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Nguyn Ph Khnh

606

Bi tp t luyn

Bi tp 1. Vit phng trnh ng trn ( )C , bit: a. i qua ( )A 3; 4 v cc hnh chiu ca A ln cc trc ta .

b. C tm nm trn ng trn ( ) ( )2 214

C : x 2 y5

+ = v tip xc vi hai ng

thng 1 : x y 0 = v 2 : x 7y 0 = .

c. i qua cc im H, M, N . Bit ( ) ( )A 0; 2 ,B 2; 2 , ( )C 4; 2 v H l chn ng cao k t B, M, N ln lt l trung im ca AB, AC.

d. Tip xc vi hai trc ta Ox, Oy ng thi tip xc ngoi vi ( )C :

( ) ( )2 2x 6 y 2 4 + = .

Bi tp 2. Vit phng trnh ng trn ( )C : a. C tm nm trn ng thng 4x 5y 3 0 = v tip xc vi cc ng thng:

2x 3y 10 0, = 3x 2y 5 0 + = .

b. Qua im ( )A 1; 5 tip xc vi cc ng thng 3x 4y 35 0,+ = 4x 3y 14 0+ + = .

c. Tip xc vi cc ng thng: 3x 4y 35 0,+ = 3x 4y 35 0, = x 1 0 = .

d. C tm M nm trn d : x y 3 0 + = , bn knh bng 2 ln bn knh ng trn

( ) 2 2C' : x y 2x 2y 1 0+ + = v tip xc ngoi vi ng trn ( )C' . e. Tip xc vi hai trc ta Ox,Oy ng thi tip xc ngoi vi ng trn

( )C' : ( ) ( )2 2x 6 y 2 4 + =

Bi tp 3. Vit phng trnh ng trn ( )C a. i qua 3 im A, B, ( )M 0;6 . Trong A, B l giao im 2 ng trn

( ) 2 21C : x y 2x 2y 18 0+ = v ( )2C : ( ) ( )2 2

x 1 y 2 8+ + = .

b. i qua hai im ( ) ( )A 2;1 , B 4; 3 v c tm thuc ng thng + =: x y 5 0 . c. i qua hai im ( ) ( )A 0; 5 ,B 2; 3 v c bn knh =R 10 . d. i qua hai im ( ) ( )A 1;0 ,B 2;0 v tip xc vi ng thng =d : x y 0 . e. i qua ( )A 1;1 ,O v tip xc vi + =d : x y 1 2 0 .

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Nguyn Ph Khnh

607

Bi tp 4. Trong mt phng to cc vung gc Oxy,

a. Cho im ( )A 0; 2 v ng thng d : x 2y 2 0 + = . Tm trn ng thng d hai im B,C sao cho tam gic ABC vung B v AB 2BC= .

b. Cho ng thng =d : x 3y 4 0 v ng trn ( ) 2 2C : x y 4y 0+ = . Tm M thuc d v N thuc ( )C sao cho chng i xng qua ( )A 3;1 .

c. Cho ng trn ( ) ( ) ( )2 2 25C : x 2 y 49

+ = v ng thng + =d : 5x 2y 11 0.

Tm im C trn d sao cho tam gic ABC c trng tm G nm trn ng trn

( )C bit ( ) ( )A 1; 2 ,B 3; 2 . d. Cho im ( )A 1;14 v ng trn ( )C c tm ( )I 1; 5 v bn knh R 13= . Vit phng trnh ng thng d i qua A ct ( )C ti M,N sao cho khong cch t M n AI bng mt na khong cch t N n AI .

e. Cho tam gic ABC c ng cao AH : x 3 3 0 = , phng trnh 2 ng

phn gic trong gc B v gc C ln lt l : x 3y 0 = v x 3y 6 0+ = . Vit

phng trnh cc cnh ca tam gic, bit bn knh ng trn ni tip tam gic ABC bng 3 .


a. Cho ng trn ( )C : ( ) ( )2 2x 1 y 1 4 + = v ng thng : =x 3y 6 0 . Tm ta im M nm trn , sao cho t M v c hai tip tuyn MA, MB ( A, B l tip im) tha ABM l tam gic vung.

b. Cho ng thng + =x yd 1: 0 v ng trn ( )C c phng trnh + + =2 2x y 2x 4y 0 . Tm im M thuc ng thng d sao cho t M k c

hai ng thng tip xc vi ng trn ti A v B , sao cho = 0AMB 60 .

c. Cho ng trn ( ) 2 2C : x y 1+ = . ng trn ( )C' tm ( )I 2; 2 ct ( )C ti hai im A, B sao cho =AB 2 . Vit phng trnh ng thng AB .

d. Cho hai im ( ) ( )A 2;1 ,B 0; 5 , ng trn ( ) ( )2 2x 1 y 3 5+ = v ng thng d : x 2y 1 0.+ + = T im M trn d k hai tip tuyn ME,MF n ( )C ( E,F l hai tip im). Bit ABEF l mt hnh thang, tnh di on EF.

e. Cho ng trn ( )C : 2 2x y 8x 2y 0+ = v im ( )A 9;6 . Vit phng trnh

ng thng qua A ct ( )C theo mt dy cung c di 4 3 .

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Nguyn Ph Khnh

608

Bi tp 6. Trong mt phng vi h ta Oxy,

a. Cho ng trn ( )C : ( ) ( )2 2x 1 y 1 10 + = . ng trn ( )C' tm ( )I ' 2; 5 ct ( )C ti hai im A,B sao cho AB 2 5= . Vit phng trnh ng thng AB .

b. Cho im ( )I 2;4 v hai ng thng 1d : 2x y 2 0, = 2d : 2x y 2 0+ = . Vit phng trnh ng trn tm I ct 1d ti hai im A, B v ct 2d ti hai im

C,D sao cho 16 5

AB CD5

+ = .

c. Cho tam gic ABC cn ti C, nh ( )B 3; 3 , ng trn ni tip tam gic

ABC c phng trnh: 2 2x y 2x 8 0+ = . Lp phng trnh cc cnh ca tam

gic ABC . Bit rng nh C c tung dng.

d. Cho im ( )M 2;1 v hai ng thng 1d : 2x y 7 0, + = 2d : x y 1 0+ + = . Vit phng trnh ng trn ( )C c tm nm trn 1d , i qua im M v ct 2d ti

hai im phn bit A, B sao cho AB 6 2= .

Bi tp 7. Trong mt phng vi h ta Oxy,

a. Cho ng trn ( )C : ( ) ( )2 2x 1 y 2 9 + + = v ng thng d : 3x 4y m 0 + = . Tm m trn d c duy nht mt im P m t c th k c hai tip tuyn PA, PB ti ( )C ( A, B l cc tip im) sao cho tam gic PAB u.

b. Cho tam gic ABC c ( )A 5; 2 , ( )B 3; 4 . Bit din tch tam gic ABC bng 8 v bn knh ng trn ngoi tip bng 2 5 . Tm ta im C c honh dng.

c. Cho tam gic ABC c nh A nm trn ng thng : x 2y 1 0, + + = ng

cao BH c phng trnh x 1 0,+ = ng thng BC i qua im ( )M 5;1 v tip xc vi ng trn ( ) 2 2C : x y 8+ = . Xc nh ta cc nh ca tam gic ABC bit cc nh B, C c tung m v on thng BC 7 2= .

d. Cho ng trn ( )C : ( )22x y 3 4+ = v mt ng trn ( )C ct ( )C ti hai im phn bit A,B. Gi s ng thng AB c phng trnh l x y 2 0,+ =

hy vit phng trnh ca ng trn ( )C c bn knh nh nht. e. Cho ng trn: ( )C : 2 2x y x 4y 2 0,+ = ( ) ( )A 3; 5 ,B 7; 3 . Tm M thuc

ng trn ( )C sao cho 2 2MA MB+ t gi tr nh nht.

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Nguyn Ph Khnh

609

Bi tp 8. Trong mt phng to cc vung gc Oxy

a. Cho ABC c 3 7

M ;2 2

v 1 5

N ;2 2

ln lt l trung im ca BC v AC .

Lp phng trnh ng trn ngoi tip ABC d : x 1

4y 2 t

3

=

= +

l ng phn

gic trong ca BAC .

b. cho ng trn ( )K : + =2 2x y 4 v hai im ( ) ( ) A 0;2 , B 0; 2 . Gi ( ) C,D C A,B l hai im thuc ( )K v i xng vi nhau qua trc tung. Bit

rng giao im E ca hai ng thng AC, BD nm trn ng trn

( ) + + =2 21K : x y 3x 4 0, hy tm ta ca E . c. Cho tam gic ABC vung ti A . nh ( )B 1;1 , ng thng AC c phng trnh: 4x 3y 32 0+ = , trn tia BC ly im M sao cho BC.BM 75= . Tm nh C

bit bn knh ca ng trn ngoi tip tam gic AMC bng 5 5

2.


a. Cho h ng cong ( )mC : ( )2 2x y 2mx 2 m 1 y 1 0+ + + = . nh m ( )mC l ng trn tm tp hp tm cc ng trn khi m thay i.

b. Cho ng trn ( )C : ( ) ( )2 2x 1 y 2 4 + + = . M l im di ng trn ng thng d : x y 1 0+ = . Chng minh rng t M k c hai tip tuyn 1 2MT , MT ti

( )C ( 1 2T , T l tip im ) v tm to im M , bit ng thng 1 2T T i qua im ( )A 1; 1 .

c. Vit phng trnh ng trn ( )C qua ( )A 1; 3 v tm ca ng trn ( )C' : 2 2x y 1+ = . Bit ( )C ct ( )C' ti B,C sao cho din tch tam gic ABC bng 2,7.

d. Cho ng thng d : 2x 4y 15 0+ = v hai ng trn c phng trnh ln lt

l ( ) ( ) ( )2 21C : x 1 y 2 9 , + = ( ) ( )2 2

2C : x 1 y 1+ + = . Tm M trn ( )1C v N trn ( )2C sao cho MN nhn ng thng d l ng trung trc v N c honh m.


