[ nguoithay.org ] tong hop bai tap ve duong tron(1)

5/21/2018 [ Nguoithay.org ] Tong Hop Bai Tap Ve Duong Tron(1)

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Trang dy hc y video wwww.nguoithay.com

B(2;-1)

A

C

x+2y-5=0

3x-4y+27=0

H

K

Bi 1Trong mt phng vi htoOxy, cho ng trn (C) : 2 2x y 2x 8y 8 0 . Vit phngtrnh ng thng song song vi ng thng d: 3x+y-2=0 v ct ng trn theo mt dy cung c dbng 6.Trang dy hc y video wwww.nguoithay.com

Hng dn: ng thng d' song song vi d : 3x+y+m=0

- IH l khong cch tI n d' :3 4 1

5 5

m mIH

- Xt tam gic vung IHB :2

2 2 25 9 164

ABIH IB

2

19 ' :3 19 0116 1 20

21 ' :3 21 025

m d x ymm

m d x y

Bi 2 Trong mt phng vi hta Oxy. Cho ng trn (C) : 012422 yxyx v ng thng d : 01 yx . Tm nhng im M thuc ng thng d sao cho tim M kc

Trang dy hc y video wwww.nguoithay.comHng dn:- M thuc d suy ra M(t;-1-t). . Nu 2 tip tuyn vung gc vinhau th MAIB l hnh vung ( A,B l 2 tip im ). Do

AB=MI= IA 2 =R 2 = 6 2 2

3 .

- Ta c : 2 2 22 2 2 8 2 3MI t t t

- Do :

12 2

2

2 2; 2 12 8 12 2

2 2; 2 1

t Mt t

t M

.

* Ch : Ta cn cch khc

- Gi d' l ng thng qua M c hsgc k suy ra d' cphng trnh : y=k(x-t)-t-1, hay : kx-y-kt-t-1=0 (1) .

- Nu d' l tip tuyn ca (C) ktM th d(I;d')=R2

2 26

1

k kt t

k

2 2 2 2 22 2 6 1 4 2 2 2 2 4 2 0t k t k t t k t t k t t

- Tgithit ta c iu kin :

2

2 2 2

2

2

4 2 0

' 4 2 4 2 4 0

4 2 14 2

t t

t t t t t

t tt t

- 1 22 2 1 22

1 2

2 6 1

' 19 0 2 ;2

12

tk k

t t t k k M

k kt

Bi 3 Trong mt phng vi htoOxy, cho ng trn (C) c phng trnh: 2 2 4 3 4 0x y x Tia Oy ct (C) ti A. Lp phng trnh ng trn (C), bn knh R = 2 v tip xc ngoi vi (C) ti AHng dn: - (C) c I( 2 3;0 ), R= 4 . Gi J l tm ng trn cn tm :

M

x+y+1=0

A

B

I(2;1)


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J(a;b) 2 2

' : 4C x a y b

-Do (C) v (') tip xc ngoi vi nhau cho nn khong cch IJ

=R+R' 2

2 2 22 3 4 2 6 4 3 28a b a a b

- V A(0;2) l tip im cho nn : 2 2

0 2 4 2a b

- Do ta c h:

22 2 2

2 222

2 3 36 4 3 24

4 02 4

a b a a b

a b ba b

- Gii htm c : b=3 v a= 2 2

3 ' : 3 3 4C x y .

* Ch : Ta c cch gii khc .- Gi H l hnh chiu vung gc ca J trn Ox suy ra OH bng a v JH bng b

- Xt cc tam gic ng dng : IOA v IHJ suy ra :4 2

3 2

IJ 6 2 3

IA IO OA

IH HJ ba

- Ttstrn ta tm c : b=3 v a= 3 .Bi 4. TrongmtphngtoOxy chongtrn (C) : x2+ y2+ 4x6y + 9 = 0 vim M( 1; -

8).Vitphngtrnhngthng d qua M saocho d ct (C) tihaiim A,B phnbitmdintch tam gic Atgitrlnnht.Hng dn:

Bi 5. Vi I ltmcangtrn (C).Cho A(1 ; 4) vhaingthng b : x + y3 = 0 ; c : x + y9 = 0.Tmim B trnb ,im C trn c saocho tam gic ABC vungcnti A.


Hng dn:Bi 6

Bi 7

2

2

4

10 5 5

Hng

d n:* D dng xc nh c nh C i xngvi A qua tm I(1,-2) => C(0;2)* Do din tch BC bng 4 suy ra

d(B;AC)=4

5 . B l giao im ca

ng thng song song vi AC v cch

AC 1 khong bng4

5; vi ng trn

(C).Kt qu ta c 4 im B c ta l(0.00, 0.00);; (2.00,4.00) .......

I

H

E

C(0;-4)

A(2;0)

I

O

I(-2 ;0)

A(0;2)

y

x


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Bi 8 Trong mt phng vi hta Oxycho ng trn hai ng trn2 2( ) : 2 2 1 0,C x y x y 2 2( ') : 4 5 0C x y x cng i quaM(1; 0). Vit phng trnh

ng thng quaMct hai ng trn ( ), ( ')C C ln lt tiA, B sao choMA= 2MB


Hng dn: * Cch 1.

- Gi d l ng thng qua M c vc t chphng 1

; :x at

u a b d y bt

- ng trn 1 1 1 2 2 2: 1;1 , 1. : 2;0 , 3C I R C I R , suy ra :

2 2 2 2

1 2: 1 1 1, : 2 9C x y C x y

- Nu d ct 1C ti A : 2

2 2 2

2 2 2 2

2 2

02 2

2 0 1 ;2

t Mab b

a b t bt Ab

a b a bt a b

- Nu d ct 2C ti B : 2

2 2 2

2 2 2 2

2 2

06 6

6 0 1 ;6

t Ma ab

a b t at Baa b a bt

a b

- Theo githit : MA=2MB 2 24 *MA MB

- Ta c :

2 22 22 2

2 2 2 2 2 2 2 2

2 2 6 64

ab b a ab

a b a b a b a b

2 22 2

2 2 2 2

6 : 6 6 04 36

4. 36 6 : 6 6 0

b a d x yb a

b a b a d x ya b a b

* Cch 2.

- Sdng php vttm I tsvtk= 1

2 . ( Hc sinh tlm )

Bi 9Trong hta Oxy, cho hai ng trn c phng trnh 2 21 : 4 5 0C x y y v

2 22 : 6 8 16 0.C x y x y Lp phng trnh tip tuyn chung ca 1C v 2 .C


10

8

6

4

2

2

4

6

8

10

12

25 20 15 10 5 5 10

Hng dn:* ng trn (C) c tm H(1;-2); bn knhR=5 tip xc vi ng thng (d) ti A'(4;2)* Tam gic ABC c trc tm H, hai nh B vC thuc (d) th A' l chn ng cao thuc BCv A thuc (C) nn AA' l ng knh vA(-2;-5)*

do trung im F ca AB thuc (C) nn

HF//=

1

2A'B =>A'B=10 .T y ta tm cta ca B= (12;-4)* Do C thuc (d) nn ta ca C tha mn

h thc:CA' =tA'B v CH . AB =0 => C 0;5( ).Ta cc nh ca tam gic l :

A=(-2;-5);B= (12;-4);C=(0;5)

C

B

F

E

A

A'

H


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Hng dn: : - Ta c :

2 2 22

1 1 1 2 2 2: 2 9 0;2 , 3, : 3 4 9 3; 4 , 3C x y I R C x y I R

- Nhn xt : 1 2 19 4 13 3 3 6I I C khng ct 2C

- Gi d : ax+by+c =0 ( 2 2 0a b ) l tip tuyn chung , thth : 1 1 2 2, , ,d I d R d I d R

2 2

2 2 2 2

2 2

23 1

3 4 22 3 4

2 3 4 3 4 23 43 2

b c

a b c b cb c a b ca b

b c a b c a b c b ca b c a b a b

a b

2

3 2 2 0

a b

a b c

. Mt khc t(1) : 2 2 22 9b c a b

- Trng hp : a=2b thay vo (1) :

2 2 2 2 2 2 2 2

2 3 5

42 9 4 41 4 0. ' 4 41 45

2 3 5

4

b

b cb

b c b b b bc c c c cc

b

- Do ta c hai ng thng cn tm :

1

2 3 5 2 3 5: 1 0 2 2 3 5 2 3 5 4 0

2 4d x y x y

1

2 3 5 2 3 5: 1 0 2 2 3 5 2 3 5 4 0

2 4d x y x y

- Trng hp :2 3

2

b ac

, thay vo (1) : 2 2

2 2

2 32

23 2

b ab

b a a ba b

2 2 2 2

0, 202

2 3 4 0 44 , 6

33 6

ab a cb c

b a a b b ab aa a b a c

b c

- Vy c 2 ng thng : 3: 2 1 0d x , 4: 6 8 1 0d x y

Bi 10Vit phng trnh tip tuyn chung ca hai ng trn :(C1) : (x - 5)

2+ (y + 12)

2= 225 v (C2) : (x1)

2+ ( y2)2= 25

Hng dn: - Ta c (C) vi tm I(5;-12) ,R=15. (C') c J(1;2) v R'=5. Gi d l tip tuyn chung cphng trnh : ax+by+c=0 ( 2 2 0a b ).

