU'Ong hQP ~ dt (C) tlili hai di~m phan bi~t *Thi dl}7. Cho iIuOng tMng ~:1n(+y-2m-l=O ill duong trim (C) : (x_l)2 +(y - 2)2 = 4. a) Tim m di ~ cat (C) tai hai diim phdn biet A va B sao cho d(J dai doan thang AB ngan nhdt. b) Tim quy tich trung diem H cua doan thang AB khi duong thang ~ thay d6i. Lai giai. a) Duong trim (C) co tam I (1;2) va ban kinh R = 2. Ta thfty ~ luon di qua di€m M(2;1) va 1M = J2 < R nen M a trong (C), suy ra ~ .luon c~t (C) tai hai diem phan biet A,B. G9i H Ia trung diem cua doan AB va d(J,~) = IH. DQ dai doan AB ngan nhat khi IH dai nhat. Do IH l AB va M E AB nen IH ~ 1M => IH max = 1M. Khi ~ .L 1M , suy ra m = -1. V~y PT ~: x - y -1 = 0. b) Di€m H la trung di€m cua doan AB nen IH .L HM. Do do H nam tren duong trim (e') co dirong kinh la 1M. PT dirong trim (e'): (x-~J +~-~J =~.O *Thi du 8. Xet hai s6 thuc x, y thoa man x 2 +4x +y2 - 5 = O. Tim gia tri Ian nhdt, va gia tri nho nhdt cua biiu thiec T = 3x + 4y. Lai giai. Tren mat phang toa dQ Oxy, Ifty di€m M(X,Y)E~:3x+4y-T=0. Hai s6 x,y thoa man x 2 + 4x + y2 - 5 = 0 nen M(X,Y)E(C):X2+4x+y2-5=0 la dirong tron tam 1(-2;0) va ban kinh R = 3. MIa diem chung cua (e) va t:. ¢::>d(l,~) ~R ¢::> 16 + TI ~ 15 ¢::> -21 ~ T < 9. D~ng thirc xay ra khi ~ Ia tiep tuyen cua (C). Nhtr v~y max(T) = 9 khi { X2 +4x+ y2 -5 = 0 ¢::> {x =-~ 3x +4y - 9 = 0 12 y=- 5 rnin(T) = -21 khi { X2 +4x+ y2 -5 ~0 ¢::> {x = _1: 3x+4y+21=0 y=-g.D 5 *Thi dl}9. (Ciiu VI.h.1 Kh6iA 2009) Cho Quang tron (C): X2+y2+4x+4y+6=;0 va duong thang ~: x + my - 2m + 3 = O. G9i I la tam duong tron (C). Tim m d~ ~ dt (C) tai hai di~m ph an biet A, B sao cho dien tich tam giac lAB Ian nhdt. Lili giiii. Duong tron (C) co tam 1(-2;-2) va 1 R2 R = J2. Ta co SlAB =-IA.IB.sinAIB~- =1. 2 2 loAN HOC.• • •. !'••••••••••••••••••••••••••••••••••••••••••••••••••• ~SIln/slif5 'CTuOitJ;e 'V www.VNMATH.com Yay SlAB IOn nh~t khi IA1JB=>d(I,tl) = ~ =1 . ~2 1-2-2m-2m+31 8 ¢::> =1=>m=O hoac m= D Jl +nr- . 15 *Thi du 10. (Cau VI.h.1 KhaiD20JI) Cho diim A(O;I) va duang iron (C): X2 +y2 - 2x +4y - 5 = O. n« phuong trinh duong thdng tl eftt (C) tai hai diim M va N saG cho tam giac AMN vuong can tai A. Loi giai. Duong trim (C) co tam 1(1;-2) ve R = J1O; IA = (0;-2). Ta co IM= IN va AM = AN => AI 1 MN nen PT tl: y = m . G9i hai giao diem M (Xl;m), N (X2;m). Khi do Xl va X2 la nghiem cua PT X2 - 2x + m? +4m - 5 = 0 (3) De PT (3) co hai nghiem Xl;X2 phan biet thi m 2 +4m-6<0 (4) AM 1 AN ¢::> AM.AN = 0 ¢::> (Xl -1)( X2-1) +m 2 = 0 ¢::> XlX2- (XI + X2) + 1+ m? = 0 Ap dung dinh li Viete d6i voi PT (3), SUY ra 2m 2 + 4m - 6 = 0 ¢::> m = - 3 hoac m = 1, thoa man (4). V~yphuongtrinh tl: y = 1 hoac tl: y = -3.0 *Thi du 11. (Cau VI.a.1 Khai A 2010) Cho hai duong thdng d,: 13.x + y = 0 va d 2 :13.x - y = O. G9i (1) fa duang tron ti~p xuc vai d l tai A, eftt d 2 tai hai diem B, C saG cho tam giac ABC vuong tai B. n« phuong trinh cua (1), bi~t rdng tam giac ABC co dien , h b ~ 13 'd';: A i h ' h ~~ d. tic ang 2 va iem co oan ao uong. un gidi. Ta co A E d, => A( a ;-aJ3) (a> 0). Tir AC.l a. => AC: x-13y-4a = 0; C = d 2 (\ AC => c( -2a; - 2J3a); BA.l d 2 =>BA:x+13y+2a=0; B~d, nBA=> +~;_ a~} Ta co S ABC =! BA.BC = 13 ¢::> 13a.3a = 13 2 2 =>a= ~ =>A(~;-I} C(~;-2} Duong tron (1) co tam 1(- 2~ ;-%) (Ila trung diem cua A C) va ban kinh R = IA = 1. Tir do PT cua duong tron (1) la (x+ 2~)' +(Y+H ~LD .:. Truimg hQ1> tl khong dt (C) *Thi du 12. Cho duong trim (C):r+y2-2x -2y -7 = 0 va duong thdng tl:3x+4y+13=0. Tim gia tri tan nhdt va gia tri nho nhdt cua khoang each tir mot aiim M tren (C) a~n duong thdng tl. Loi giiii. Duong tron (C) co tam 1(1; 1)va ban kinh R = 3; d(I,tl)=4>R nen dirong thang tl khong d.t duong tron (C). Ta vi~t PT tiep tuyen cua (C) va song song voi tl. Co hai tiep tuyen la tll :3x+4y+8=0 voi tiep diem M, ( 4 ;-7) va ~ :3x+4y- 22=0 55' r ,~ d'~ M(1417) VO'I tiep tern 2 5;5 . Khi d6 d(tl,tll)=I,d(tl,tl2)=7. Voi diem M thuoc (C) suy ra l=d(tl,tll):S;d(M,tl):S;d(tl,tl2)=7. Nhu vay , (14 17) max d(M, tl) = 7 khi M =- M2 5;5 . , ,(~~) mmd(M,tl) =1 khi M =- MJ 5;5 .0 ToAN HOC •••••.•••••~,.••••.•.••••••••••• A ~StVllsiif5 'CTuOitte ~ www.VNMATH.com Nh~n xet. Hai diem MJ va M2 la giao diem cua (C) va dirong thang d qua tam I, vuong goc voi f:: ; maxd(M,f:: ) = d(I,f:: )+ R, mind(M,f:: ) = d(I,/),.)-R. *Thi d1}13.Cho duong trim (Q:r+j-a-t4y+l=<l va duang thdng thay d6i f:: : 1m+y-2m-2=0. Tim m d~ khodng each nho nh6t tir di~m M th QC (C) din duang thdng f:: dat gia tri fern nh6t. Liri gidi: (h. 3) N Hinh 3 Duong trim (C) co tam 1(1;-2) va R = 2. GQi h la khoang each nho nhat ill M dSn f:: Ta th~y f:: luon di qua di~m N(2;2) a ngoai duong tron (C). • Truong h9P f:: cat (C) thi h = O. • Truong h9P f:: khong clit (C) GQi H la chan dirong vuong goc ha ill I xuong f:: , A la giao diem cua doan IH va (C) thi h=HA. R5 rang la h dat gia tri Ian nhat