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,Nguyn Ph Khnh

610

a. Cho ng trn ( )C : 2 2x y 4x 2y 3 0+ + = . T im ( )A 5; 3 k c 2 tip tuyn vi ng trn ( )C . Vit phng trnh ng thng i qua 2 tip im.

b. Cho ng trn ( )C : 2 2x y 4+ = v ng thng ( )d : x y 4 0+ + = . Tm im A thuc ( )d sao cho t A v c 2 tip tuyn tip xc ( )C ti M, N tho mn

din tch tam gic AMN bng 3 3 .

Bi tp 11. Trong mt phng to cc vung gc Oxy, cho ABC c ( )A 1;1 , trc tm ( )H 31;41 v tm ( )I 16; 18 ng trn ngoi tip ABC . Hy tm ta cc nh B,C .

Bi tp 12. Trong mt phng to cc vung gc Oxy, cho ng trn

( ) 2 2C : x y 2x 4y 0+ + = v ng thng d : x y 0 = . Tm ta cc im M trn ng thng d , bit t M k c hai tip tuyn MA,MB n ( )C ( A,B l

cc tip im) v ng thng AB to vi d mt gc vi 3

cos10

= .

Bi tp 13. Trong mt phng to cc vung gc Oxy, cho ng trn ( )C :

( ) ( )2 2x 1 y 1 9 + + = c tm I . Vit phng trnh ng thng i qua ( )M 6; 3 v ct ng trn ( )C ti hai im phn bit A, B sao cho tam gic

IAB c din tch bng 2 2 v AB 2> .


( ) 2 2C : x y 2x 4y 4 0+ + = c tm I v ng thng : + + =2x my 1 2 0 . Tm m din tch tam gic IAB l ln nht.


( ) ( ) ( )2 2C : x 1 y 1 25 + + = v ( )M 7; 3 . Vip phng trnh ng thng qua M ct ( )C ti A, B sao cho MA 3MB= .


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,Nguyn Ph Khnh

611

a. Cho ng trn ( )C c phng trnh : + + =2 2x y 2x 6y 6 0 v im ( )M 3;1 . Gi 1 2T ,T l cc tip im ca cc tip tuyn k t M n ( )C . Vit

phng trnh ng thng i qua 1 2T ,T .

b. Cho ng trn ( )C : 2 2x y 4x 2y 15 0+ + = Gi I l tm ng trn ( )C . ng thng i qua ( )M 1; 3 ct ( )C ti hai im A v B . Vit phng trnh ng thng bit tam gic IAB c din tch bng 8 v cnh AB l cnh ln nht.

Bi tp 17. Trong mt phng to cc vung gc Oxy, cho tam gic ABC c

trc tm H . Bit ng trn ngoi tip tam gic HBC l 2 2x y x 5y 4 0+ + = ,

H thuc ng thng : 3x y 4 0 = , trung im AB l ( )M 2; 3 . Xc nh to cc nh ca tam gic.

Bi tp 18. Trong mt phng to cc vung gc Oxy, cho im ( )A 1;0 v

cc ng trn ( )C : 2 2x y 2+ = v ( ) 2 2C' : x y 5+ = . Tm ta cc im B v C ln lt nm trn cc ng trn ( )C v ( )C' tam gic ABC c din tch ln nht.


( ) ( ) ( )2 2C : x 1 y 2 25 + = . T ( )E 6; 2 v hai tip tuyn EA, EB (A, B l tip im) n (C). Vit phng trnh ng thng AB .


( ) ( )2 2C : x 1 y 2 + = v hai im ( )A 1; 1 , ( )B 2; 2 . Tm ta im M thuc ng

trn ( )C sao cho din tch tam gic MAB bng 12

.


( )C : ( ) ( )2 2x 2 y 1 10 + = . Tm ta cc nh ca hnh vung MNPQ, bit M trng vi tm ca ng trn ( )C , hai nh N, Q thuc ng trn ( )C , ng thng PQ i qua E( 3;6) v Qx 0> .

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,Nguyn Ph Khnh

612

Bi tp 22. Trong mt phng to cc vung gc Oxy, cho ng thng

: x + y + 2 = 0 v ng trn ( )C : 2 2x y 4x 2y 0+ = . Gi I l tm v M thuc ng thng . Qua M k tip tuyn MA,MB . Tm M sao cho din tch t gic MAIB bng 10 .

Bi tp 23. Trong mt phng to cc vung gc Oxy, cho ng trn ( )C :

( ) ( )2 2x 1 y 2 25 + = . a. Vit phng trnh tip tuyn ca ( )C : Ti im ( )M 4;6 Xut pht t im ( )N 6;1 b. T ( )E 6; 3 v hai tip tuyn EA,EB ( A,B l tip im) n ( )C . Vit

phng trnh ng thng AB .

Bi tp 24. Trong mt phng to cc vung gc Oxy, cho tam gic ABC c

nh ( )A 3; 7 , trc tm l ( )H 3; 1 , tm ng trn ngoi tip l ( )I 2;0 . Xc nh to nh C , bit C c honh dng.


a. Cho hai ng thng 1d : 3x y 0+ = v 2d : 3x y 0 = . Gi ( )T l ng trn tip xc vi 1d ti A , ct 2d ti hai im B v C sao cho tam gic ABC vung

ti B . Vit phng trnh ca ( )T , bit tam gic ABC c din tch bng 32

v

im A c honh dng.

b. Cho ng trn ( ) + + =2 2C : x y 2x 4y 0 v ng thng =d : x y 0 . Tm ta cc im M trn ng thng d , bit t M k c hai tip tuyn MA, MB

n ( )C ( A, B l cc tip im) v khong cch t im ( )N 1; 1 n AB bng 3

5.

Bi tp 26. Trong mt phng to cc vung gc Oxy, cho cho im ( )A 1; 4 .

Tm hai im M,N ln lt nm trn hai ng trn

( ) ( ) ( )2 21C : x 2 y 5 13 + = v ( )2C : ( ) ( )2 2

x 1 y 2 25 + = sao cho tam gic

MAN vung cn ti A .

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Nguyn Ph Khnh

613


a. Cho cc ng trn ( ) ( )2 211

C : x 1 y2

+ = v ( )2C : ( ) ( )2 2

x 2 y 2 2 + = . Vit

phng trnh ng thng d tip xc vi ng trn ( )1C v ct ng trn

( )2C theo dy cung c di 2 2 .

b. Cho ng trn ( )C : ( ) ( )2 2x 1 y 1 9 + + = c tm I . Vit phng trnh ng thng i qua ( )M 6; 3 v ct ng trn ( )C ti hai im phn bit A, B sao cho tam gic IAB c din tch bng 2 2 v AB 2.>

c. Cho ng trn ( )C : 2 2x y 4x 4y 1 0+ = ng thng d : y mx m 1= + . ng thng d ct ( )C ti hai im A, B . Tip tuyn ti A v B ct nhau ti P . Xc nh cc gi tr ca m bit P thuc ng thng d' : x 3y 9 0+ + = .

Bi tp 28. Trong mt phng Oxy, cho ng trn ( )C : ( ) ( )2 2x 1 y 2 5 + = . a. Vit phng trnh ng thng d i qua im ( )M 3; 1 v ct ng trn ( )C ti hai im A,B sao cho AB 2.= b. Vit phng trnh ng thng 1d i qua ( )N 2;1 sao cho 1d ct ng trn ( )C ti hai im C, D c di nh nht.

Bi tp 29. Trong mt phng ta Oxy,

a. Cho hnh vung ABCD, c cnh AB i qua im ( )M 3; 2 , v Ax 0> . Tm ta cc nh ca hnh vung ABCD khi ng trn

( ) ( ) ( )2 2C : x 2 y 3 10 + = ni tip ABCD .

b. Cho tam gic ABC, c ( )A 2, 2 , ( )B 4,0 , ( )C 3; 2 1 v ( )C l ng trn ngoi tip tam gic. ng thng d c phng trnh 4x y 4 0+ = . Tm trn d

im M sao cho tip tuyn qua M tip xc vi ( )C ti N tha mn NABS t gi tr ln nht?

c. Cho ng trn ( )C : ( ) ( )2 2x 1 y 2 1 + + = v ng thng ( ) : 2x y 1 0 + = . Tm im A thuc ng thng ( ) sao cho t A k c cc tip tuyn

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,Nguyn Ph Khnh

614

AB, AC ( B,C l cc tip im ) n ng trn ( )C ng thi din tch tam gic ABC bng 2,7 .

d. Cho ng trn ( )C : 2 2x y 2x 4y 4 0+ = c tm I v im ( )M 3;0 . Vit phng trnh ng thng , bit ct ( )C ti hai im phn bit A ,B sao cho t gic ABIM l hnh bnh hnh.


a. Cho ng trn ( ) ( ) ( )2 2C : x 4 y 6 5. + = im ( ) ( )A 2; 5 ,B 6; 5 nm trn ( )C . nh C ca tam gic ABC di ng trn ng trn ( )C . Tm ta trc tm H ca tam gic ABC bit H nm trn ng thng ( )d : x y 1 0 + = .

b. Cho 2 ng trn ( ) 2 2C : x y 9+ = v ( )C' : 2 2x y 18x 6y 65 0+ + = . T im M thuc ( )C' k 2 tip tuyn vi ( )C , gi A,B l cc tip im. Tm ta im M bit AB 4,8= .

c. Cho tam gic u ABC . ng trn ( )C ni tip tam gic ABC c phng

trnh l ( ) ( )2 2x 1 y 2 5 + = , ng thng BC i qua im 7M ;2 .2

Xc nh

ta im A .

d. Cho 2 ng trn ( ) 2 21C : x y 13+ = v ( ) ( )2 2

2C : x 6 y 25 + = . Gi A l giao

im ca ( )1C v ( )2C vi Ay 0< . Vit phng trnh ng thng i qua A v ct ( )1C , ( )2C theo 2 dy cung c di bng nhau.


a. Cho ng trn ( )C : 2 2 2x y 2x 2my m 24 0+ + = c tm I v ng thng : mx 4y 0.+ = Tm m bit ng thng ct ng trn ( )C ti 2 im phn bit A,B tho mn din tch IAB 12= .

b. Cho tam gic ABC c trc tm H thuc ng thng 3x y 4 0, = bit

ng trn ngoi tip tam gic HBC c phng trnh : 2 2x y x 5y 4 0 ,+ + = trung im cnh AB l ( )M 2; 3 . Tm ta 3 nh

tam gic ?.