- Khi ta c : 2 2 2 2

5 12 2, 15 1 , , 5 2a b c a b ch I d h J d a b a b

- T(1) v (2) suy ra :5 12 3 6 3

5 12 3 25 12 3 6 3

a b c a b ca b c a b c

a b c a b c

9

32

2

a b c

a b c

. Thay vo (1) :2 22 5a b c a b ta c hai trng hp :

- Trng hp : c=a-9b thay vo (1) : 2 2 2 2 22 7 25 21 28 24 0a b a b a ab b


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Suy ra :

14 10 7 14 10 7 175 10 7: 0

21 21 21

14 10 7 14 10 7 175 10 7: 0

21 21 21

a d x y

a d x y

- Trng hp : 2 2 2 2 232 1 : 7 2 100 96 28 51 0

2c a b b a a b a ab b . V nghim . (

Ph hp v : 16 196 212 ' 5 15 20 400IJ R R . Hai ng trn ct nhau ) .

Bi 11 Cho ng trn (C): x2+ y22x6y + 6 = 0 v im M (2;4)Vit phng trnh ng thng i qua M ct ng trn ti 2 im A v B, sao cho M l trung im caAB


Hng dn: - ng trn (C) : 2 2

/( )1 3 4 1;3 , 2, 1 1 4 2 0M Cx y I R P M

nm trong hnh trn (C) .

- Gi d l ng thng qua M(2;4) c vc t chphng 2; :4

x atu a b d y bt

- Nu d ct (C) ti A,B th : 2 2 2 2 21 1 4 2 2 0 1at bt a b t a b t ( c 2 nghim t )

V vy iu kin : 2 2 2 2 2' 2 3 2 3 0 *a b a b a ab b

- Gi 1 1 2 22 ;4 , 2 ;4A at bt B at bt M l trung im AB th ta c h:

1 2 1 2

1 2

1 2 1 2

4 4 00

8 8 0

a t t a t t t t

b t t b t t

. Thay vo (1) khi p dng vi t ta c :

1 2 2 22

2 40 0 : : 6 01 1

a b x yt t a b a b d d x ya b

Bi 12 Trong mt phng vi hta Oxy, cho ng trn (C): x2+ y2- 2x - 2my + m2- 24 = 0 c tmv ng thng : mx + 4y = 0. Tm m bit ng thng ct ng trn (C) ti hai im phn bit A,Btha mn din tch tam gic IAB bng 12.

Hng dn: - (C) : 2 2

1 25 (1; ), 5x y m I m R .

- Nu d : mx +4y=0 ct (C) ti 2 im A,B th

2 2

2 2

4

16 42 24 0 1

16 4

my x

m mx x m

- iu kin : 2' 25 0m m R . Khi gi1 1 2 2; , ;

4 4

m mA x x B x x

2 2 2

2 2

2 1 2 1 2 12

16 258

16 4 16

m m mAB x x x x x x

m

- Khong cch tI n d =2 2

4 5

16 16

m m m

m m

- Tgithit :2 2

22 2

51 1 25 25. .8 . 4 5 12

2 2 1616 16

mm mS AB d m

mm m


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2

22 2 2

2

255 3 25 25 9 16

16

mm m m m

m

- Ta c mt phng trnh trng phng , hc sinh gii tip .Bi 13 Vit phng trnh ng trn i qua hai im A(2; 5), B(4;1) v tip xc vi ng thng cphng trnh 3x y + 9 = 0.Hng dn:Gi M l trung im AB suy ra M(3;3 ) . d' l ng trung trc ca AB th d' c phng trnh : 1.(x-3)-

2(y-3)=0 , hay : x-2y+3=0 .- Tm I ca (C) nm trn ng thng d' cho nn I(2t-3;t) (*)

- Nu (C) tip xc vi d th 3 2 3 9 5

10,

210 10

t t th I d R t R

. (1)

- Mt khc : R=IA= 2 2

5 2 5t t . (2) .

- Thay (2) vo (1) : 2 2 2 2105 2 5 4 5 30 50 10

2t t t t t t

2 6 34

12 2 06 34

tt t

t

. Thay cc gi trt vo (*) v (1) ta tm c ta tm I v bn knh R

ca (C) .* Ch : Ta c thsdng phng trnh (C) : 2 2 2 2 0x y ax by c ( c 3 n a,b,c)- Cho qua A,B ta to ra 2 phng trnh . Cn phng trnh th3 sdng iu kin tip xc ca (C) v d khong cch ttm ti d bng bn knh R .Bi 14 Cho ng trn (C): x2+ y22x + 4y + 2 = 0.Vit phng trnh ng trn (C') tm M(5, 1) bit (C')

t (C) ti cc im A, B sao cho 3AB .Hng dn: - ng trn (C) :

2 2

1 2 3 1; 2 , 3x y I R .

- Gi H l giao ca AB vi (IM). Do ng trn (C') tm Mc bn knh R' = MA . Nu AB=

3 IA R , th tam gic

IAB l tam gic u , cho nn IH= 3. 3 3

2 2 ( ng cao

tam gic u ) . Mt khc : IM=5 suy ra HM=3 7

52 2

.

- Trong tam gic vung HAM ta c2

2 2 249 3 13 '4 4 4

ABMA IH R

- Vy (C') : 2 2

5 1 13x y .

Bi 15 Trong mt phng vi h ta Oxy cho ng trn (C) c phng trnh (x-1)2

+ (y+2)2

= 9 v ng thng d: x + y + m = 0. Tm m trn ng thng d c duy nht mt im A m t k c hatip tuyn AB, AC ti ng trn (C) (B, C l hai tip isao cho tam gic ABC vung.Hng dn:- (C) c I(1;-2) v bn knh R=3 . Nu tam gic ABCvung gc ti A ( c ngha l tA kc 2 tip tuynti (C) v 2 tip tuyn vung gc vi nhau ) khi ABIC l hnh vung . Theo tnh cht hnh vung ta c

IA= IB 2 (1) .- Nu A nm trn d th A( t;-m-t ) suy ra :

I M

A

B

H

I(1;-2)

B

C

A

x+y+m=0


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2 2

1 2IA t t m . Thay vo (1) :

2 2

1 2 3 2t t m

2 22 2 1 4 13 0t m t m m (2). trn d cng 1 im A th (2) c ng 1 nghim t , t ta c iu

kin : 22 10 25 0 5 0 5m m m m .Khi (2) c nghim kp l :

1 2 0 1 5 1 3 3;82 2

mt t t A

Bi 16 Trong mt phng toOxy cho hai ng thng (d1) : 4x - 3y - 12 = 0 v (d2): 4x + 3y - 12 = 0Tm totm v bn knh ng trn ni tip tam gic c 3 cnh nm trn (d1), (d2), trc Oy.

Hng dn: - Gi A l giao ca 1 24 3 12 0

, : 3;0 Ox4 3 12 0

x yd d A A

x y

- V (BC) thuc Oy cho nn gi B l giao ca 1d vi Oy : cho x=0 suy ra y=-4 , B(0;-4) v C l giao ca

2d vi Oy : C(0;4 ) . Chng tB,C i xng nhau qua Ox , mt khc A nm trn Ox v vy tam gic AB

l tam gic cn nh A . Do tm I ng trn ni tip tam gic thuc Ox suy ra I(a;0).

- Theo tnh cht phn gic trong :5 5 4 9

4 4 4

IA AC IA IO OA

IO AO IO IO

4 4.3 4

9 9 3

OAIO . C ngha l I(

4;0

3)

- Tnh r bng cch : 5 8 51 1 15 1 1 18 6

. .5.32 2 2 2 2 15 5

AB BC CAS BC OA r

r r

.

Bi 17 Trong mt phng Oxy , cho hai ng trn : 2 21 : 13C x y v 2 2

2 : 6 25C x y ct

nhau ti A(2;3).Vit phng trnh ng thng i qua A v ct 1 2,C C theo hai dy cung c di bnhau

Hng dn:

- Tgithit : 1 2: 0;0 ,

13. ; 6;0 , ' 5C I R C J R

- Gi ng thng d qua A(2;3) c vc t chphng 2

; :3

x atu a b d

y bt

- d ct 1C ti A, B : 2 2 2 2 22 2

22 3

3 2 2 3 0

13

x ata b

y bt a b t a b t ta b

x y

2 2 2 2

2 3 3 2;

b b a a a bB

a b a b

. Tng td ct 2C ti A,C th ta ca A,C l nghim ca h:

2 2 2 22 2 2 2 2 2

2 2

22 4 3 10 6 2 3 8 3

3 ;

6 25

x ata b a ab b a ab b

y bt t Ca b a b a b

x y

- Nu 2 dy cung bng nhau th A l trung im ca A,C . T ta c phng trnh :

2 2 22

2 2 2 2

20 ; :

2 3 310 6 24 6 9 0

3 3; // ' 3;2

2 2

xa d

b ab y ta ab ba ab

a b a ba b u b b u


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Suy ra :2 3

:3 2

x td

y t

. Vy c 2 ng thng : d: x-2=0 v d': 2x-3y+5=0

Bi 18 Trong mt phng vi hta Oxy, cho im E(-1;0) v ng trn( C ): x

2+ y

28x4y16 = 0.

Vit phng trnh ng thng i qua im E ct ( C ) theo dy cung MN c di ngn nht.

