Sự tương giao đường thẳng và đường tròn

Sự tương giao đường thẳng và đường tròn

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Bai toan vi su' tuong giao gii i a duong

thdng va duong t r im thuang x udt hien

trong cdc di thi tuyen sinh vao Dai h9C , Cao

ac i ng va c o th i tcng dun g d i gia i 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)

I

Hinh 1

N~u tir di~m M ngoai (C) ke hai tiep tuyen

MA v a ME d~n duong trim (C) (h 1) thi

• L 'lM A l = L 'lMEl, 1M2 =R 2+AM - =R 2+BM2.

• A 11M tai trung di~m H cua AB.

• SI AMB=2SI A M=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, SI PQ= ~ R 2sinPIQ

( G V THPT T hanh N han , TP H 6 C hi Minh )

• IH l PQ , R 2 =IH 2 +HP 2 =I H 2 +P Q 2

4

• DQ dai doan PQ IOn nhelt khi L 'l di qua tam

1 cua dirong tron (C)

.: Truong hQ'Ph>R.L 'l va (C) khong codi~m

chung

• Tir mot diem belt ki tren L 'l luon ke duoc hai

• Lely diem K n~m tren duong tron (C) thi

h - R ~ d(K , L 'l) ~ h+R

II cAc THi DV MINH HQA : Truo g hQ1>L' lti~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 ct it 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 g c toa

dQ0va ban kinh R = fi

Ti~p tuyen L 'l qua A(a ; O) , B(O ; b)(a > O , b > O )

coP'T x + y =l<=>bx+a y -ab=O

a b

Ta co d( O ,L 'l) =R d , J labl = fi

a 2 +b 2

<=>ab =~2(a2 +b2) ~ 2M => ab ~ 4, nen

S OAB= !OA.GB =!ab ~ 2 Dang thirc xay ra

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 ' l 1:X-y=O, L ' l 2:X-7y=O Xac dinh toa d(J tam K va tinh ban kinh cua duong iron (Cl), biit rang duang iron (Cl)

tiip xuc vai cac duong thdng L 'll , L 'l 2 va tam

K thuoc duang iron(C)

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Lai giai. GQi K(a;b)E(C)~(a-2)2 +b2=~;

( C]) ti~p XUC fl],fl2 ~ la ~bl = la ~bl

"\12 5 2

Tir'd'0ta co, {5(a-2) 2 +5b 2 =4

5la-bl =la-7bl·

Gifti M nay ta duoc (a; b) = ( ~; ~ ). Ban

kinh dirong trim (C]) la R =la ~bl =2J2 0

"\12 5

*Thi d1}.3 (e l lu V.a.l Kh8i B 2006)

Cho duong trim (C): x2 +y2 -2x-6y+6=0

va diim M(-3; 1) G9i TJ va T2 fa cac tiep

diim cua cac tiip tuyen ke tit M din (C)

Viit phuong trinh duong thdng TJT2.

Liri giai Duong tron (C) c6 tam 1(1;3) ban

kinh R =2,IM = 215 >R nen Mnfun ngoai (C)

Ta c6 MI; =MI; =.J MJ2 - R 2 =4 nen T ; ,T 2

thuoc dirong tron (c,) c6 tam M va ban kinh

R'=4.

PT dirong tron (C') la r + y +6x-2 y -6=0.

Tir d6 (C) ciit (C,) tai hai di~m T;, T

Xet he PT {X2+y2 - 2x - 6 y +6=0

~ 2x+ y-3 =0 (1) Do T; ,T2 la giao diem

T;, T thoa man d~ng thirc (1)

Do d6 PT duong thang T;T 2 la 2x+y-3=0.0

v a di i m A(2; 1). DU'Cmg thdng d t ha y d J i di

qua A ci it (C) tai hai di i m T ] va hHai tiep

tu ye n c ua (C) tai hai di i m T] va T 2 c dt nhau

tai di i m M Tim qu y ti c h c ua di i m M

Liri giai: Duong tron (C) c6 tam 1(1; 2) ban

kinh R = 2, lA = J2 < R nen di~m A (ytrong

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 =.JMJ 2 _R 2 la

x2 +y2 - 2xox- 2yoY +2xo +4y o -1 =O Tuong tir Till du 3, ta tim diroc PT T;T 2 :

(2xo - 2) x + (2yo -4)y - 2 x o -4 yo +2=O

Do duong thang T;T 2 di qua A (2; 1) nen (2xo -2).2+(2Yo-4).1-2xo -4 Yo +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

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 T fa hai tiip di i m) va duong thdng T;T 2 luon di qua mot diim c ddinh.

~ A dai b ~ 815

ao ai ang .

c) Tim vi tri di i m M d i di e n ti c h tu giac OT;MT 2 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 illM 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·rem; co; dinh. H(2 2) 3 ; 3

b) Tac6 S O T lM 7 =l T;T2 GM=0T;.MJj

2

= R JM0 2 - R 2 Suy ra

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4 J 5 O ' M =2 J M0 2 - 4=> OM =2J5

5

V~y co hai diem M ,(4;2) va M 2(2;4)

c) Ta co S 01\ M =OT j M Y ; =R JM0 2 - R 2

= 2 J M0 2 - 4 Tir do S 0 \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 200 7 )

Cho duong trim (C): (x_l)2 +(y+2)2 =9

va dut mg t h dng t : : 3x - 4y +m =O Tim m

di tre n t :.co duy nhdt mot diim P ma tir do

co thi ke duac hai tiep tuyen PA, P B d~n

duong trim (C) (A, B la cac tiep diim) sao

cho tam giac P AB 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 Ib n kinh R ' =6