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,Nguyn Ph Khnh

615

c. Cho ng trn ( )C : 2 2x y 2x 4y 2 0+ + + = .Gi ( )C' l ng trn c tm

( )I 5;1 v ct ng trn ( )C ti 2 im M,N sao cho MN 5= .Hy vit phng trnh ca ( )C' . d. Cho tam gic ABC c nh ( )A 1;1 , trc tm ( )H 1; 3 , tm ng trn ngoi tip ( )I 3; 3 . Xc nh ta cc nh B, C, bit rng B Cx x .<


a. Cho ng thng ( )d : x y 1 0 + = v ng trn ( ) 2 2C : x y 2x 4y 4 0+ + = . Tm im M thuc ng thng ( )d sao cho qua M k c cc tip tuyn MA,MB n ng trn vi A, B l cc tip im ng thi khong cch t

im 1

N ;12

n ng thng i qua AB l ln nht.

b. Cho ng trn ( )C : ( ) ( )2 2x 1 y 2 16+ + = v ng thng c phng trnh 3x 4y 5 0.+ = Vit phng trnh ng trn ( )C c bn knh bng 1 tip xc ngoi vi ( )C sao cho khong cch t tm I ca n n l ln nht

c. Cho tam gic ABC ni tip ng trn ( ) ( ) ( )2 2C : x 1 y 1 10 + = . im ( )M 0; 2 l trung im cnh BC v din tch tam gic ABC bng 12 . Tm ta

cc nh ca tam gic ABC.

d. Cho 3 im ( )M 2, 1 , ( )N 3;2 , ( )P 3;4 v ng trn ( )C :

( ) ( )2 2x 1 y 2 25 + + = . Gi ( )d qua M ct ( )C ti A,B sao cho IABS t gi tr ln nht. Hy xc nh ta ( )E d sao cho 2 2EN EP+ t gi tr nh nht, vi I l tm ng trn

Hng dn gii

Bi tp 1.a. Gi A1, A2 ln lt l hnh chiu ca A ln hai trc Ox, Oy

suy ra ( ) ( )1 2A 3;0 , A 0; 4

Gi s ( ) 2 2C : x y 2ax 2by c 0+ + = .

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,Nguyn Ph Khnh

616

Do ( )1 2A,A ,A C nn ta c h:

3a6a 8b c 25 2

6a c 9 b 2

8b c 16 c 0

= + =

+ = =

+ = =

.

Vy phng trnh ( )C : 2 2x y 3x 4y 0+ = .

b. Gi ( )I a; b l tm ca ng trn (C), v ( )1I C nn: ( )2 2 4a 2 b

5 + = ( )

Do ( )C tip xc vi hai ng thng 1 2, nn ( ) ( )1 2d I, d I, =

a b a 7b

b 2a,a 2b2 5 2

= = =

b = 2a thay vo ( ) ta c c: ( )2 2 24 16a 2 4a 5a 4a 05 5

+ = + = phng

trnh ny v nghim

a = 2b thay vo ( ) ta c: ( )2 2 4 4 82b 2 b b ,a5 5 5

+ = = = .

Suy ra ( )14

R d I,5 2

= = . Vy phng trnh ( )2 2

8 4 8C : x y

5 5 25

+ =

.

c. Ta c ( ) ( ) ( )M 1;0 ,N 1; 2 ,AC 4; 4 =

. Gi ( )H x; y , ta c:

( ) ( )

( )( )

4 x 2 4 y 2 0 x 1BH AC H 1;1y 14x 4 y 2 0H AC

+ + = =

=+ =

Gi s phng trnh ng trn: 2 2x y ax by c 0+ + + + = .

Ba im M, N, H thuc ng trn nn ta c h phng trnh :

a c 1 a 1

a 2b c 5 b 1

a b c 2 c 2

= =

+ = = + + = =

.

Phng trnh ng trn: 2 2x y x y 2 0+ + = .

d. ng trn ( )C c tm ( )I 6; 2 , bn knh R = 2.

Gi ( ) ( ) ( )2 2 2C' : x a y b R' + = th ( )C' c tm ( )I ' a; b , bn knh R. V ( )C' tip xc vi Ox, Oy nn suy ra

( ) ( ) a bd I',Ox d I',Oy a b R 'a b

== = = =

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,Nguyn Ph Khnh

617

Hn na (C) tip xc vi Ox, Oy v tip xc ngoi vi (C) nn (C) nm bn phi trc Oy, do a > 0.

TH1: ( ) ( ) ( )2 2 2a b R C' : x a y a a= = + = V ( )C' tip xc ngoi vi ( )C nn: II ' R R'= +

( ) ( )2 2a 6 a 2 2 a + = + a 2 = hoc a 18= Trng hp ny c 2 ng trn l :

( ) ( ) ( )2 2'1C : x 2 y 2 4 + = v ( ) ( ) ( )2 2' 22C : x 18 y 18 18 + = . TH2: ( ) ( ) ( )2 2 2a b R C' : x a y a a= = + + = Tng t nh trng hp 1, ta c : II ' R R ' a 6= + =

Vy trng hp ny c 1 ng trn l ( ) ( ) ( )2 2'3C : x 6 y 6 36 + = . Tm li , c 3 ng trn tha cn tm l :

( ) ( ) ( ) ( )2 2 2 2 2x 2 y 2 4, x 18 y 18 18 + = + = v ( ) ( )2 2x 6 y 6 36 + = .

Bi tp 2.a. ( ) ( )2 2 81x 2 y 1 ,13

+ = ( ) ( )2 2 25x 8 y 713

+ + + =

b. ( ) ( )2 2x 2 y 1 25, + = 2 2 2

202 349 185x y

49 49 49

+ + =

c. 2 2 2

35 40 32x y ,

3 3 3

+ =

( )2 2x 5 y 16, + = ( )2 2x 15 y 256+ + =

d. ng trn ( )C' c tm ( )I ' 1;1 , bn knh R ' 1= . Gi I l tm v R l bn knh ca ng trn ( )C , ta c R 2R' 2= = v

( )I d I a;a 3 + V ( )C v ( )C' tip xc ngoi vi nhau nn II ' R R' 3= + =

( ) ( )2 2 2a 1 a 2 9 a a 2 0 a 1 + + = + = = hoc a 2= .

( ) ( ) ( ) ( )2 2a 1 I 1; 4 C : x 1 y 4 4= + =

( ) ( ) ( ) ( )2 2a 2 I 2;1 C : x 2 y 1 4= + + = . e. ng trn ( )C' c tm ( )I ' 6; 2 , bn knh R ' 2= .

Gi ( ) ( ) ( )2 2 2C : x a y b R + = th ( )C c tm ( )I a; b , bn knh R .

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,Nguyn Ph Khnh

618

V ( )C tip xc vi Ox,Oy nn suy ra ( ) ( )d I,Ox d I,Oy a b R' a b= = = = hoc a b= Hn na ( )C v ( )C' tip xc ngoi v nm bn phi trc Oy , do a 0> .

TH1: ( ) ( ) ( )2 2 2a b R C : x a y a a= = + =

V ( )C v ( )C' tip xc ngoi nn : ( ) ( )2 2II ' R R ' a 6 a 2 2 a= + + = + a 2 = hoc a 18=

Trng hp ny c 2 ng trn l :

( ) ( ) ( )2 21C : x 2 y 2 4 + = v ( ) ( ) ( )2 2 2

2C : x 18 y 18 18 + = .

TH2: ( ) ( ) ( )2 2 2a b R C : x a y a a= = + + =

Tng t nh trng hp 1, ( ) ( )2 2II ' R R ' a 6 a 2 2 a= + + + = + a 6 =

Vy, trng hp ny c 1 ng trn l ( ) ( ) ( )2 23C : x 6 y 6 36 + = . Tm li , c 3 ng trn tha cn tm l :

( ) ( ) ( ) ( )2 2 2 2 2x 2 y 2 4, x 18 y 18 18 + = + = v ( ) ( )2 2x 6 y 6 36 + = .

Bi tp 3.a. Ta giao im ca ( )1C v ( )2C l nghim ca h:

( ) ( )

2 2 2 2

2 2 2 2

x y 2x 2y 18 0 x y 2x 2y 18 0

x y 2x 4y 3 0x 1 y 2 8

+ = + =

+ + =+ + =

2 2x y 2x 2y 18 0

152x y

2

+ =

+ =

( )2

15y 2x

293

5x 24x 0 4

= +

+ + =

Gi 1 2x ,x l hai nghim ca ( ) , suy ra 1 115

A x ; 2x ,2

+

2 2

15B x ; 2x

2

+

.

Suy ra ( ) ( )2 22 1 2 1 2 1 2111

AB 5 x x 5 x x 4x x5

= = + =

Gi M l trung im AB , suy ra

1 2M

M 1 2

x x 12x 12 272 5 M ;

5 1015 27y x x

2 10

+= =

= + + =

.

Phng trnh ng thng AB : 4x 2y 15 0 + =

Phng trnh ng trung trc ca on AB : x 2y 3 0+ = .

Gi I l tm ca ng trn ( )C , suy ra ( )I I 2a 3; a +

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,Nguyn Ph Khnh

619

Mt khc: ( ) ( ) ( ) ( )22 2 22 2 10a 27AB 111d I,AB IM 2a 3 a 6

4 20 20

++ = + = + + +

a 1 =

Suy ra ( )I 5; 1 , bn knh R IM 5 2= = . Vy, phng trnh ca ( )C : 2 2(x 5) (y 1) 74 + + = .

b. Gi ( ) 2 2C : x y 2ax 2by c 0+ + =

V ( )C i qua A,B nn ta c: + = + =

4a 2b c 5

8a 6b c 25 ( )1

Mt khc: ( )C c tm ( )I a; b thuc + = + =: x y 5 0 a b 5 0 ( )2

T ( )1 v ( )2 ta c h : + = = + = =

+ = =

4a 2b c 5 a 0

8a 6b c 25 b 5

a b 5 0 c 5

Vy phng trnh ( ) 2 2C : x y 10y 5 0+ + = . c. Gi ( )I a; b l tm ca ng trn ( )C .