Hng dn: - 2 2

: 4 2 36 4;2 , 6C x y I R

- Nhn xt : P/(M,C)=1+8-16=-7


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T(3) v (4) chng t(AB) c phng trnh l : 0 03 3 1 1 25 5x x y y - Theo githit th (AB) qua C(0;1) suy ra :

0 0 0 03 3 2 1 25 3 2 14 0(6)x y x y

- Kt hp vi (*) ta c h:0

0 0

0 0 0

13 22 6 0 16

; 1163 2 14 0 3

3

yx y

Mx y x

Bi 20 Trong mp Oxy, cho ng trn (C): x2

+ y

2

6x + 2y + 6 = 0 v im P(1;3).a.Vit phng trnh cc tip tuyn PE, PF ca ng trn (C), vi E, F l cc tip im.b.Tnh din tch tam gic PEF.Hng dn: - (C):

2 2

3 1 4 3; 1 , 2x y I R

- Gisng thng qua P c vc t php tuyn

; : 1 3 0n a b d a x b y Hay : ax+by-(a+3b)=0 (*).

- d l tip tuyn ca (C) th khong cch ttm I n d bng bn knh :

2 2 2 2

3 3 2 42 2a b a b a b

a b a b

2 2 2 22 4 3 0a b a b ab b

0 1 0 1 0

4 3 0 4 41 3 0 3 4 6 0

3 3

b a x x

b a bb a a x a y x y

-Ta c : PI=2 5 , PE=PF= 2 2

20 4 4PI R .Tam gic IEP ng dng vi IHF suy ra :

IF 2 5 IF 2 45 ,

IH 2 5 5 5 5

EP IP EPIH EH

EH IE

2 8 1 1 8 8 322 5 EF.PH=

2 2 55 5 5 5EPFPH PI IH S

Bi 21 Trong mpOxy, cho 2 ng thng d1: 2x + y 1 = 0, d2: 2x y + 2 = 0. Vit pt ng trn (C) ctm nm trn trc Ox ng thi tip xc vi d1v d2.

Hng dn: - Gi I(a;0) thuc Ox . Nu (C) tip xc vi 2 ng thng th : 1 2

1

, ,

,

h I d h I d

h I d R

2 1 2 21

5 5

2 12

5

a a

aR

. T(1) : a=1

4, thay vo (2) : R=

2

25 1 5:

10 4 100C x y

Bi 22 Cho ng trn (C): x2+ y22x 4y + 3 = 0. Lp pt ng trn (C) i xng vi (C) qua thng : x 2 = 0

Hng dn: Ta c (C): 2 2

1 2 2 1;2 , 2x y I R

I(3;-1)E

F

P(1;3)

O

x

y

H


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- Gi J l tm ca (C') th I v J i xng nhau qua d : x=2 suy ra J(3;2) v (C) c cng bn knh R . Vy

(C'): 2 2

3 2 2x y i xng vi (C) qua d .

Bi 23 Trong mt phng vi hta Oxy, cho ng thng d: x + y 3 = 0 v 2 im A(1; 1), B(3; Tm ta im M thuc ng thng d sao cho khong cch tM n ng thng AB bng 1.Hng dn: - M thuc d suy ra M(t;3-t) . ng thng (AB) qua A(1;1) v c vc t chphng

1 1

4; 3 : 3 4 4 04 3

x yu AB x y

- Theo u bi : 3 4 3 4 1 8 55

t tt

3 3;0

13 13; 10

t M

t M

* Ch :

ng thng d' song song vi (AB) c dng : 3x+4y+m=0 . Nu d' cch (AB) mt khong bng 1 th

h(A,d')=13 4

15

m

2 ' :3 4 2 0

12 ' :3 4 12 0

m d x y

m d x y

. Tm giao ca d' vi d ta tm c M .

Bi 24 Trong htrc 0xy, cho ng trn (C): x2+y2 -8x+12=0 v im E(4;1). Tm toim M trntrc tung sao cho tM kc 2 tip tuyn MA, MB n (C), vi A,B l cc tip im sao cho E thucng thng ABHng dn: - ng trn (C) :

2 24 4 2;0 , 2x y I R

- Gi M(0;a) thuc Oy . 1 1 2 2; , ;A x y B x y C - Tip tuyn ti A v B c phng trnh l :

1 1 2 24 4 4 , 4 4 4x x y y x x y y

- tha mn 2 tip tuyn ny cng qua M(0;a)

1 1 2 14 0 4 4 , 4 0 4 4x y a x y a .Chng t(AB) c phng trnh : -4(x-4)+ay=4- Nu (AB) qua E(4;1) : -4(0)+a.1=4 suy ra : a=4Vy trn Oy c M(0;4 ) tha mn .Bi 25 Viet phng trnh ng tron (C ) co ban knh R = 2 tiep xuc vi truc hoanh va co tam I namtren ng thang (d) : x + y 3 = 0.Hng dn: - Tm I nm trn d suy ra I(t;3-t) . Nu (C) tip xc vi Ox th khong cch tI n Ox

bng bn knh R=2 :

1

2

5 5; 23 2

3 2 3 2 1 1;2

t It

t t t I

-Nh vy c 2 ng trn : 2 2 2 2

1 2: 5 2 4 , : 1 2 4C x y C x y .

Bi 26 Trong mat phang Oxy cho ng tron (C) co phng trnh :x2+ y22x 6y + 6 = 0.

a. Viet phng trnh ng thang i qua M(2 ; 4) cat ng tron (C) tai 2 iem A, B sao cho M la trungiem oan AB.b. Viet phng trnh tiep tuyen cua (C) sao cho tiep tuyen ay song song vi ng thang co phng tr: 2x + 2y 7 = 0.

y

xI(4;0)O

1 E(4;1)

A

B

M

d'


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c. Chng to ng tron (C) va ng tron (C ) : x2+ y24x 6y + 4 = 0 tiep xuc nhau. Viet phngtrnh tiep tuyen chung cua chung tai tiep iem

Hng dn: - (C) : 2 2

1 3 4 1;3 , 2x y I R .

a. Gi A(x;y) thuc (C) suy ra 2 2

1 3 4x y (1) , B i xng vi A qua M suy ra B(4-x;8-y) .

m bo yu cu bi ton th B thuc (C) : 2 2

3 5 4x y (2).

- T(1) v (2) ta c h:

2 2 2 2

2 2 2 2

1 3 4 2 6 6 0 3

6 10 30 0 43 5 4

x y x y x y

x y x yx y

- Ly (3) -(4) ta c phng trnh : 4x+4y-24=0 , hay : x+y-6=0 . chnh lng thng cn tm .b. Gi d' l ng thng // vi d nn n c dng : 2x+2y+m=0 (*) . d' l tip tuyn ca (C) th :

4 2 82 6

, ' 2 8 4 28 4 2 8

mmh I d m

m

c. (C'): 2 2

2 3 9 ' 2;3 , ' 3x y I R

- Ta c : II'=1 , R'-R=1 . Chng thai ng trn tip xc trong vi nhau .

- Tm ta tip im :

2 2 2 2

2 2 2 2

1 3 4 2 6 6 02 2 0 1

4 6 4 02 3 9

x y x y x yx x

x y x yx y

.

Thay vo phng trnh u ca h: 22 6 9 0 3 0 3 1;3y y y y M .

- Tip tuyn chung qua M v vung gc vi IJ suy ra d': 1(x-1)=0 hay : x-1=0 .Bi 27 Trong mat phang Oxy cho (E) co phng trnh : 4x2+ 9y2= 36.a. . Cho 2 ng thang (D) : ax by = 0 va (D) : bx + ay = 0 (a2+ b2> 0). Tm giao iem E, F cua (D)vi (E) va giao iem P, Q cua (D) vi (E). Tnh dien tch t giac EPFQ theo a, b.b. Chng minh rng MPFQ lun ngoi tip mt ng trn cnh ? Vit phng trnh ng trn c.c. Cho iem M(1 ; 1). Viet phng trnh ng thang i qua M va cat (E) tai hai iem A, B sao cho Mtrung iem cua oan thang ABHng dn: a. Hai ng thng (D) v (D') vung gc nhau .

- (D) giao vi (E) ti E,F c ta l nghim ca h:

2

22 2 4 9 364 9 36

ax-by=0

byy

x y a

byx

a

2 2 2 2 2 2 2 2

6 6 6 6; , ;

9 4 9 4

9 4 9 4

b a b aE F

a b a b a b a b

- Tng t(D') ct (E) ti P,Q vi ta l nghim:

2

22 2 4 9 364 9 36

ax+by=0

by yx y a

byx

a

2 2 2 2 2 2 2 2

6 6 6 6; , ;

9 4 9 4 9 4 9 4

b a b aP Q

a b a b a b a b

- Tnh din tch tam gic EPFQ ;Bi 28 Trong mat phang toa o Oxy, cho ho ng thang phu thuoc tham so :(x 1)cos+ (y 1)sin1 = 0

a. Tm tap hp caciem cua mat phang khong thuoc bat ky ng thang nao cua ho.


