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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 .
Bi tp 5. Trong mt phng to cc vung gc Oxy,
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.
Bi tp 9. Trong mt phng to cc vung gc Oxy,
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.
Bi tp 10. Trong mt phng to cc vung gc Oxy,
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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> .
Bi tp 14. Trong mt phng to cc vung gc Oxy, cho ng trn
( ) 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.
Bi tp 15. Trong mt phng to cc vung gc Oxy, cho ng trn
( ) ( ) ( )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= .
Bi tp 16. Trong mt phng to cc vung gc Oxy,
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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.
Bi tp 19. Trong mt phng to cc vung gc Oxy, cho ng trn
( ) ( ) ( )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 .
Bi tp 20. Trong mt phng to cc vung gc Oxy, cho ng trn
( ) ( )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
.
Bi tp 21. Trong mt phng to cc vung gc Oxy, cho ng trn
( )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.
Bi tp 25. Trong mt phng to cc vung gc Oxy,
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
Bi tp 27. Trong mt phng to cc vung gc Oxy,
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.
Bi tp 30. Trong mt phng ta Oxy,
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.
Bi tp 30. Trong mt phng ta Oxy,
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 .<
Bi tp 31. Trong mt phng ta Oxy,
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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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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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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