Ta c phng trnh ( ) ( ) ( )2 2C : x a y b 10 + = . Do ( )A,B C nn ta c h

= =+ + = + + = = + =+ + =

=

2 2 2 2

2 2

a 1

b 2a b 10b 15 0 a b 10b 15 0a 34a 4b 12 0a b 4a 6b 3 0b 6

Vy c hai ng trn tha yu cu bi ton l:

( ) ( )+ + =2 2x 1 y 2 10 v ( ) ( ) + =2 2x 3 y 6 10 . d. Gi s ng trn ( )C c phng trnh l + + =2 2: x y 2ax 2by c 0

Do ( )A,B C nn ta c: + = + =

1 2a c 0

4 4a c 0.

( )C tip xc vi d nn suy ra ( )( )

= = + 2 2a b

d I, d R a b c2

( ) + + =2 2a b 2ab 2c 0 3

T ( ) ( ) ( )1 , 2 , 3 ta c 3 1a ,b ,c 22 2

= = = hoc 3 7

a ,b ,c 22 2

= = = .

Vy, c hai ng trn tha yu cu bi ton l:

+ + =2 2x y 3x y 2 0 v + + + =2 2x y 3x 7y 2 0 .

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,Nguyn Ph Khnh

620

e. Gi ( ) 2 2C : x y 2ax 2by c 0+ + = l ng trn cn tm

V ( )C i qua = =

c 0O,A

a b 1 ( )1

Do ( )C tip xc vi ( )( ) + = =d : x y 1 2 0 d I, d R

( ) +

= + 2 2a b 1 2

a b c 22

T ( )1 v ( )2 gii h thu c = = =a 0,b 1,c 0 hoc = = =a 1, b 0,c 0 . Vy c hai ng trn tha mn l : + =2 2x y 2y 0 v + =2 2x y 2x 0 .

Bi tp 4.a. Ta c AB d nn AB c phng trnh : 2x y 2 0+ = .

Ta im B l nghim ca h : x 2y 2 0 2 6

B ;2x y 2 0 5 5

+ = + =

.

Suy ra 2 5 AB 5

AB BC5 2 5

= = = .

Phng trnh ng trn tm B, bn knh 5

BC5

= l:

2 22 6 1

x y5 5 5

+ =

.

Ta im C l nghim ca h : 2 2x 2y 2 0 x 0, y 1

4 72 6 1 x ,yx y5 55 5 5

+ = = = = = + =

Vy, 2 6

B ;5 5

, ( )C 0;1 hoc 2 6B ;5 5

, 4 7

C ;5 5

tha yu cu bi ton .

b. V ( )M d M 3m 4; m + . Do N i xng vi M qua A nn ( )N 2 3m; 2 m

V ( )N C nn ( ) ( ) ( )2 2 22 3m 2 m 4 2 m 0 10m 12m 0 + = = 6

m 0,m5

= =

Vy c hai cp im tha yu cu bi ton: ( ) ( )M 4;0 ,N 2; 2 v

38 6 8 4M ; , N ;

5 5 5 5.

c. Ta c: C d nn ta c ta

11 5cC c;

2

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,Nguyn Ph Khnh

621

Ta trong tm 4 11 5;

3 6

+

cGc

. Do G nm trn ng trn ( )C nn ta c

phng trnh: ( ) ( )2 2 2c 2 5c 13 25 29c 114c 85 0

9 36 9

++ = + + = c 1, =

85c

29= .

Vy c hai im C tha yu cu bi ton l: ( )

1 285 372

C 1;8 , C ;29 29

.

d. Cch 1: ( )2 2

A/ CP AM.AN AI R 466 0= = = >

, suy ra A nm ngoi ng

trn. Hn na ( )2 2

A/ CP 2AM 2MN 466 MN 233= = = = .

Bi ton tr thnh: V it phng trnh ng thng qua A ct ng trn ( )C theo dy cung MN 233= .

Cch 2: Gi s ( )M x; y v M thuc ng trn nn ta c:

( ) ( )2 2x 1 y 5 169 + + = V M l trung im ca AN nn ta c: ( )N 2x 1; 2y 14+

im N thuc ng trn nn ta c: ( ) ( )2 22x 2y 9 169+ = .

Ta c h: ( ) ( )( ) ( )

2 2

2 2

x 1 y 5 169

2x 2y 9 169

+ + = + =

e. ( )I 3; 3 l ta tm ng trn ni tip tam gic ABC . Vit phng trnh BC i qua im ( )B b;c v vung gc vi AH , ta B cn tm tha B d : x 3y 0 = v ( )d I; BC r 3= =

Bi tp 5.a. ng trn ( )C c tm I(1; 1), bn knh R = 2. V ABM vung v IM l ng phn

gic ca gc AMB nn 0AMI 45= Trong tam gic vung IAM , ta c:

IM 2 2= , suy ra M thuc ng trn

tm I bn knh R ' 2 2= . Mt khc M nn M l giao im

B M

AI

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,Nguyn Ph Khnh

622

ca v ( )I,R ' . Suy ra ta ca M l nghim ca h :

( ) ( ) ( ) ( )2 2 2 2x 3y 6 0 x 3y 6

x 1 y 1 8 3y 5 y 1 8

= = +

+ = + + =

2

y 1,x 3x 3y 69 3

y ,x5y 14y 9 05 5

= = = + = =+ + =

Vy, c hai im ( )1 23 9

M 3; 1 ,M ;5 5

tha yu cu bi ton.

b. ng trn c tm ( )I 1; 2 v bn knh: =R 5 . Tam gic AMB l tam gic u v MI l phn gic gc AMB nn = 0IMA 30

Do : = = =20

IAMI 2 5 IM 20

sin 30

Do M d nn suy ra ( )0 0M x ; x 1+

Khi ta c: ( ) ( )2 22 20 0 0 0 0MI x 1 x 1 20 x 9 x 3, x 3= + + = = = = Vy c 2 im M tha mn iu kin bi ton: ( ) ( )M 3; 4 ,M 3; 2 c. Ta c + = = 2 2 2OA OB AB 2 OAB vung ti O . Mt khc OI l ng trung trc ca on thng AB nn A,B thuc cc trc to . Vy:

( ) ( )A 1; 0 ,B 0;1 , phng trnh ng thng + =AB : x y 1 0 ( ) ( )A 1;0 ,B 0; 1 , phng trnh ng thng + + =AB : x y 1 0 .

e. Ta tm ng trn l ( )I 4;1 ;bn knh R 17= Gi l ng thng qua A v ct ng trn ti M,N phng trnh ca

c dng l: ( )y k x 9 6= + .

Gi H l trung im MN ,ta c: ( )2

2 MNIH R 17 12 5 d I;2

= = = =

2

k 2 y 2x 124k 1 9k 65 1 1 21

k y xk 1 2 2 2

= = + = = = ++

Bi tp 6. a. ng trn ( )C c tm ( )I 1;1 , bn knh R 10= . di II ' 3 5= Gi H l giao im ca II ' v AB , suy ra H l trung im AB nn AH 5= .

Do II ' AB nn ta c: 2 2IH IA AH 5= =

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,Nguyn Ph Khnh

623

TH 1: H thuc on II '

I 'H 2 5 = 1

IH II'3

=

( ) ( )H HIH x 1; y 1 , II ' 3; 6= =

Ta c: H H

H H

x 1 1 x 0

y 1 2 y 1

= =

= =

( )H 0; 1 . V AB i qua H v

nhn ( )1n II ' 1; 23

= =

lm VTPT

H

B

A

I

I'

Phng trnh AB l: x 2y 2 0+ + = .

TH 2: H khng nm trong on II ' , suy ra 1

I 'H 4 5 IH II '4

= =

Hay H H

H H

3 1x 1 x 1 14 4 H ;

3 1 4 2y 1 y

2 2

= =

= =

.

Phng trnh 3

AB : x 2y 04

+ + = .

b. Gi R l bn knh ng trn cn tm v F,G ln lt l hnh chiu vung

gc ca I trn 1d v 2d . D thy 2 5 6 5

IF , IG5 5

= = .

Li c: 2 2 2 2 2 24 36

FB R IF R , GD R IG R5 5

= = = =

Theo bi ton: ( )16 5 16 5AB CD 2 FB GD R5 5

+ = + =

d. K IH AB AH 3 2 = . 1I d nn ( )I x;7 2x+

Li c: R IM IA= = v tam gic IAH vung ti H nn c: 2 2 2IM IH AH= +

Trong ( )18 3x

IH d I;d2

+= =

Bi tp 7. a. ng trn (C) c tm v bn knh ln lt l: ( )I 1; 2 ;R 3 = .

Do tam gic PAB u nn 0API 30= IP 2IA 2R 6= = = . Suy ra P thuc vo ng trn (C) c tm I v bn knh R = 6.

d

300

B

I

A

P

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,Nguyn Ph Khnh

624

M P d nn P chnh l giao im ca ng thng d v ng trn ( )C' Suy ra trn d c duy nht im P tha mm yu cu bi ton khi v ch khi ng thng d tip xc vi ng trn ( )C' ti P, hay l

( )d I,d 6= m 19,m 41 = = . b. Ta c phng trnh AB : x y 7 0+ + =

Gi M l trung im AB, ta ( )M 4; 3 . . Phng trnh ng trung trc AB l: x y 1 0 + = .

Gi ( )C c;d v c 0> l ta cn tm. Theo bi ton, ta c: ( )AB.d C; AB 16= c d 7 8 + + = ( )1

Gi I l tm ng trn ngoi tip, suy ra: ( )I x; x 1+ v IA R 2 5= = 2x 8x 7 0 x 7 + + = = hoc x 1=

TH1: ( )x 7 I 7; 6= .

Phng trnh ng trn ( )C ngoi tip ABC : ( ) ( )2 2x 7 y 6 20+ + + =

( )C C nn c : ( ) ( )2 2c 7 d 6 20+ + + = , trng hp ny khng tha v c 0> TH2: ( )x 1 I 1;0= .