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b. Chng minh moi ng thang cua ho eu tiep xuc vi mot ng tron co nh.Hng dn: b. Gi 0 0;I x y l im cnh . Khong cch tI n d c gi trl :

0 0 0 0

2 20 0

1 os + y -1 sin 1 1 0 111 1;1

1 0 11sin os

x c x xI

y yc

- Vi kt qutrn chng td lun tip xc vi ng trn (C) c tm I v bn knh bng 1 ( Khng ph

thuc vo . (C): 2 2

1 1 1x y

Bi 29 Cho hai im A(1;1), B(4;-3) v ng thng d : x-2y-1=0.a. Tm ta im C trn d sao cho khong cch tC n ng thng AB=6( HKB-04)b. Tm ta trc tm v tm ng trn ngoi tip tam gic OAB ?( HKA-2004)Hng dn:

a/ (AB) qua A(1;1) c 1 1

3; 4 : 4 3 7 03 4

x yu AB AB x y

- C thuc : x-2y-1=0 suy ra C(2t+1;t ) do : 4 2 1 3 7

6 11 3 305

t tt

1

2

3 7;3

27 43 27;

11 11 11

t C

t C

b/ - ng thng qua O vung gc vi AB c phng trnh : 3x-4y=0.- ng thng qua B v vung gc vi OA c phng trnh : (x-4)+(y+3)=0.- ng thng qua A v vung gc vi OB c phng trnh : 4(x-1)-3(y-1)=0hay : 4x-3y-1=0

- Vy ta trc tm H l nghim :

43 4 1 03 4 04 37

1 0 1 ;3 7 7

4 3 1 0 4 3 1 0 7

x xx y x

x y y x H

yx y x y

- Gisng trn ngoi tip tam gic (C): 2 2 2 2 0x y ax by c - (C) qua O(0;0) suy ra c=0 (1)

- (C) qua A(1;1) suy ra : 2-2a-2b=0 , hay : a+b=1 (2)- (C) qua B(4;-3) suy ra : 25-8a+6b=0 , hay : 8a-6b=25 (3)

- T(2) v (3) ta c h:

31 171

1 1 14 14

8 6 25 8 6(1 ) 25 31 31

14 14

b ba b b a

a b a aa a

- Vy (C) : 2 2 31 17 07 4x y x y

Bi 30 Trong mt phng Oxy cho im M(2;-1) v ng trn 2 21 : 9C x y (1) .Hy vit phng

trnh ng trn 2 :C c bn knh bng 4 v ct ng trn 1C theo dy cung qua M c di nhnh.

Hng dn: Gi 2 :C c tm I'(a;b) suy ra :

2 2 2 2 2 2

2 : 16 2 2 16 0 1C x a y b x y ax by a b

Ly (1) -(2) ta c : 2 22 2 7 0ax by a b ( chnh l ng thng trc ng phng )Dy cung ca hai ng trn nm trn ng thng ny .


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V dy cung qua M(2;-1) ln ta c : 2 22 24 2 7 0 2 1 12a b a b a b

Bi 31 . Trong mt phng vi hta Oxycho hai ng thng : 3 8 0x y , ' :3 4 10 0x y imA(-2 ; 1). Vit phng trnh ng trn c tm thuc ng thng , i qua imAv tip xc vng thng .Hng dn: Tm I ca ng trn thuc nn I(-3t8; t)

Theo yc th k/c t I n bng k/c IA nn ta c 2 22 2

3( 3 8) 4 10( 3 8 2) ( 1)

3 4

t tt t

Gii tip c t = -3Khi I(1; -3), R = 5 v pt cn tm: (x 1) + (y + 3) = 25.Bi 32 . Trong mt phng vi hta Oxycho ng trn hai ng trn :

2 2( ) : 2 2 1 0,C x y x y 2 2( ') : 4 5 0C x y x cng i quaM(1; 0).Vit phng trnh ng thng quaMct hai ng trn ( ), ( ')C C ln lt tiA, B sao choMA= 2MB.Hng dn: + Gi tm v bn knh ca (C), (C) ln lt lI(1; 1) ,I(-2; 0) v 1, ' 3R R , ng

thng (d) quaM c phng trnh 2 2( 1) ( 0) 0 0, ( 0)(*)a x b y ax by a a b .+ GiH, Hln lt l trung im caAM, BM.

Khi ta c: 2 2 2 22 2 ' ' 'MA MB IA IH I A I H 2 2

1 ( ; ) 4[9 ( '; ) ]d I d d I d ,

.IA IH

2 2

2 2

2 2 2 2

94 ( '; ) ( ; ) 35 4. 35

a bd I d d I d

a b a b

2 22 2

2 2

3635 36

a ba b

a b

Dthy 0b nn chn6

16

ab

a.

Kim tra iu kin IA IH ri thay vo (*) ta c hai ng thng thom n.Bi 33 . Trong mt phng vi hta Oxy, cho ng trn (C): x2+ y2- 2x - 2my + m2- 24 = 0 c tI v ng thng : mx + 4y = 0. Tm m bit ng thng ct ng trn (C) ti hai im phn bit Atha mn din tch tam gic IAB bng 12.Hng dn:

ng trn (C) c tm I(1; m), bn knh R = 5.Gi H l trung im ca dy cung AB.Ta c IH l ng cao ca tam gic IAB.

IH =2 2

| 4 | | 5 |( , )

16 16

m m md I

m m

22 2

2 2

(5 ) 2025

16 16

mAH IA IH

m m

Din tch tam gic IAB l 12 2 12SIAB IAHS

2

3

( , ). 12 25 | | 3( 16) 16

3

m

d I AH m mm

Bi 34 . Cho ng trn (C): x2+ y22x + 4y + 2 = 0.

Vit phng trnh ng trn (C') tm M(5, 1) bit (C') ct (C) ti cc im A, B sao cho 3AB .Hng dn:

Phng trnh ng trn (C): x2+ y22x + 4y + 2 = 0 c tm I(1,2) 3R ng trn (C') tm M ct ng trn (C) ti A, B nn AB IM ti trung im H ca on AB.

Ta c2

3

2

ABBHAH . C 2 v tr cho AB i xng qua tm I.

Gi A'B' l v tr th 2 ca AB, Gi H' l trung im ca A'B'

I

A BH

5


14/29

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Ta c:

2

2 2 3 3IH ' IH IA AH 3

2 2

,

2 2MI 5 1 1 2 5

Vy c 2 ng trn (C') tha ycbt l: (x 5)2+ (y1)2= 13 hay (x5)2+ (y1)2= 43

Bi 35 Trong (Oxy) cho ng trn (C): 2 2 3

2x y v 2:P y x . Tm trn (P) cc im M m t

kc 2 tip tuyn n (C) v 2 tip tuyn to vi nhau mt gc0

60 Hng dn: Gi M 20 0 0 0;x y P y x . d l ng thng tip tuyn ca (P) ti M th d c phng

trnh : 0 0 0 01

2 02

y y x x x y y x . d l tip tuyn ktM n (C) th iu kin cn v l

Bi 36 Trong (Oxy) cho . thng d: 3x-y+5=0 v ng trn (C): 2 2 2 6 9 0x y x y . Tm im Mthuc (C) v im N thuc d sao cho MN c di nhnht ?

Hng dn: (C) : 2 2

1 3 1 1;3 , 1x y I R

- Gi d' //d th d': 3x-y+m=0 . d' tip xc vi (C) ti M ( M l im cch d nhnht ) , khi :

6

10 ' :3 6 10 03 3; ' 1 6 10

10 6 10 ' :3 6 10 0

m d x ymh I d R m

m d x y

GisN' thuc d ta lun c : 2 2'M N M N . Du bng ch xy

ra khi N' trng vi N . Vy ta chcn lp ng thng quaI(-1;3) v vung gc vi d suy ra ng thng

1 3:

3

x t

y t

. Khi ct d' ti 2 im :

1

3 1 3 3 6 10 010

t t t .

V 1

3 1 3 3 6 10 0

10

t t t .

Do vy ta tm c 2 im M : 13 1

1;310 10

M

, v

2

3 11 ;3

10 10M

. Tng t ct d ti N c ta l nghim :

1 31 7 29

3 ;10 10 10

3 5 0

x t

y t t N

x y

. Ta chn M bng cch tnh 1 2,M N M N , sau so snh : Nu

1 2

M N M N th M l2

M . Cn1 2

M N M N th M l1

M .

Bi 37 Trong (Oxy) cho 2 2

: 1 3 1C x y v im1 7

;5 5

M

. Tm trn (C) im N sao cho M

c di ln nht ?

Hng dn: (C) vit di dng tham s: 1 sin

1 sin ;3 ost3 ost

x tN C N t c

y c

Khi :2 2

2 26 8 12 16sin ost sin os sin ost+45 5 5 5

MN t c t c t t c

d'

d'

d:3x-y+5=0

M

M

N

N'

I(-1;3)


15/29

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12 16 12 16

sin ost+5 5 4 sin ost *5 5 20 20

t c t c

. V :

2 212 16

120 20

,

12 3 16 4os ;sin =

20 5 20 5c th (*) trthnh : 5 4sin 5 4 1t

Du ng thc xy ra khi : sin 1 22

t t k

Do vy :3 3 2

sin sin os = 1 sin 12 5 5 5t c x t

Tng t:4 4 19 2 19

ost=cos sin 3 ost=3+ ;2 5 5 5 5 5

c y c N

Bi 38 Trong (Oxy) cho hai im 2 3;2 , 2 3; 2A B a/ Chng ttam gic OAB l tam gic ub/ Chng minh rng tp hp cc im M sao cho : 2 2 2 32MO MA MB l mt ng trn (C).c/ Chng t(C) l ng trn ngoi tip tam gic OAB.