Tren ~ co duy nhat di€m P thoa man yeu d

bai toan ¢ > d ( J , ~ ) =6=> m=19 ; m =-41.0

phan bi~t

ill duong trim (C) : (x_l)2 +(y - 2)2 =4

a) Ti m m di ~ cat (C) tai h i diim p dn

biet A va B sao cho d (J dai doan t hang AB

ngan nhdt.

b) Tim quy tich trung diem H cua doan

thang A B k hi duong t h ang ~ t h ay 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 1 M = J2 <R nen M atrong (C),

su ra ~ luon c~t (C) tai hai diem phan biet

A, B.

d ( J , ~) =IH DQ dai doan AB ngan nhat khi

IH dai nhat Do IH -.l A B va M E A B nen

IH ~ 1M = IHm ax =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 + ~ -~=~.OJ

x2+4 x +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 O xy, Ifty

di€m M(X , Y)E~:3x+4y-T=0.

Hai s6 x ,y thoa man x 2 +4 x +y2 - 5= 0

nen M (X,Y )E(C):X 2 +4 x + y2 -5=0 la dirong trontam 1(-2;0) vaban 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+ y 2 -5 =0¢ : :> {x =-~

y

=-5

rnin(T) =-21 khi

{

X 2 +4x+ y 2 -5 ~ 0¢ >{x = _1:

3 x +4 y +21=0 y =-g D

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Cho Quang tron (C): X2+y2+4x+ 4 y+6= ; 0

va duong t h ang ~: 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 lA B Ian nhdt.

Lili giiii Duong tron (C)co tam 1(-2;-2) va

R = J2 Ta co SlA =-IA IB.sinAIB~- =1

loAN HOC • • • !'••••••••••••••••••••••••••••••••••••••••••••••••••• ~SIln / sli f 5

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Yay SlAB IOn nh~t khi IA1JB=>d(I,tl) = ~ =1

1-2-2m-2m+31 8

Jl +nr- 15

*Thi du 10.( Cau VI h 1 KhaiD 2 0JI )

C h di i m A (O ;I ) va duang iron (C):

X2+y2 - 2 x +4 y - 5=Ọ n« phuon g t r i nh

duong thdng tl e ftt (C) tai hai di i m M v a N

s aG c ho tam gia c A MN vuon g c an tai Ạ

Loi giaị Duong trim (C) co tam 1(1;-2)

veR = 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

De PT (3) co hai nghiem Xl;X2 phan biet thi

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

2m2 +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 ạ1 Khai A 2010)

C ho hai duong thdng d,: 13 x +y =0 va

d 2 : 1 3 x - y =Ọ G9 i (1) f a duang tron ti p

x u c vai d l tai A, e f t d 2 tai ha i diem B , Cs a G

c ho tam giac ABC vuong tai B. n«phuong

trinh cua (1), bi ~ t rdng tam gia c ABC c o di e n

, h b ~ 13 ' d' ;: A i h ' h ~~ d

ti ang 2 v a iem c o oan a uon g.

ungidị Ta co A E d, => Ăa ;-aJ3) (a> 0).

Tir AC ạ=>AC: x-13y-4a =0;

C=d 2 (\ AC => c( -2a; - 2J3a);

BA d 2 =>BA:x+13y+2a=0;

B~d, nBA=> +~;_ ã}

Ta co SABC =! BẠBC = 13 ¢::>13ạ3a =13

=>a= ~ =>Ẵ;-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

.: Truimg hQ1>t khong dt (C)

*Thi du 12.C h o duong trim (C):r+y2-2x

-2y -7 =0 va duong thdng tl : 3x+4y+13=0

Tim gia tri ta n n dt va gia tri nho nhdt cua

duon g thdn g t

Loi giiiị Duong tron (C) co tam 1(1; 1)va

ban kinh R = 3; d(I,tl)=4>R nen dirong thang tl khong d.tduong tron (C)

Ta vĩt 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)

VÓI 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

max d(M, tl)=7 khi M =-M2 5;5

ToAN HOC •••••.••••• ~ ••••.•.••••••••••• , A ~ S tVl l sii f5

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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

GQih la khoang each nho nhat ill M dSn f::

Ta th~y f:: luon di qua di~m N(2; 2) angoai

duong tron (C)

• Truong h9P f:: c at (C) thi h=O

• Truong h9P f:: khong clit (C)

GQi H la chan dirong vuong goc ha illI 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 redo suy ra m = ~ 0

4

*Thi d1}14 Xet cac s6 thuc a, b, c, d thoa

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): x 2 +y2-2x+2y-23=0 co tam 1(1;-1)

va R = 5; di~m B (c ;d) thuoc duong thang

f:: :3x-4 y +23=0 thi T= (a- c/ + ( b -d ) 2 =AIJ 2.

Ta co d(I,f:: )=6> R nen f:: khong c~t (C)

T dat gia tri nho nh~t khi doan A B ngan nhat.

Theo Thi du lZita co minA B , = 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)

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'P doan 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-ay 2 - 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 illgiac MAlB co dien tich bang 10

4.Cho duong tron (C):X2+y2 -8 x +6 y +21=0

va dirong thang d: X +Y-1 =0 Xac dinh toa

dQ cac dinh hinh vuong A BCD ngoai tiep duong tron (C), biSt rang diem A nam tren d

5. Cho dirong tron ( C ):r+y- 4 x-6y- 1 =O Tim toa dQ diem M thuoc duong thang :

d : 2 x - Y+3=0 sao cho MI = 2R , trong do I

la tam va RIa ban kinh cua duong tron (C)

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