Phng trnh ng trn ( )C ngoi tip ABC : ( )2 2x 1 y 20+ + =

( )C C nn c : ( )2 2c 7 d 20+ + = ( )2 Ta im C l nghim ca h phng trnh ( )1 v ( )2

( ) ( )22 22c d 7 8 c d 1 c d 15 c 3

d 2c 7 d 20c 7 d 20

+ + = + = + = =

= + + =+ + =

Vy, ta C cn tm l ( )C 3; 2 . c. Gi im ( )0B 1; y , t vit c phng trnh ng thng BC l: ( )( ) ( )0y 1 x 5 6 y 1 0 + =

BC tip xc vi ( )C ( )d I;BC R = ( )

( )0

20

5 y 1 62 2

y 1 36

=

+

20 017y 26y 295 0 + = , kt hp BC 7 2= , ta tm c 0y 5=

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,Nguyn Ph Khnh

625

Vy, ( ) ( ) ( )B 1; 5 C 8; 12 , A 23; 12 d. ng trn ( )C ct ( )C ti hai im phn bit A, B nn AB l 1 dy cung ca ng trn ( )C , khi ng knh nh nht ca ng trn ( )C chnh l AB .

e. ( )C c tm 1I ; 22

. Hn na: 2

2 2 2ABMA MB 2MN2

+ = +

2 2MA MB+ nh nht khi MN nh nht, iu ny xy ra khi M l giao im ca ng thng IN v ( )C ( )M 2;0 .

Bi tp 8. Gi N' l im i xng ca N qua phn gic trong gc A 3 5N' ;2 2

Phng trnh AB i qua N' nhn vect ch phng MN

c phng trnh:

x y 1 0 + = . Ta A tha h ( )

x 1

4y 2 t A 1; 2

3x y 1 0

=

= +

+ =

. T y, tm c

( ) ( )B 3; 4 ,C 0; 3 . ng trn: 2 2

3 7 5x y

2 2 2

+ =

b. V C,D thuc ng trn ( )K m li i xng vi nhau qua trc tung nn ta 2 im c dng l: ( ) ( ) C a;b , D a;b ( )a,b 0 Ta c: + =2 2a b 4 ( )1 . Phng trnh ng thng: ( ) ( ) =AC : b 2 x a y 2 0, ( ) ( )+ + + =BD: b 2 x a y 2 0

Ta im E l nghim ca h: ( ) ( )( ) ( )

= =

+ + + = =

2axb 2 x a y 2 0 b

4b 2 x a y 2 0y

b

V ( ) 1E K nn c:

+ =

2

2

a 16 a4 6 4 0

b bb + =2 24a 4b 6ab 16 0 ( )2

T ( )1 v ( )2 suy ra = =28a 6ab 0 4a 3b c. Cch 1: To nh ( )A 5; 4 . Gi E l giao im ca ng trn ngoi tip ca

tam gic AMC vi BA th ta c BA.BE BM.BC 75= =

( v M nm trn tia BC ),

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Nguyn Ph Khnh

626

tm c to ca E l ( )E 13; 10 . Tam gic AEC vung ti A nn C l giao

ca ng trn tm E , bn knh r 5 5= vi ng thng AC . To ca C l

nghim ca h: ( ) ( ) ( )22 24x 3y 32 0

x 13 y 10 5 5

+ =

+ =

( )C 8;0 hoc ( )C 2;8 .

Cch 2: Gi I l tm ng trn ngoi tip tam gic AMC.

V B nm ngoi ng trn ( )I nn ta c: BM.BC BM.BC=

( )1

Ta c: ( )( )2 2

B/ IP BM.BC BI R= =

vi 5 2

R2

= ( )2

T ( )1 v ( )2 suy ra 2 2 2 425BI R 75 BI4

= =

Phng trnh AB : 3x 4y 1 0 + = v tm c ( )A 5; 4

Gi ( )I x; y ta c: 2

2

125AI

4425

BI4

=

=

13I ; 2

2

7I ;6

2

Phng trnh ng trung trc IN ca AC AC IN N = ( )C 8;0 hoc ( )C 2;8

Cch 3: T M dng MK BC, ( )K AB Gi I l trung im KC I l tm ng trn ngoi tip tam gic AMC ( Do t gic AKCM ni tip )

Ta c ABC ng dng MBK nn : AB BC

AB.BK MB.BC 75MB BK

= = =

Phng trnh ng thng AB qua im ( )B 1;1 v c VTPT ( )3; 4 : 3x 4y 1 0 + = .

V A l giao im ca AB v AC nn ( )A 5;4 AB 5 BK 15 = = AK 10 = 2 2AC 4R AK 5 = =

Gi 32 4t

C t; AC3

v ( )

22 20 4t

AC 5 t 5 253

= + =

t 2 = hoc t 8=

Vy, ( )C 8;0 hoc ( )C 2;8

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Nguyn Ph Khnh

627

Bi tp 9. a. ( )2 2x y 2mx 2 m 1 y 1 0+ + + = c a m, b m 1, c 1= = =

( )mC l ng trn th ( )22 2 2a b c m m 1 1 0+ = + >

22m 2m 0 m 0 > < hoc m 1> .

Tm x m

I : x y 1 0y m 1

= + + =

= . iu kin:

m 0 x 0

m 1 x 1

< > > <

Vy, tp hp tm I l ng thng x y 1 0+ + = vi x 0

x 1

> <

b. ng trn ( )C c tm ( )I 1; 2 bn knh r 2=

M nm trn d nn ( )M m; m 1+ ( ) ( ) ( )2 2 2IM m 1 m 3 2 m 1 8 = + + = + + V IM 2> nn M nm ngoi ( )C , do qua M k c 2 tip tuyn ti ( )C .

Gi J l trung im IM nn ta im m 1 m 1

J ;2 2

+

. ng trn ( )T ng

knh IM c tm J bn knh 1IM

r2

= c phng trnh ( )T l:

( )22 2 2 m 1 8m 1 m 1x y

2 2 4

+ + + + =

T M k c 2 tip tuyn 1 2MT ,MT n ( )C , nn 1 2T , T l hai giao im ca ( )C v ( )T .

Ta 1 2T , T tha mn h: ( ) ( )

( )

2 2

22 2

x 1 y 2 4

2 m 1 8m 1 m 1x y

2 2 4

+ + = + + + + =

( )1 2T T : ( ) ( )m 1 x m 3 y m 3 0+ + + + = V ( )1 2A T T nn c: m 1 m 3 m 3 0+ + = ( )m 1 M 1; 2 =

c. Gi ( )I a; b l ta tm ca ( )C c bn knh ( ) ( )2 2R a 1 b 3= +

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Nguyn Ph Khnh

628

( )C ct ( )C' ti B,C nn c h: ( ) ( ) ( ) ( )

2 2

2 2 2 2

x y 1

x a y b a 1 b 3

+ =

+ = +

9ax by a 3b 0

2 + + = ( )BC ( )

2 2

9d A; BC

2 a b =

+

Hn na, ( )

ABC2S 6BC OI5d A; BC

= =

Gi H l giao im ca OI v BC BC 3

BH OI2 5

= =

Hn na: 2BOI1 3

IB IO S IK.OB IK OI2 5

= = = vi K l trung im OB

Xt IKO vung ti K ta c : 2 2 2 25

KI OK OI OI2

+ = = hoc 25

OI18

=

Nu 2 25 5

OI AI2 2

= = , ta c h:

( ) ( )

2 2

2 2

25a b

425

a 1 b 34

+ =

+ =

d. Nu ta gi ( )M a; b v ( )N c;d th ta c bn n s cn phi tm ra .

( )d l ng trung trc MN nn c ( )

dMN.n 0

I d

=

, trong I l trung im MN .

Hn na ( ) ( )1M a; b C v ( ) ( )2N c;d C

Ta c h:

( ) ( )( )( ) ( )

( )

( ) ( )( )

2 22 2

2 22 2

a 1 b 2 9a 1 b 2 9

c 1 d 1c 1 d 1

15a 2d2 a c 4 b d 0 2

15a c 2 b d 15 0 c 2b2

+ = + = + + = + + = = + =

+ + + = =

Bi tp 10. a. ng trn ( )C c tm ( )I 2; 1 v bn knh R 2 2= . Gi 1 2T ,T l 2 tip im m tip tuyn qua A k n ( )C . Nhn xt: hai tip im 1 2T ,T cng nhn on IA di 1 gc vung, nn 1T , 2T

thuc ng trn ng knh IA . Vy, ng trn ( )C v ng trn ng knh IA c 2 im chung 1 2T ,T . Gi ( )C' l ng trn ng knh IA .

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Nguyn Ph Khnh

629

( ) ( )M x; y C' IM AM IM.AM 0 =

( )1 ( )IM x 2 ; y 1 ,= +

( )AM x 5 ; y 3=

( ) ( )( ) ( )( ) 2 21 x 2 x 5 y 1 y 3 0 x y 7x 2y 7 0 + + + = + + =

1T , 2T tha h 2 2

2 2

x y 4x 2y 3 03x 4y 10 0

x y 7x 2y 7 0

+ + = + =

+ + =

Vy, 1 2T T : 3x 4y 10 0+ = .

b. im ( )A d A a; 4 a .t MAN 2 , OA x 0= = >

Ta c: OM 2

sin ,OA OA

= = AM

cosOA

= 2

2

4 x 4s in2

x

=

( )2

2AMN 2

1 4 x 4S x 4

2 x

= . Vi AMNS 3 3= ( )

32 44 x 4 27x =

2x 16 x 4 = =

Vi ( )22OA 4 a 4 a 4= + + = a 4 = hoc a 0= Vy, ta im A cn tm ( )A 4;0 hoc ( )A 0; 4

Bi tp 11. Gi A' l im i xng ca A qua tm ( )I A' 33; 37 Ta thy, BHCA' l hnh bnh hnh nn HA' ct BC ti trung im M ca BC ,

khi ( )BC : 3x 4y 5 0 + = Phng trnh ng trn ngoi tip tam gic ABC :

( ) ( )2 2x 16 y 18 650 + + =

Ta B,C l nghim h phng trnh: ( ) ( )2 23x 4y 5 0

x 16 y 18 650

+ =

+ + =

Vy, ( )B 3; 1 , ( )C 5; 5 hoc ngc li l ta cn tm.