Hng dn: a/ Ta c : 2

22

3 2 4, 4, 4OA OB AB . Chng tOAB l tam gic u .

b/ Gi M(x;y) th ng thc githit cho tng ng vi biu thc :Ta c : 2 2 2 2 2 2 2 2 2, 4 3 4 16, 4 3 4 16MO x y MA x y x y MB x y x y

2 2 2 2 2 2 2 8 332 3 3 8 3 32 32 0

3MO MA MB x y x x y x

2 2

24 3 4 3

3 3x y

. Chng tl ng trn (C) c tm

4 3 4 3;0 ,

3 3I R

c/ Thay ta O,A,B vo (1) ta thy tha mn , chng t(C) l ng trn ngoi tip tam gic OAB.Bi 39 Trong mt phng ta Oxy cho ng trn 2 2( ) : 1C x y , ng thng ( ) : 0d x y m .Tm m ( )C ct ( )d ti A v B sao cho din tch tam gic ABO ln nht.Hng dn: *(C) c tm O(0;0) , bn knh R=1 *(d) ct (C) ti hai im phn bit ( ; ) 1d O d

*Ta c1 1 1

. .sin .sin2 2 2

OABS OAOB AOB AOB T din tch tam gic AOB ln nht khi v ch

khi 090AOB 1

( ; )2

d I d 1m

Bi 40 Trong mt phng vi htoOxy, cho ng trn (C) c phng trnh: 2 2 4 3 4 0x y x Tia Oy ct (C) ti A. Lp phng trnh ng trn (C), bn knh R = 2 v tip xc ngoi vi (C) ti A.

Hng dn: A(0;2), I(-2 3 ;0), R= 4, gi (C) c tm I Pt ng thng IA :

2 3

2 2

x t

y t

, 'I IA => I

2 3 ;2 2t t ), 1

2 ' '( 3;3)2

AI I A t I (C): 2 2

3 3 4x y

Bi 41 Cho ng trn (C): x2+ y22x6y + 6 = 0 v im M (2;4)a) Vit phng trnh ng thng i qua M ct ng trn ti 2 im A v B, sao cho M l trung

im ca AB.b) Vit phng trnh cc tip tuyn ca ng trn, bit tip tuyn c hsgc k = -1.Hng dn: a. (C) : I(1; 3), R= 2, A, B ( )C , M l trung im AB => IM AB ng thng d c

tm l g thng AB; d i qua M c vect php tuyn l IM

=> d: x + y - 6 =0


16/29

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b. g thng tip tuyn c dng : y = - x + mx + ym =0 (d)d tip xc vi (C) ( ; ') 2d I d R

4 2 2

4 2 2

m

m

Pt tip tuyn :

(4 2 2) 0

(4 2 2) 0

x y

x y

Bi 42 Trong mt phng vi hta Oxy, cho ng trn (C): x2+ y2- 2x - 2my + m2- 24 = 0 c tmv ng thng : mx + 4y = 0. Tm m bit ng thng ct ng trn (C) ti hai im phn bit A,Btha mn din tch tam gic IAB bng 12.Hng dn: ng trn (C) c tm I(1; m), bn knh R = 5.Gi H l trung im ca dy cung AB.Ta c IH l ng cao ca tam gic IAB.

IH =2 2

| 4 | | 5 |( , )

16 16

m m md I

m m

22 2

2 2

(5 ) 2025

16 16

mAH IA IH

m m

Din tch tam gic IAB l 12 2 12SIAB IAHS

2

3

( , ). 12 25 | | 3( 16) 16

3

m

d I AH m m

m

Bi 43 Vit phng trnh ng trn i qua hai im A(2; 5), B(4;1) v tip xc vi ng thng cphng trnh 3x y + 9 = 0.

Hng dn:Gi ;I a b l tm ng trn ta c h

2 2 2 2

2

2 2

2 5 4 1 (1)

3 9; 2 5 210

a b a bIA IB

a bIA d I a b

1 2 3a b thvo (2) ta c 2 12 20 0 2 10b b b b *) vi 2 22 1; 10 : 1 2 10b a R C x y

*)vi 2 2

10 17; 250 : 17 10 250b a R C x y

Bi 44 Cho ng trn (C): x2+ y22x + 4y + 2 = 0. Vit phng trnh ng trn (C') tm M(5, 1) bi

(C') ct (C) ti cc im A, B sao cho 3AB .

Hng dn: Phng trnh ng trn (C): x2+ y22x + 4y + 2 = 0 c tm I(1,2) 3R ng trn (C') tm M ct ng trn (C) ti A, B nn AB IM ti trung im H ca on AB.

Ta c2

3

2

ABBHAH C 2 vtr cho AB i xng qua tm I. Gi A'B' l vtr th2 ca AB ;

Gi H' l trung im ca A'B' ; Ta c:

2

2 2 3 3IH ' IH IA AH 3

2 2

Ta c: 2 2

MI 5 1 1 2 5 v2

7

2

35HIMIMH ;

3 13MH ' MI H ' I 5

2 2

Ta c: 134

52

4

49

4

3MHAHMAR

22221

434

172

4

169

4

3'MH'H'A'MAR

22222

Vy c 2 ng trn (C') tha ycbt l: (x5)2+ (y1)2= 13hay (x5)2+ (y1)2= 43

I

A BH

5


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Bi 45 Trong mt phng vi h ta Oxy cho ng trn (C) c phng trnh (x-1)2+ (y+2)2= 9 v ng thng d: x + y + m = 0. Tm m trn ng thng d c duy nht mt im A m t k c hatip tuyn AB, AC ti ng trn (C) (B, C l hai tip im) sao cho tam gic ABC vung.Hng dn: T ph ng trnh chnh tc ca ng trn ta c tm I(1;-2), R = 3, t A k c 2 tip tuy

AB, AC ti ng trn v ACAB => t gic ABIC l hnh vung cnh bng 3 23IA

7

56123

2

1

m

mm

m

Bi 46 Trong mt phng toOxy cho hai ng thng (d1) : 4x - 3y - 12 = 0 v (d2): 4x + 3y - 12 = 0Tm totm v bn knh ng trn ni tip tam gic c 3 cnh nm trn (d1), (d2), trc Oy.Hng dn: Gi A l giao im d1v d2ta c A(3 ;0) ;Gi B l giao im d1vi trc Oy ta c B(0 ; - 4)Gi C l giao im d2vi Oy ta c C(0 ;4) ; Gi BI l ng phn gic trong gc B vi I thuc OA khi ta c I(4/3 ; 0), R = 4/3Bi 47 Cho im A(-1 ;0), B(1 ;2) v ng thng (d): x - y - 1 = 0. Lp phng trnh ng trn i qua

im A, B v tip xc vi ng thng (d).Hng dn: Gisphng trnh cn tm l (x-a)2+ (x-b)2= R2V ng trn i qua A, B v tip xc vi d nn ta c hphng trnh

2 2 2

2 2 2

2 2

(1 )

(1 ) (2 )

( 1) 2

a b R

a y R

a b R

2

0

1

2

a

b

R

Vy ng trn cn tm l: x2+ (y - 1)2= 2

Bi 48 Vit phng trnh tip tuyn chung ca hai ng trn :(C1) : (x - 5)

2+ (y + 12)

2= 225 v (C2) : (x1)

2+ ( y2)2= 25

TH2 : 5A12B + C = -3(A + 2B + C)4A 3B

C2

, thay vo (2) ta c : 96A2+ 28AB + 51B2=

Phng trnh ny v nghim . ( xong 420)Bi 49Trong hto Oxy cho hai ng thng 032: yxd v 053: yx . Lp phng

trnh ng trn c bn knh bng5

102, c tm thuc d v tip xc vi .

Hng dn: Tm ng trn thuc d nn c dng );32( aaI ng trn tip xc vi nn

RId ),(5

102

10

2

a2;6 aa

Vi 6a ta c )6;9(I suy ra phng trnh ng trn:5

8)6()9( 22 yx

Hng dn: ng trn (C1) c tm I1(5 ; -12) bn knh R1= 15 , ng trn (C2) c tm I2(1 ; 2) bnknh R1= 5 . Nu ng thng Ax + By + C = 0 (A

2+ B

2 0) l tip tuyn chung ca (C1) v (C2) th

khong cch tI1v I2n ng thng ln lt bng R1v R2, tc l :

2 2

2 2

5A 12B C15 1

A B

A 2B C

5 2A B

T(1) v (2) ta suy ra : | 5A12B + C | = 3| A + 2B + C |

Hay 5A12B + C = 3(A + 2B + C)TH1 : 5A12B + C = 3(A + 2B + C) C = A9B thay vo (2) :

|2A7B | = 5 2 2A B 2 221A 28AB 24B 0 14 10 7A B

21

Nu ta chn B= 21 th sc A = - 14 10 7 , C = 203 10 7

Vy c hai tip tuyn :(- 14 10 7 )x + 21y 203 10 7 = 0


18/29

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vi 2a ta c )2;7( I ,suy ra phng trnh ng trn:5

8)2()7( 22 yx

Vy c hai ng trn thomn l:5

8)6()9( 22 yx v

5

8)2()7( 22 yx .

Bi 50 Trong hto Oxy cho ng thng 022: yxd v ng trn 1)1()1(: 22 yxC

Lp phng trnh cc tip tuyn ca ng trn )(C bit tip tuyn to vi ng thng dmt gc 045 .

Hng dn:ng trn c tm )1;1(I bn knh 10R

Gi ),( ban l vect php tuyn ca tip tuyn )0( 22 ba ,

v ng thng to vi ng thng d mt gc bng 045 nn2

1

5.