khi H == N . Khi do, h = IN - R = JU - 2 va luc nay f:: 1. IN re do suy ra m = ~ .0 4 *Thi d1}14. Xet cac s6 thuc a, b, c, d thoa man a2+b2-2a+2b-23=0 va 3c-4d=-23. Tim gia tri nho nh6t cua bi~u thuc T= (a-c)2 +(b-dr Lai giiii. Xet diem A (a; b) thuoc duong tron (C): x2+y2-2x+2y-23=0 co tam 1(1;-1) va R = 5; di~m B (c ;d) thuoc duong thang f:: :3x-4y+23=0 thi T= (a-c/ +(b-d)2 =AIJ2. Ta co d(I,f:: ) = 6 > R nen f:: khong c~t (C). T dat gia tri nho nh~t khi doan AB ngan nhat. Theo Thi du l Zi ta co minAB,=d(I,f:: )-R=l. Khi do B la hinh chieu cua I len f:: , A leigiao diem cua doan thang IB va duong tron (C). -13 19 A Tir do a = -2; b = 3,' c = -' d = Yay 5' 5 . minT = 1.0 BAIT~P 1. Cho duong tron (C): X2+ y2 - 6x -10 = O. Duong thang d qua diem A dt (C) tai hai di~mM, N. a) ViSt phuong trinh duong th~ngd trong cac tnrong hQ'Pdoan MN nho nhat, Ian nhat. b) Tim quy tich trung diem H cua doan MN. A ,{1m+3 y +m+3=0 2. Cho h~ phirong tnnh X2+ y2 _ 2x-15 = 0 (m la tham so). a) Chung minh h~ da cho co hai nghiem phan biet, b) GQi (Xl; Yl )va (X2; Y2) la hai nghiem cua he. Tim gia tri Ian nh~t, gia tri nho nh~t cua bi~u tlnrc F = (Xl - X2)2 +(Yl - Y2)2. 3. Cho duong thang f:: : X + Y + 2 = 0 va duong tron (C): X2 -ta y2 - 4x - 2 Y = O. GQi Ila tam cua (C), M la d~~m thuoc dirong thang f:: Qua Mke cac tiep tuyen MA va ME dSn (C) (A, B la cac tiSp diem). Tim toa dQ diem M, biSt rang ill giac MAlB co dien tich bang 10. 4. Cho duong tron (C):X2 +y2 -8x+6y+21=0 va dirong thang d: X +Y -1 = 0 . Xac dinh toa dQ cac dinh hinh vuong ABCD ngoai tiep duong tron (C), biSt rang diem A nam tren d. 5. Cho dirong tron(C):r+y-4x-6y-12=O. Tim toa dQ diem M thuoc duong thang : d: 2x - Y + 3 = 0 sao cho MI = 2R, trong do I la tam va RIa ban kinh cua duong tron (C). lr<:>~~ tic:><: ~: • ~ ~~~~ , Cfu6itr;e V www.VNMATH.com . SIPQ = ~ R2 sinPIQ. NGUYEN TUAN LAM (GV THPT Thanh Nhan, TP H6 Chi Minh) • IH l PQ, R2 = IH2 + HP2 = IH2 + PQ2 . 4 • DQ dai doan PQ IOn nhelt khi L'l di qua tam 1 cua dirong tron (C). .:. Truong. Duong tron (C) c6 tam 1(1; 2) ban kinh R = 2, lA = J2 < R nen di~m A (y trong duong tron (C), suy ra dirong thang d qua A luon ciit (C) tai hai di~m T] va T 2 . Gift su M (xo;Yo), PT dirong tron. Duong tron (C) c6 tam 1(1;3) ban kinh R = 2 ,IM = 215 > R nen M nfun ngoai (C). Ta c6 MI; = MI;. =.J MJ2 - R2 = 4 nen T;,T2 thuoc dirong tron (c,) c6 tam M va ban kinh R'=4. PT dirong tron