Bi tp 12. ng trn ( )C c tm ( )I 1; 2 , bn knh R 5= . Gi ( )M m; m v ( )0 0T x ; y l tip im v t M n ( )C .

Khi , ta c( )

( )( ) ( )( )0 0 0 02 20 0 0 0

x 1 x m y 2 y m 0IT.MT 0

T C x y 2x 4y 0

+ + ==

+ + =

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Nguyn Ph Khnh

630

( ) ( ) ( ) ( )2 20 0 0 0

0 02 20 0 0 0

x y m 1 x m 2 y m 0m 1 x m 2 y m 0

x y 2x 4y 0

+ + = + + + =

+ + =.

Suy ra phng trnh ( ) ( )AB : m 1 x m 2 y m 0 + + + = .

Mt khc AB to vi d mt gc vi 3

cos10

= nn ta c:

( ) ( )2 2

2 2

m 1 m 2 35 2m 2m 5 m m 0

102 m 1 m 2

= = + + + =

+ +

m 0,m 1 = =

Th li ta thy c hai trng hp ny ta u IM R= hay ( )M C . Vy, khng c im M tha yu cu bi ton.

Bi tp 13. ng trn ( )C c tm ( )I 1; 1 , bn knh R 3= . Gi H l trung im ca AB

Suy ra AIB1

IH AB S HI.AB 2 22

= =

4 2

ABHI

= . Hn na: 2 2 2AH HI IA+ =

2

2 22

AB 8HI 9 HI 9

4 HI + = + =

H

IA

B

4 2HI 1 AB 4 2

HI 9HI 8 0HI 2 2 AB 2

= = + =

= =

V i qua M nn phng trnh c dng: ax by 6a 3b 0+ + =

( ) 2 22 2

7a 4b 4 12HI 1 d I, 1 1 15b 56ab 48a 0 b a, b a

3 5a b

= = = + = = =

+

Vy : 3x 4y 6 0 + + = hoc : 5x 12y 6 0 + = l ng thng cn tm.

Bi tp 14.

IAB1 9 9

S IA.IB.sin AIB sin AIB2 2 2

= =

Suy ra IAB9

maxS2

= khi v ch khi

sin AIB 1= 0AIB 90 = . Gi H l hnh chiu ca I ln khi

H

IA

B

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,Nguyn Ph Khnh

631

0 03

AIH 45 IH IA.cos 452

= = =

Ta c ( ) 22

1 2m 3d I; IH m 8m 16 0 m 4

22 m

= = + + = =

+

Vy, vi m 4= tha mn yu cu bi ton.

Bi tp 15. D thy M nm ngoi ng trn, gi h l khong cch t I n ng thng cn tm

Ta c: 2 2 2

2 2 2

h MB R 25h 4

h 4MB IM 52

+ = = =

+ = =( ) ( )M d d : a x 7 b y 3 0 + =

V I cch d mt khong bng h nn 2 2

6a 4b4 a 0

a b

= =

+hoc

12a b

5=

Suy ra y 3= hoc 12x 5y 69 0 = .

Bi tp 16. a. ng trn ( )C c tm ( )I 1; 3 v bn knh =R 2 . Do = >IM 2 5 R nn im M ngoi ng trn ( )C . Gi ( )0 0T x ; y l tip im ca tip tuyn k t M .

Ta c: ( ) ( )

=

T CT C

MT IT MT.IT 0 , trong :

( )( )

= +

=

0 0

0 0

MT x 3; y 1

IT x 1; y 3

Do : ( )( ) ( )( )

+ + =

+ + =

2 20 0 0 0

0 0 0 0

x y 2x 6x 6 0

x 3 x 1 y 1 y 3 0

+ + = + =

+ + =

2 20 0 0 0

0 02 20 0 0 0

x y 2x 6x 6 02x y 3 0

x y 2x 4y 0 ( )

Ta cc tip im 1 2T ,T tha mn ng thc ( ) . Vy, phng trnh ng thng i qua 1 2T ,T l: + =2x y 3 0 .

b. ng trn ( )C c tm ( )I 2; 1 , bn knh R 2 5.= Gi H l trung im

AB . t ( )AH x 0 x 2 5 .= < < Khi ta c 2

1IH.AB 8 x 20 x 8 x 4

2= = = hoc x 2= (khng tha

AB IA< ). Suy ra AH 4 IH 2.= =

Phng trnh ng thng qua M : ( ) ( ) ( )2 2a x 1 b y 3 0 a b 0 + + = + >

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,Nguyn Ph Khnh

632

Ta c ( ) ( )2 2

a 2bd I, AB IH 2 2 a 3a 4b 0

a b

+= = = =

+.

Vy c hai ng thng tha mn l y 3 0+ = v 4x 3y 5 0.+ + =

Bi tp 17. Tm 1 5I ;2 2

ca ng trn v ( )H 2; 2 l giao im ca v ng

trn HBC . Gi ( )N a; b l trung im ca BC HA 2IN=

( )A 2a 1; 2b 3 + ( )B 3 2a;9 2b .

V ( )B HBC nn ( ) ( ) ( ) ( )2 23 2a 9 2b 3 2a 5 9 2b 4 0 + + =

2 22a 2b 5a 13b 23 0 + + = ( )1 .

Ta c c ( )BN 3a 3; 3b 9=

v BN AH nn BN.AH 0=

.

( )( ) ( )( ) 2 22a 1 3a 3 2b 5 3b 9 0 2a 2b 3a 11b 16 0 + = + + = ( )2

T ( )1 v ( )2 gii ra c 1b 3 a2

= = hoc 5

b a 12

= = .

Vi ( )b 3 B 2; 3 M= (loi). Vi ( ) ( ) ( )5b B 1; 4 ,A 3; 2 ,C 1;12

=

Bi tp 18. Gi s ABC c din tch ln nht. Khi CO AB, BO AC (V nu

khng, chng hn CO khng vung gc vi AB th tn ti im C' thuc

ng trn ( )C' sao cho C'O AB v ( ) ( )d C'; AB d C; AB> C' AB CABS S > ) Suy ra ABC c din tch ln nht th O l trc tm ca tam gic do B CBC OA x x = .

Ta c ( ) ( ) ( ) ( )B B B C B B B CAB x 1; y , OC x ; y , OB x ; y , AC x 1; y

( )B B B CCO AB x x 1 y y 0 + = ( )1 Li c ( ) ( )B C , C C' suy ra 2 2B Bx y 2+ = ( )2 v 2 2C Bx y 2+ = ( )3

T ( )1 , ( )2 v ( )3 suy ra B B5 5

x 1, x2

= = (loi)

Ta c ( ) ( )22A B C B B B1 1

S x x y y S 1 x 7 2x2 4

= =

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Nguyn Ph Khnh

633

Nu B B5 5

x 1 S 3,x S 32

= = = < nn S ln nht khi v ch khi

B Cx x 1= = khi ta c ta xc nh c ( ) ( )B 1;1 , C 1; 2 hoc ( ) ( )B 1; 1 , C 1; 2 l im cn tm.

Bi tp 19. Gi ( )A a; b . Ta c:

( ) ( ) ( )( )( ) ( )( )

2 2 2 2

2 2

A C a b 2a 4b 20 0a 1 b 2 25

IA.NA 0 a 1 a 6 b 2 b 1 0 a b 5a 3b 4 0

+ = + =

= + + = + + =

7a b 16 0 + + =

T ta suy ra c A : 7x y 16 0 + + = .

Tng t ta cng c c B AB AB : 7x y 16 0 + + = .

Bi tp 20. Ta c AB 10= v ( ) ( )MAB1 1 1

S d M,AB .AB d M,AB2 2 10

= = =

Li c ( )AB 1; 3=

nn ( )n 3; 1=

l VTPT ca ng thng AB

Suy ra phng trnh ( ) ( )AB : 3 x 1 y 1 0 + = hay 3x y 4 0 = .

Gi ( ) ( ) ( )2 2M a; b C a 1 b 2 + =

Khi ( )3a b 41 1

d M,AB 3a b 4 110 10 10

= = =

Ta c h phng trnh: ( ) ( )2 22 2a 1 b 2 a 1 b 2

3a b 4 1 3a b 4 1

+ = + =

= =

hoc

( )2 2a 1 b 2

3a b 4 1

+ =

=

( )2 2a 1 b 2b 3a 5

+ =

= hoc ( )

2 2a 1 b 2

b 3a 3

+ =

=

( ) ( )2 2

a 1 3a 5 2

b 3a 5

+ =

= hoc ( ) ( )

2 2a 1 3a 3 2

b 3a 3

+ =

=

25a 16a 12 0

b 3a 5

+ =

= hoc

25a 10a 4 0

b 3a 3

+ =

=

12 4a ,a

5 5b 3a 5

= =

=

hoc 5 5

a5

b 3a 3

= =

. Vy c bn im tha iu kin bi ton l:

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Nguyn Ph Khnh

634

1 2 3 412 11 4 13 5 5 3 5 5 5 3 5

M ; , M ; , M ; v M ;5 5 5 5 5 5 5 5

+

.

Bi tp 21. Ta c ( )M 2;1 v EQ l tip tuyn ca ( )C . Phng trnh EQ c dng: ( ) ( )a x 3 b y 6 0+ + =

V ( )d M,EQ 10= nn c:

( ) ( )2 2 22 2

5a 5b10 5a 5b 10 a b

a b

= = +

+

2 23a 10ab 3b 0 + = a 3b = hoc b 3a= - a = 3b, ta c phng trnh EQ : 3x y 3 0+ + = . Khi ta Q l nghim ca

h: 2 2 x 1(x 2) (y 1) 10

y 03x y 3 0

= + =

=+ + = . Trng hp ny ta loi v Qx 0> .

- b = 3a, ta c phng trnh EQ : x 3y 15 0+ = . Khi ta Q l nghim

ca h: ( ) ( ) ( )2 2 x 3x 2 y 1 10 Q 3; 4

y 43x y 3 0

= + = =+ + =

.