2

22

ba

ba

ab

ba

3

3

Vi ba 3 , phng trnh tip tuyn c dng )(03 cyx

RId );( 1010

4

c

14

6

c

c

Vi ab 3 , phng trnh tip tuyn c dng )(03 cyx

RId );(

1010

2

c

12

8

c

c

Vy c bn tip tuyn cn tm l: ;063 yx 0143 yx ; ;083 yx 0123 yx .Bi 51 Trong mp vi htrc toOxy, cho ng trn (C): x2+ y26x2y + 5 = 0 v ng thng(3x + y3 = 0. Lp phng trnh tip tuyn vi ng trn(C),bit tp tuyn khng i qua gc toO vhp vi ng thng (d) mt gc 450

Hng dn:C) c tm I(3;1), R = 5 ; Tip tuyn (): ax + by + c = 0

( , ) 5

2os(d, ) =

2

d I d

c

==> 1: 2xy10 = 0; 2: x + 2y10 = 0

Bi 52 Cho ng trn (C): x2

+ y

2

6x2y + 1 = 0. Vit phng trnh ng thng (d) i qua M(0;2) ct (C) theo mt dy cung c di l = 4.Hng dn:d1: 2x + y2 = 0; d2: x2y + 4 = 0Bi 53 Trong mat phang vi he toa o ecac vuong goc Oxy, xet tam giac ABC vuong tai A, phng trnng thang BC la 3 3 0x y , cac nh A va B thuoc truc hoanh va ban knh ng tron noi tie

bang 2. Tm toa o trong tam G cua tam giac ABC.Hng dn:I l tm ng trn ngoi tip ABC ==> yI= 2

BI: y = tan300(x1) ==> y =

1 1

3 3x ==> 1 1 2 3.x

TH1: Neu A va O khac pha oi vi B 1 1 2 3x . ==> A(3 2 3 ;0)

==> 17 4 3 6 2 3

;3 3

G

TH2:Neu A va O cung pha oi vi B 1 1 2 3.x

==> A( 1 2 3

==> 24 3 1 6 2 3

;3 3

G

Bi 54 Trong mt phng vi hto ,Oxy cho tam gic ABCc )6;4(A , phng trnh cc ng thcha ng cao v trung tuyn ktnh Cln lt l 0132 yx v 029136 yx . Vit phtrnh ng trn ngoi tip tam gic ABC.Hng dn:


19/29

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- Gi ng cao v trung tuyn k t Cl CHvCM. Khi CHc ph ng trnh 0132 yx ,CM c ph ng trnh .029136 yx

- T h ).1;7(029136

0132

C

yx

yx

)2,1( CHAB

unCHAB .

- T h )5;6(029136

0162M

yx

yx

; (8; 4).B

Gi s ph ng trnh ng trn ngoi tip

.0: 22 pnymxyxABC VA,B, Cthuc ng trn nn

0750

04880

06452

pnm

pnm

pnm

72

6

4

p

n

m

.

Suy ra pt ng trn: 0726422

yxyx hay.85)3()2( 22 yx

Bi 55 Trong mt phng vi hta Oxy, cho ng trn ( C) cphng trnh 84222 yxyxv ng thng ( ) c phng trnh : 0132 yx . Chng minh rng ( ) lun ct ( C) ti hai imphn bit A, B . Tm toim M trn ng trn ( C) sao cho din tch tam gic ABM ln nht.

Hng dn: ng trn (C) c tm I(-1; 2), bn knh R = 13 .

Khong cch tI n ng thng ( ) l13

9),( Id < R ; Vy ng thng ( ) ct (C) ti hai im A

B phn bit. Gi M l im nm trn (C), ta c ),(.2

1 MABM dABS

Trong AB khng i nn ABMS ln nht khi ),( Md ln nht.

Gi d l ng thng i qua tm I v vung gc vi ( ).PT ng thng d l 3x + 2y - 1 = 0Gi P, Q l giao im ca ng thng d vi ng trn (C). ToP, Q l nghim ca hphng trnh

0123

084222

yx

yxyx

5,3

1,1

yx

yx P(1; -1); Q(-3; 5) ; Ta c

13

4),( Pd ;

13

22),( Qd

Ta thy ),( Md ln nht khi v chkhi M trng vi Q. Vy ta im M (-3; 5).

Bi 56 Trong mt phng vi hta Oxy, cho ng trn ( C): 024222 yxyx . Vit phn

trnh ng trn ( C') tm M (5, 1) bit ( C') ct ( C) ti hai imA , B sao cho 3AB .

Hng dn: Phng trnh ng trn (C): x2

+ y2

2x + 4y + 2 = 0 c tm I(1,2), bk 3R ng trn (C') tm M ct ng trn (C) ti A, B nn AB IM ti trung im H ca on AB.

Ta c2

3

2

ABBHAH

Gi Trng hp 1:C 2 vtr cho AB i xng qua tm I. Gi A'B' l vtr th2 ca AB

Gi H' l trung im ca A'B' ; Ta c:

2

2 2 3 3IH ' IH IA AH 3

2 2

M(6; 5)

A(4; 6)

C(-7; -1)

B(8; 4)

H


20/29

www.nguoithay.o


Ta c: 2 2

MI 5 1 1 2 5 v2

7

2

35HIMIMH ;

3 13MH ' MI H ' I 5

2 2

Ta c: 134

52

4

49

4

3MHAHMAR

22221

434

172

4

169

4

3'MH'H'A'MAR

22222

Vy c 2 ng trn (C') tha ycbt l: (x5)2+ (y1)2= 13 hay (x5)2+ (y1)2= 43

Bi 57 Trong mt phng vi htoOxycho tam gic ABCc tm ng trn ngoi tip (4;0)I ,ng cao v ng trung tuyn ktA c phng trnh ln lt l 2 0x y v 2 3 0x y

Tm tocc nh , , .A B C

Hng dn: Ta Al nghim ca h2 0 1

(1;1)2 3 0 1

x y xA

x y y

.

GiMl trung im caBC.DoIl tm ng trn ngoi tip ABC

/ /IM BC IM AH (1;1)IM AHn n

Pt ng thngIMl 1( 4) 1( 0) 0 4 0x y x y .DoMl trung im caBCnnM thuc trung tuyn ktA

Ta Ml nghim ca h4 0

(5; 1)2 3 0

x yM

x y

.

(1;1)BC AHBC AH n u .

PtBCl 1( 5) 1( 1) 0 6 0x y x y . Ta c 2 2(1 4) 1

10IA

Phng trnh ng trn ngoi tip tam gicABCl 2 2( 4) 10x y .

Ta BCl nghim ca h2 2 3( 4) 10

36 0

xx y

yx y

hoc7

1

x

y

.

Vy (3; 3)B th (7;1)C , hoc (7;1)B th (3; 3)C

Bi 58 Trong mt phng to Oxy cho ng trn (C) c ph ng trnh 2 2 2 4 20 0x y x y . T im M(2; 4) k cc tip tuyn n ng trn (C), gi cc tip im l T1v T2. Vit ph ng trnh ng thng T1T2.

Hng dn: ng trn c tm I(1; -2), bn knh R = 5. C 2 2 2(2 1) (4 2) 37 5IM IM =R Gi

T(x; y) l mt tip im , c ( 2; 4)MT x y , ( 1; 2)IT x y c

2 2. 0 3 2 6 0 (1)MT IT x y x y

T(C) nn 2 2 2 4 20 0 (2)x y x y

(1)(2) 6 14 0x y =>T thuc ng thng d c ph ng trnh x + 6y14 = 0 Do vai tr ca T1v

T2nh nhau nn d l ng thng i qua T1T2.Bi 59 Trong mt phng to Oxy cho ng trn (C) tm I c phng trnh 2 2 2 4 20 0x y x y . Vitph ng trnh ng thng i qua im M(8; 0), ct ng trn (C) ti hai im A, B sao cho tam gic IAB c dintch ln nht.Hng dn: ng trn c tm I(1; -2), bn knh R = 5.


21/29

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21 1 . .sin .sin2 2

IABS IA IB I R I suy ra IAB c din tch ln nht khisinI= 1 0 90I , IAB vung c

suy ra5

( , ) ( , )

2 2

Rd I AB d I ng thng qua A(8; 0) c ph ng trnh : a(x8) +by = 0,

2 2 0a b 2 2

5 | 7 2 | 5( ; )

2 2

a bd I

a b

2 2... 73 56 17 0a ab b a=b hoc 73a = -17b

+) nu a = b chn a = b = 1, ng thng c pt: x + y - 8 =0

+) nu 73a = -17b chn a = 17, b = -73, ng thng c pt: 17x -73y136 = 0Bi 60 Trong mt phng vi hta Oxy, cho ng trn ( C) c phng trnh 084222 yxyx v

ng thng ( ) c phng trnh : 0132 yx . Chng minh rng ( ) lun ct (C) ti hai im phn bit A

. Tm toim M trn ng trn ( C) sao cho din tch tam gic ABM ln nht.

Hng dn: ng trn (C) c tm I(-1; 2), bn knh R = 13 .

Khong cch tI n ng thng ( ) l13

9),( Id < R ;Vy ng thng ( ) ct (C) ti hai im A, B phn

bit.Gi M l im nm trn (C), ta c ),(.2

1 MABM dABS Trong AB khng i nn ABMS ln nht khi

),( Md ln nht.Gi d l ng thng i qua tm I v vung gc vi ( ).PT ng thng d l 3x + 2y - 1 = 0

Gi P, Q l giao im ca ng thng d vi ng trn (C). ToP, Q l nghim ca hphng trnh:

0123

084222

yx

yxyx

5,3

1,1

yx

yx P(1; -1); Q(-3; 5)

Ta c

13

4),( Pd ;

13

22),( Qd Ta thy ),( Md ln nht khi v chkhi M trng vi Q.

Vy ta im M (-3; 5).hay l ng thng i qua M v vung gc vi AM. ; PT ng thng : x + y - 2 = 0

Bi 61 Trong mt phng vi hta Oxy, cho ng trn ( C): 024222

yxyx . Vit phngtrnh ng trn ( C') tm M (5, 1) bit (C') ct ( C) ti hai imA , B sao cho 3AB .

Hng dn: Phng trnh ng trn (C): x2+ y22x + 4y + 2 = 0 c tm I(1,2), bk 3R

ng trn (C') tm M ct ng trn (C) ti A, B nn AB IM ti trung im H ca on AB.

Ta c2

3

2

ABBHAH ; Gi

Trng hp 1: C 2 vtr cho AB i xng qua tm I.Gi A'B' l vtr th2 ca AB ; Gi H' l trung im ca A'B'

Ta c:

2

2 2

3 3

IH ' IH IA AH 3 2 2

Ta c: 2 2

MI 5 1 1 2 5 v2

7

2

35HIMIMH ;

3 13MH ' MI H ' I 5

2 2

Ta c: 134

52

4

49

4

3MHAHMAR

22221

434

172

4

169

4

3'MH'H'A'MAR

22222

Vy c 2 ng trn (C') tha ycbt l: (x5)2+ (y1)2= 13 hay (x5)2+ (y1)2= 43


22/29

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Bi 62 Trong mt phng ta Oxy, cho im 2;1A v ng trn (C): 2 2

1 2 5.x y Vit phn

trnh ng thng d qua A ct ng trn (C) ti hai im phn bit B, C sao cho on thng BC ngn nht.Hng dn: Kim tra im A ta thy A nm trong ng trn (C).

Khi PA/(C)= 2 2. . 3AB AC AB AC IA R . Suy ra AB.AC=3.

Theo BT AM-GM ta c 2 . 2

3BC AB AC AB AC .

ng thc xy ra khi A l trung im ca BC. ng thng d l qua A(2;1) nhn (1; 1)IA l vect ph

tuyn. Vy phng trnh ng thng d l x-y-1=0.Bi 63 Cho ABC c din tch bng 3/2; A(2;3), B(3;2), trng tm G (d) 3xy8 =0. tm bn kinh ng

trn ni tip ABC.

Hng dn: Cho ABC c din tch bng 3/2; A(2;3), B(3;2), trng tm G (d) 3xy8 =0. tm bn kinhng trn ni tip ABC.

C(a; b) , (AB): xy5 =0 d(C; AB) =5 2

2

ABCa b S

AB

8(1)

5 32(2)

a ba b

a b

Trng tm G 5 5;3 3

a b (d) 3ab =4 (3); (1), (3)C(2; 10) r = 3

2 65 89

S

p

(2), (3) C(1;1) 32 2 5

Srp

Bi 64 Trong mt phng vi hto ,Oxy cho tam gic ABCc )6;4(A , phng trnh cc ng thng cha

ng cao v trung tuyn ktnh Cln lt l 0132 yx v 029136 yx . Vit phng trnh

ng trn ngoi tip tam gic ABC.

Hng dn:- Gi ng cao v trung tuyn k t Cl CHv CM. Khi

CHc ph ng trnh 0132 yx ,

CM c ph ng trnh .029136 yx - T h

).1;7(029136

0132

C

yxyx

- )2,1( CHAB

unCHAB

0162: yxABpt .

- T h )5;6(029136

0162M

yx

yx

).4;8(B

- Gi s ph ng trnh ng trn ngoi tip .0: 22 pnymxyxABC

VA,B, Cthuc ng trn nn

075004880

06452

pnmpnm

pnm

72

6

4

pn

m

.

Suy ra pt ng trn: 0726422 yxyx hay .85)3()2( 22 yx

Bi 65 Trong mt phng vi hta Oxy, cho hai im A(1 ; 2), B(1 ; 6) v ng trn(C): (x - 2)

2+ (y - 1)

2= 2. Lp phng trnh ng trn (C) qua B v tip xc vi (C) ti A.

Hng dn: (C) c tm I(2 ; 1) v phng trnh ca ng thng AI: x + y3 = 0.Pt ca (C) : x2+ y2+ 2ax + 2by + c = 0 c tm I(-a ; -b)A(1 ; 2), B(1 ; 6) thuc (C) v tm I thuc ng thng AI. Ta c hphng trnh:

M(6; 5)A(4; 6)

C(-7; -1)

B(8; 4)H


23/29

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03

37122

542

ba

cba

cba

, gii hc a = 1, b = -4, c = 9; Pt ca (C) : x2+ y2+ 2x8y + 9 = 0

Bi 66 Trong mt phng vi htrc ta Oxy cho A(4;3), ng thng (d) : xy2 = 0 v (d): x + y 4 =

ct nhau ti M. Tm ( ) ( ')B d v C d sao cho A l tm ng trn ngoi tip tam gic MBC.

Hng dn:

Tm ta M l giao ca d v d.Tm ta H,K l hnh chiu ca A trn d v dT tm c B thuc d v C thuc d l ccim i xng ca M qua H,K

Bi 67 Trong mt phng (Oxy), cho ng trn (C ): 2 22x 2y 7x 2 0 v hai im A(-2; 0), B(4; 3). Vitphng trnh cc tip tuyn ca (C ) ti cc giao im ca (C ) vi ng thng AB.


Hng dn:+ ng trn(C ) :2

2 2 2 2 27 7 652x 2y 7x 2 0 x y x 1 0 x y

2 4 16

(C ) c tm7

I ;04

v bn knh65

R4

+ ng thng AB vi A(-2; 0) v B(4; 3) c phng trnhx 2 y x 2

y6 3 2

, hay :

+ Giao im ca (C ) vi ng thng AB c ta l nghim hPT2

2 2 2 x 2 5x(x 2) 02x 2y 7x 2 0 2x 2 7x 2 0x 0; y 12

x 2x 2x 2; y 2x 2

22

2

y =y =y =

Vy c hai giao im l M(0; 1) v N(2; 2)

+ Cc tip tuyn ca (C ) ti M v N ln lt nhn cc vect7

IM ;14

v

1IN ;2

4

lm cc vect php

tuyn , do cc tip tuyn c phng trnh ln lt l :

7

(x 0) 1(y 1) 0 7x 4y 4 04

, hay :

1

(x 2) 2(y 2) 0 x 8y 18 04

, hay :

Bi 68 Trong mt phng Oxy, cho im 1;3A nm ngoi (C): 2 2 6 2 6 0x y x y .Vit phng trnh ng thng d qua A ct (C) ti hai im B v C sao cho AB=BC

K

H

C

B

A(4;3)

d'

d

M


24/29

www.nguoithay.o


Hng dn: Theo yu cu bi ton , ,A B C thng hng v AB=BC.Gi2 1

( ; ), ( ; )2 1

m aB a b C m n

n b

.

Do B, C nm trn (C) nn2 2

2 2

3

6 2 6 0 1

56 2 6 0

1

a

a b a b b

mm n m n

n

hoc

7

5

15

9

5

13

5

a

b

m

n

.

Vy c hai ng thng tha mn yu cu bi ton l x+y-4=0 v 7x+y-10=0.

Bi 69 Cho hai ng trn : (C1) : x2+ y24x +2y4 = 0 v (C2) : x

2+ y2-10x -6y +30 = 0

c tm ln lt l I, Ja/.Chng minh (C1) tip xc ngoi vi (C2) v tm ta tip im H .

b/.Gi (d) l mt tip tuyn chung khng i qua H ca (C1) v (C2) . Tm ta giao im K ca (d) v ng thnIJ . Vit phng trnh ng trn (C) i qua K v tip xc vi hai ng trn (C1) v (C2) ti H .


Hng dn: a).Chng minh (C1) tip xc ngoi vi (C2) v tm ta tip im H .(C1) c tm I( 2; -1) v bn knh R1= 3 . (C2) c tm J(5;3) v bn knh R=2.

Ta c: IJ2= ( 52)2+ ( 3 + 1)2= 25 IJ = 5 = R1+ R2Suy ra (C1) v (C2) tip xc ngoi vi nhau . Ta tip im H c xc

nh bi :

HI H J H

I H J HH

19x2 x x 3 x x 52HI 3HJ

72 y y 3 y yy

5

b).Gi (d) l mt tip tuyn chung khng i qua H ca (C1) v (C2) . Tm ta giao im K ca (d) v ng thIJ . Vit phng trnh ng trn (C) i qua K v tip xc vi hai ng trn (C1) v (C2) ti H .

C : 2KI 3KJ

I K J K K

KI K J K

2 x x 3 x x x 11

y 112 y y 3 y y

ng trn (C) qua K , tip xc vi (C1) , (C2) ti H nn tm E ca (C) l trung im ca KH :37 31

E ;5 5

.

Bn knh (C) l EH = 6; Ph ng trnh ca (C) l :

237 31

x y 365 5

Bi 70 Trong mt phng Oxy cho ng trn (C) tmI(-1; 1), bn knhR=1,Ml mt im trn

( ) : 2 0d x y . Hai tip tuyn quaMto vi (d) mt gc 450tip xc vi (C) tiA, B. Vit phng trnh

thngAB.