Ta c ( )P 15 3x; x v ( ) ( )2 2QP MQ 12 3x 4 x 10 x 3,x 5= + = = = - x = 3, ta c P(6; 3), suy ra tm ca hnh vung ( )I 4; 2 nn ( )N 5;0 - x = 5, ta c ( )P 0; 5 , suy ra tm ca hnh vung ( )I 1; 3 nn ( )N ( 1; 2) .

Vy c hai b im tha yu cu bi ton:

( ) ( ) ( ) ( )M 1; 2 ,N 5; 0 ,P 6; 3 ,Q 3; 4 v ( ) ( ) ( ) ( )M 2;1 ,N 1; 2 ,P 0; 5 ,Q 3; 4 .

Bi tp 22. ng trn ( )C c tm ( )I 2;1 , bn knh R 5 AI 5= = .

MAI AIBM1 1

S S 5 MA.IA 52 2

= = =

MA 2 5 = . Suy ra 2 2 2IM IA AM 25= + = .

M M nn ( )M m; m 2 ,

suy ra ( )22 2IM 25 m 2 (m 3) 25= + + =

2m m 6 0 m 3,m 2 + = = = .

Vy ( )M 2; 4 v ( )M 3;1 l hai im cn tm.

P

N

M

Q

E

B

I

A

M

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Nguyn Ph Khnh

635

Bi tp 23. ng trn ( )C c tm ( )I 1; 2 , bn knh R 5=

a. Tip tuyn i qua M v vung gc vi IM nn nhn ( )IM 3; 4=

lm vect

php tuyn, phng trnh tip tuyn l: ( ) ( )3 x 4 4 y 6 0 3x 4y 36 0 + = + = . Gi l tip tuyn cn tm. Do i qua N nn phng trnh c dng

: ( ) ( )a x 6 b y 1 0 ax by 6a b 0+ + = + + = , 2 2a b 0+ ( )

Ta c: ( ) 2 22 2

7a bd I, R 5 7a b 5 a b

a b

+ = = + = +

+

( ) ( )2 2 27a b 25 a b + = + 2 224a 14ab 24b 0 + = 2

a a 324 12 24 0 a b

b b 4

+ = =

hoc 4

a b3

=

3

a b4

= thay vo ( ) ta c: 3 7bx by b 0 3x 4y 14 04 2

+ + = + + = .

4

a b3

= thay vo ( ) ta c: 4 bx by 9b 0 4x 3y 27 03

+ = + = .

Vy, c hai tip tuyn tha yu cu bi ton l: 3x 4y 14 0+ + = v

4x 3y 27 0 + = .

b. Gi ( )A a; b . Ta c: ( ) ( ) ( )

( )( ) ( )( )

2 2A C a 1 b 2 25

IA.NA 0 a 1 a 6 b 2 b 3 0

+ = = + + =

2 2

2 2

a b 2a 4b 20 0

a b 5a 5b 0

+ =

+ + =7a b 20 0 + =

T ta suy ra c A : 7x y 20 0 + = .

Tng t ta cng c c B AB AB : 7x y 20 0 + = .

Bi tp 24. Cch 1: Gi ( )M x; y l trung im ca BC , D l im i xng vi A qua O .

Ta c BH CD,CH BD nn t gic BDCH l hnh bnh hnh nn M l trung

im HD

T suy ra, ( )( ) ( )

0 2 2 x x 2AH 2MI M 2; 3

y 36 2 y

= = =

= =

Nn ng thng BC qua M c ( )AH 0;6=

l vtpt c phng trnh l :

y 3 0+ = .

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,Nguyn Ph Khnh

636

Gi ( )C a; 3 , do IA IC=

( ) ( ) ( )2 2 225 7 a 2 3 + = + + 2a 4a 61 0 a 2 65 + = =

( )C 2 65; 3 + Cch 2. ng trn ngoi tip tam gic ABC

c phng trnh : ( )2 2x 2 y 74+ + = . Phng trnh AH : x 3= , do BC AH BC : y m = ( m 7 )

H

M

I

A

BC

Ta B, C l nghim ca phng trnh : ( )2 2x 2 m 74+ + = 2 2x 4x m 70 0 + + = ( )

V ( ) c hai nghim, trong c t nht mt nghim dng nn m 70<

Khi : 2 2B 2 74 m ; m , C 2 74 m ; m +

V 2BH AC AC.BH 0 m 4m 21 0 m 3 = + = =

Vy, ( )C 2 65; 3 + l ta cn tm.

Bi tp 25.a. V ABC vung ti B nn AC l ng knh ca ( )T . Gi ( )1 2ASB d , d t= = ta c BAC ASB t= = (gc c cnh tng ng vung gc). Gi s bn knh ( )T l R ta c :

2ABC

BC.BA ACsin t.ACcos tS 2R sin t cos t

2 2= = = .

Mt khc ( )

( ) ( ) ( )2 2 223. 3 1. 1 1

cos t t2 3

3 1 3 1

+ pi= = =

+ +

.

Suy ra 2ABC3

S R2

= t c R 1= .

Do 1 2A d , C d nn ( ) ( )A a; a 3 ,C c; c 3 thm na vecto ch phng ca 1d l ( )1u 1; 3

c phng vung gc vi AC

nn:

( )1AC.u 0 c a 3 c a 0 c 2a= + = =

.

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Nguyn Ph Khnh

637

Mt khc ( ) ( )22

AC 2R 2 c a 3 c a 2 = = + + = 2 a 3 2 = v a 0> nn

3a

3= .

Tm ng trn l trung im ca AC l :

( )a c 3 a 3 3a 3 3I ; c a ; ;2 2 2 2 6 2

+ = =

.

Vy, phng trnh ca ( )T l 2 2

3 3x y 1.

6 2

+ + + =

Cch 2: Ta c 1d tip xc vi ( )T c ng knh l AC nn 1AC d T gi thit ta c : 0 0 0 0AOx 60 ,BOx 120 AOB 60 , ACB 30= = = =

Nn 2 2ABC1 3 3 3

S AB.BC AB AB AB 12 2 2 2

= = = =

V ( )2A d A x; 3x ,x 0, > 2 2 1OA .AB A ; 13 3 3

= =

4 2OC 2OA C ; 2

3 3

= =

.

ng trn ( )T ng knh AC c: 2 3 ACI ; , R 12 23

= =

.

Phng trnh ( )T :2 2

1 3x y 1

22 3

+ + + =

.

b. V ( ) M d M m; m .Gi ( )0 0A x ; y .

Khi , ta c: ( )

( ) ( ) + + == + + =

2 20 0 0 02 20 0 0 0

x y m 1 x m 2 y m 0IA.MA 0

A C x y 2x 4y 0

Suy ra ( ) ( ) + + + =0 0m 1 x m 2 y m 0 . Do , phng trnh AB l: ( ) ( ) + + + =m 1 x m 2 y m 0 .

Mt khc: ( ) = 3d N,AB5

hay

= + +2 2

m 3 3

5(m 1) (m 2)

Gii phng trnh ny ta tm c = = 58

m 0,m13

.

Ta loi =m 0 , v khi ( )M C .

Vy c mt im M tha yu cu bi ton:

58 58M ;

13 13.

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Nguyn Ph Khnh

638

Bi tp 26. Xt php quay

0(A; 90 )Q : M N v ( ) ( )'1 1C C

M ( ) ( ) ( ) ( )' '1 1 2 1M C N C N C C . Vi 0(A;90 )

Q , ta c phng trnh ( ) ( )2' 21C : x y 5 13+ = Ta im N l nghim ca h:

( ) ( )

2 2 2 2

2 2 2 2

x (y 5) 13 x y 10y 12 0

x y 2x 4y 20 0x 1 y 2 25

+ = + + =

+ = + =

2 2 2x y 10y 12 0 5y 53y 134 0

x 3y 16 x 3y 16

+ + = + =

= =

1 3 129x

10 53 129

y10

+=

+ =

hoc

1 3 129x

10 53 129

y10

=

=

Trng hp ny c hai b im: + + +

23 129 51 3 129 1 3 129 53 129M ; ,N ;

10 10 10 10

V +

23 129 51 3 129 1 3 129 53 129M ; ,N ;

10 10 10 10.

Vi 0(A; 90 )

Q , ta c phng trnh ( ) ( ) ( )2 2'1C : x 2 y 3 13 + =

Ta im N l nghim h: ( ) ( )( ) ( )

2 2

2 2

x 2 y 3 13 x 4

y 6x 1 y 2 25

+ = =

= + =

hoc x 5

y 5

=

=

Trng hp ny c hai b im: ( ) ( )M 1;7 ,N 4;6 v ( ) ( )M 0;8 ,N 5; 5 . Cch 2: Gi ( )M a; b v ( )N c;d ln lt l 2 im nm trn ng trn ( )1C , ( )2C

( )( )

( ) ( )( ) ( )

( )2 2

1

2 22

M C a 2 b 5 131

N C c 1 d 2 25

+ =

+ =

Li c: AMB vung cn ti A nn c: ( )AM.AN 0 2AM AN

=

=

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,Nguyn Ph Khnh

639

Bi tp 27a. ( )1C c tm ( )1I 1;0 v bn knh 11

R2

=

( )2C c tm ( )2I 2; 2 v bn knh 2R 2= Gi s d l ng thng cn tm v d ct ( )2C ti A, B nn d qua ( )2I 2; 2 v tip xc ( )1C .

d qua ( )2I 2; 2 , c vecto php tuyn ( )n a; b 0

c phng trnh:

( ) ( )a x 2 b y 2 0 + =

d tip xc ( )1C khi ( )11

d I ;d2

=2 2

a 2b 1

2a b

+ =

+ 2 2a 8ab 7b 0 + + =

( )( )a b a 7b 0 a b + + = = hoc a 7b= Vi a b= , suy ra d : x y 0 =

Vi a 7b= , suy ra d : 7x y 12 0 =

Vy, c 2 ng thng cn tm: x y 0, = 7x y 12 0 =

b. Cch 1: Gi H l hnh chiu ca I trn AB , suy ra Gi H l trung im ca AB hay AB 2AH= . t ( )AH x , 0 x 3 .= < <

2 4 2IAB

1 1S IA.AB 2 2 9 x 2x x 9x 8 0

2 2= = + =

x 1 = hoc x 2 2=

Vi x 1 AB 2= = khng tha.

Vi x 2 2 AB 4 2= = nhn. Suy ra IH 1=

Cch 2: ( )C c tm ( )I 1; 1 , bn knh R 3= . ng thng i qua M c dng:

( ) ( )a x 6 b y 3 0,+ + = 2 2a b 0+ > .

Gi H l hnh chiu ca I trn AB th 2

2 2 ABIH IA 8 IH 2 24

= = =

2AIB

1 1S IA.IB.sinAIB R .sinAIB

2 2= = ,

4 2 AIB 1sinAIB cos

9 2 3= = hoc

AIB 2 2cos

2 3= . D thy,

AIB IHcos

2 IA=

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Nguyn Ph Khnh

640

Kt hp gi thuyt suy ra:

( )AIB 1 AIB

cos d I; AB IA.cos 12 3 2

= = = hay

2 2

7a 4b1

a b

=

+ 2 248a 56ab 15b 0 + = ( )( )4a 3b 12a 5b 0 =

c. ( )C c tm ( )I 2; 2 , bn knh R 3= IP qua ( )I 2; 2 v vung gc vi d nn c phng trnh: x my 2m 2 0+ =

IBP vung nn c ( )2 2R IB IH.IP d I;d .IP= = =

Bi tp 28. ( )C c tm ( )I 1; 2 , bn knh R 5=

a. ( )( )2

2 ABd I; d R 2 x 3 0, 12x 5y 31 02

= = = + =

b. CD ngn nht khi ( )( )1d I; d ngn nht

Bi tp 29a. ng trn ( )C c tm ( )I 2; 3 , bn knh R 10= Gi ng thng AB i qua M , c phng trnh: ( ) ( )a x 3 b y 2 0,+ + + =

2 2a b 0+ > . ng trn ni tip ABCD nn AB tip xc vi ng trn ( )C

khi v ch khi ( )d I; AB R= 2 2

5a 5b10

a b

+ =

+ 2 23a 10ab 3b 0 + + =

( )( )a 3b 3a b 0 + + = a 3b= hoc b 3a= TH1: a 3b= chn a 3, b 1 AB : 3x y 7 0= = + = , v A AB nn ( )A a;7 3a+ v a 0>

Hn na: 2IA R 2 IA 20= = ( ) ( )2 2a 2 3a 4 20 + + = a 0 = hoc a 2= ( khng tha a 0> ).

TH2: b 3a= chn a 1, b 3 AB : x 3y 3 0= = = , v A AB nn ( )A 3 3a;a+ v a 0>

Hn na: 2IA R 2 IA 20= = ( ) ( )2 23a 1 a 3 20 + + = a 1 = ( tha ) hoc a 1= ( khng tha a 0> ).

Khi ( )A 6;1 , I l trung im ca ( )AC C 2; 5

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Nguyn Ph Khnh

641

b. Nhn thy, ABC vung ti C suy ra tm ng trn ngoi tip ABC l

( )I 3; 1 bn knh bng 2 21 1AB BC CA 22 2

= + = .

Phng trnh ng trn ( )C ngoi tip ABC : ( ) ( )2 2x 3 y 1 2 + + =

N l im ty trn ( )C nn 2 2 2

NAB1 NA +NB AB

S NA.NB 22 4 4

= = =

NABS t gi tr ln nht bng 2 khi NA NB= . N l giao im ca ng

trung trc on AB vi ( )C , nn ta N tha h:

( ) ( ) ( )( )

2 2 x 2 y 0 N 2;0x 3 y 1 2x 4 y 2 N 4; 2x y 2 0

= = + + =

= = + =

( )M m; 4 4m v NO.NM 0=

c. ng trn ( )C c tm ( )I 1; 2 , bn knh R 1=

Ta thy: 0BIC BAC 180 sin BIC sin BAC+ = = ( )1

Hn na: 2 2ABIC ABC BIC1 1

S S S IB.AB IB sin BIC AB sin BAC2 2

= + = + ( )2

T ( )1 v ( )2 , suy ra: ( ) 2 2 2 22IB.AB

2IB.AB IB AB sin BAC sin BACIB AB

= + =+

Mt khc: 3 3

2ABC 2 2 2

1 IB.AB AB 27S IB sin BIC

2 10IB AB 1 AB= = = =

+ + ( )3

T ( )3 AB 3 = hay 2 2 2IA AB IB 10= + = ( ) ( )2 2a 1 2a 3 10 + + = vi ( )A a; 2a 1+

d. ( )C c tm ( )I 1; 2 , bn knh R 3= Do ABIM l hnh bnh hnh nn AB MI ( )MI 2; 2 =

l vtcp ca

: x y m 0+ + =

Gi H l trung im AB, ta c 1 1 1

HB AB MI 8 22 2 2

= = = =

2 2IH R HB 9 2 7= = = d(I; ) 7 m 3 14 = =

Bi tp 30.a. Gi s ( )C c;d v ( )H h; h 1+ , { }c 2;6 .

Trong : ( ) ( )2 2c 4 d 6 5 + =

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Nguyn Ph Khnh

642

( ) ( ) ( ) ( )AC c 2;d 5 , AB 4;0 , BH h 6; h 4 , CH h c; h 1 d= = = = +

H l trc tm tam gic ABC nn c: BH.AC 0

CH.AB 0

=

=

( ) ( ) ( )( )( ) ( )( ) ( )

2 2h 4 d 6 5 1

h 2 h 6 d 5 h 4 0 2

+ =

+ =

Ly ( )1 tr ( )2 , ta c ( )( )d 5 d h 3 0 d 5 = = hoc d h 3= +

Vi d 5= thay vo ( )1 ta c: 2h 8h 12 0 h 2 + = = hoc h 6=

Vi d h 3= + thay vo ( )1 ta c: 22h 14h 20 0 h 2 + = = hoc h 5= b. ng trn ( )C c tm ( )O 0;0 c bn knh R 3= .

T AB 4,8 OH=1,8= v 2OA

MO 5OH

= =

Gi s M c ta ( )M a; b ta c: 2 2a b 25+ = ( )1

Hn na ( )M C' nn c: 2 2a b 18a 6b 65 0+ + = ( )2 Gii h ( )1 v ( )2 ta tm c: ( ) ( )M 5;0 , M 4; 3

c. ng trn ( )C c tm ( )I 1; 2 , bn knh R 5= . ng trn ( )C ni tip tam gic ABC nn c ( )d I; BC R= , n y ta tm c hoc BC : 2x 4y 15 0+ = hoc BC : 2x 4y 1 0 + = .

Gi J l giao ca AI vi BC. rng ABC u nn IJ BC v I l trng

tm ca ABC nn AI 2IJ=

ta A

d. ( )A 2; 3 l giao im ( )1C v ( )2C . Phng trnh ng thng i qua A c dng: ( ) ( )a x 2 b y 3 0 + + = .

ng trn ( )1C c tm ( )O 0;0 , bn knh 1R 13= ng trn ( )2C c tm ( )I 6;0 , bn knh 2R 5=

Theo gi thit, suy ra: ( ) ( )2 2 2 21 2R d O, R d I, = x 3y 7 0 + + = .

Bi tp 30a. ng trn ( )C c tm ( )I 1;m v bn knh R 5=

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Nguyn Ph Khnh

643

Ta c: ( ) 25 m

d I; 5m 16

= + + > 2 2t2

> hoc 2 2

t2+

< ( )

Phng trnh i qua hai tip im A,B c dng :

( )( ) ( )( )t 1 x 1 t 3 y 2 9 0 + + + =

Ta c: ( )2

3t 1d N; AB

2 2t 4t 10

+=

+ +.

Xt ( )2

3t 1f t

2 2t 4t 10

+=

+ + tha iu kin ( )

Ta c: ( )( )32

2t 14f ' t

2t 4t 10

+=

+ +

Vi 1

t3

th ( )f ' t 0> th hm s ( )f t ng bin trn na khong 1 ;3

+

Vi 1

t3

< th ( )f ' t 0 t 7= =

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Nguyn Ph Khnh

645

Lp bng bin thin, suy ra ( ) 5f t2

hay ( ) 5d N,AB2

ng thc xy ra khi t 7= tc ( )M 7; 6

Vy, ( )M 7; 6 l im cn tm. th gi tr ln nht bng 52

b. ng trn ( )C c tm ( )I 1; 2 ,R 4 = v im I thuc ng thng . ng trn ( )C' c tm J bn knh R ' 1= v tip xc ngoi vi ng trn ( )C suy ra qu tch ca im I l ng trn ( )K c tm I bn knh R R ' 5+ =

hay ( )K : ( ) ( )2 2x 1 y 2 25+ + = . Khong cch ca I ti l ln nht khi I l giao im ca ng thng d i qua J v vung gc vi vi ng trn ( )K . d c phng trnh : 4x 3y 10 0 + = .

Ta im I tha mn h: ( ) ( )2 2

x 2,y 64x 3y 10 07

x 2, yx 1 y 2 252

= = + = = = + + =

Vi ( ) ( ) ( )2 2I 2;6 x 2 y 6 1, + = vi ( )2

27 7I 2; x 2 y 1

2 2 + + + =

Vy, khong cch t I ti ln nht bng 5

c. Gi s : ( )B x;y th do ( )M 0; 2 l trung im ca BC nn ( )C x;4 y .

D thy B,C u thuc ( )C nn ta c h : ( ) ( )( ) ( )

2 2

2 2

x 1 y 1 10

x 1 y 3 10

+ = + + =

t y tm

c ta B,C v BC 4 2= .

Gi ( )A a;b , t gi thit suy ra a b 2 6 + = , hn na ( ) ( )A a;b C t y ta tm c ta im A .

d. ( )C c tm ( )I 1; 2 , R 5 = Phng trnh tng qut ca ( )d qua M c dng: ( ) ( )a x 2 b y 1 0 + + = vi

2 2a b 0+ > .

Din tch ( )( )IAB1

S .AB.d I; d2

= , AB c nh v ( )( )2 2

a bd I; d 2

a b

+=

+

ng thc xy ra khi a b 1 d : x y 1 0= = + = , ( ) ( )E d E t;1 t .

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