Hng dn: Dthy ( )I d . Hai tip tuyn hp vi (d) mt gc 450suy ra tam gicMABvung cn v tam gi

IAMcng vung cn . Suy ra: 2IM .


25/29

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( ) (M d M a; a+2), ( 1; 1)IM a a ,0

2 2 1 22

aIM a

a

.

Suy ra c 2 im tha mn:M1(0; 2) vM2(-2; 0).

+ ng trn tmM1bn kinhR1=1 l (C1):2 2 4 3 0x y y .Khi ABi qua giao im ca (C ) v (C1) n

AB: 2 2 2 24 3 2 2 1 1 0x y y x y x y x y .

+ ng trn tmM2bn kinhR2=1 l (C2):2 2 4 3 0x y x .Khi ABi qua giao im ca (C ) v (C2) n

AB: 2 2 2 24 3 2 2 1 1 0x y x x y x y x y .

+ KL: Vy c hai ng thng tha mn: 1 0x y v 1 0x y

Bi 71 .Trong mt phng vi hta Oxy cho ng trn (C) c phng trnh (x-1)2+ (y+2)2= 9 v ngthng d: x + y + m = 0. Tm m trn ng thng d c duy nht mt im A m t kc hai tip tuyn ABAC ti ng trn (C) (B, C l hai tip im) sao cho tam gic ABC vung.Hng dn: Tphng trnh chnh tc ca ng trn ta c tm I(1;-2), R = 3, tA kc 2 tip tuyn AB, A

ti ng trn v ACAB => tgic ABIC l hnh vung cnh bng 3 23IA

513 2 1 6

72

mmm

m

Bi 72 Vit ph ng trnh ng trn i qua hai im A( 2;5 ), B9 4; 1) v tip xc vi ng thng c ph ng tr093 yx .

Hng dn: Hai ng trn tho mn bi c phng trnh:

2501017:;1021: 22222

1 yxCyxC

Bi 73 Trong mt phng vi htoOxy, cho ng trn (C) : 2 2x y 2x 8y 8 0 . Vit

phng trnh ng thng song song vi ng thng d: 3x+y-2=0 v ct ng trn theo mt dy cung c dibng 6.

Hng dn:ng trn (C) c tm I(-1;4), bn knh R=5; Gi phng trnh ng thng cn tm l ,

=> : 3x+y+c=0, c2 (v // vi ng thng 3x+y-2=0)V ng thng ct ng trn theo mt dy cung c di bng 6=> khong cch ttm I n bng

2 25 3 4 2

4

10 13 4, 4

3 1 4 10 1

ccd I

c

(tha mn c2)

Vy phng trnh ng trn cn tm l: 3 4 10 1 0x y hoc 3 4 10 1 0x y .

Bi 74 Trong mt phng toOxy , cho ng trn ( C) : 2 2 2 6 15 0x y x y v ng thng (d) :

3 0mx y m ( ml tham s). Gi I l tm ca ng trn . Tm mng thng (d) ct (C) ti 2 im ph

bit A,B thomn chu vi IAB bng 5(2 2) .

Hng dn: tgithit ta tm c khong cch tI(1;-3) n ng thng (d) l

5 2

2, t tm ra m?

Bi 75 Trong mt phng Oxy cho ng trn (C) tmI(-1; 1), bn knhR=1,Ml mt im trn( ) : 2 0d x y . Hai tip tuyn quaMto vi (d) mt gc 450tip xc vi (C) tiA, B. Vit phng trnh thngAB.Hng dn: Dthy ( )I d . Hai tip tuyn hp vi (d) mt gc 450suy ra tam gicMABvung cn v tam gi

IAMcng vung cn . Suy ra: 2IM .

( ) (M d M a; a+2), ( 1; 1)IM a a ,0

2 2 1 22

aIM a

a

.


26/29

www.nguoithay.o


Suy ra c 2 im tha mn:M1(0; 2) vM2(-2; 0).

+ ng trn tmM1bn kinhR1=1 l (C1):2 2 4 3 0x y y .Khi ABi qua giao im ca (C ) v (C1

nnAB: 2 2 2 24 3 2 2 1 1 0x y y x y x y x y .

+ ng trn tmM2bn kinhR2=1 l (C2):2 2 4 3 0x y x .Khi ABi qua giao im ca (C ) v (C2

nnAB: 2 2 2 24 3 2 2 1 1 0x y x x y x y x y .

+ KL: Vy c hai ng thng tha mn: 1 0x y v 1 0x y .

Bi 76 Trong mat phang vi he toa o Oxy cho ng tron (C) : x2+ y2+ 4x + 4y + 6 = 0 va ng thang : xmy 2m + 3 = 0 vi m la tham so thc. Goi I la tam cua ng tron (C). Tm m e cat (C) tai 2 iem phan biA va B sao cho dien tch IAB ln nhat.

Hng dn: (C) : x2+ y2+ 4x + 4y + 6 = 0 co tam la I (-2; -2); R = 2

Gia s cat (C) tai hai iem phan biet A, B. Ke ng cao IH cua ABC, ta co

SABC=1

IA.IB.sinAIB2

= sin AIB

Do o SABCln nhat khi va ch khi sin AIB= 1 AIB vuong tai I

IH =IA

1

2

(thoa IH < R) 2

1 4m1

m 1

1 8m + 16m2= m2+ 1 15m28m = 0 m = 0 hay m =8

15

Bi 77 Trong mt phng vi htrc toOxy cho A(4; 3), ng thng (d):xy2 = 0v (d): x + y 4 = 0 ct nhau ti M. Tm B(d); C(d) sao cho A l tm ng trnngoi tip tam gic MBC.

Hng dn: M(3;1), Ly B(a; 2a)(d) C(b;4b) (d); V (d) (d)==> A l trung im BC : B(6;4), C(2;2)

Bi 78 Trong mt phng vi hta Oxycho hai ng thng : 3 8 0x y , ' :3 4 10 0x y

v

imA(-2 ; 1). Vit phng trnh ng trn c tm thuc ng thng , i qua imAv tip xc vi ngthng .Hng dn: Tm I ca ng trn thuc nn I(-3t8; t); Theo yc th k/c tI n bng k/c IA nn ta c

2 2

2 2

3( 3 8) 4 10( 3 8 2) ( 1)

3 4

t tt t ; Gii tip c t = -3 th I(1;-3) v IA=5 ..

Bi 79 Trong mpOxy , cho ng trn (C) c tm I (2;4 ), bn knh R= 2 2 v ng thng (D) c phng trn

7 0x y .Tm toim M nm trn (D) tM kti ng trn (C) hai tip tuyn

MA,MB ( A,B l hai tip im ) sao cho tam gic IAB c din tch ln nht.

Hng dn: khoang cach t I en dng thang (D) la 9 22

. Dien tch tam giac IAB la

1. . ( ,( ))

2S AB d I D . at IM=x th 9 2

2x

d=

9 2

2

K

x

M

B

A

I


27/29

www.nguoithay.o


2 2

2

3

2

8 8 8; 8 . 8 8 1 ( )

9 2

2

16 9 2'( ) 8. ; .

282. 1

9 2min ( )

2

th KA KB x S x f xx x x

vi x

co f x x

xx

Suy ra f x f

Cc bi tluynBi 80 Trong mt phng 0xychng minh rng ng trn 2 2 2 2( ) : 2 4 4 0mC x y m x my m lun tip x

vi 2 ng cnh m ta phi chr.Bi 81 Trong mt phng ta Oxy cho ng trn (C): x2+ y22x 4y 6 = 0. Gi (C) l ng trn tm I2 ; 3) v ct ng trn (C) ti hai im A, B sao cho AB = 2. Vit phng trnh ng thng AB.

Bi 82 Trong mt phng vi hta Oxy cho im A(3; 5) v ng trn (C): x2+y2+ 2x 4y 4 = 0. TA kcc tip tuyn AM, AN n (C) (M, N l tip im). Vit phng trnh MN v tnh khong cch gia hai im M, Bi 83 . Trong mt phng vi hta Oxy cho im M(4; 1) v ng trn (C): x2+y22x 3 = 0. Vit

phng trnh ng thng (d) i qua M v ct (C) theo mt dy cung c di bng 2 2 .Bi 84 Trong mt phng ta Oxy cho ng trn (C): x2+ y2+8x 6y = 0 v ng thng (d): 3x4x+10 = 0.Vit phng trnh ng thng vung gc vi (d) v ct (C) ti hai im A, B tha AB = 6.Bi 85 . Trong mt phng vi h ta Oxy, cho ng trn (C) c phng trnh: x2+ y26x + 5 = 0. Tm imM thuc trc tung sao cho qua M k c hai tip tuyn vi (C) m gc gia hai tip tuyn bng 600.

Bi 86 Trong mt phng vi htoOxy, cho ng thng d: 2 0x y v ng trn (C): 2 2 5x y . Ttoim M thuc ng thng d m qua kc hai tip tuyn MA, MB ti (C) (A, B l cc tip im) saocho tam gic MAB u.

Bi 87 Trong mt phng vi htoOxy, cho tam gic ABC ni tip ng trn (C): 2 21 2 5x y , A

0), 090ABC v din tch tam gic ABC bng 4. Tm tocc nh A, B, C.

Bi 88 Trong mt phng vi htoOxy, cho ng trn (C): 2 22 4x y . Gi I l tm ca (C).Tm to

im M c tung dng thuc (C) sao cho tam gic OIM c din tch bng 3 .

Bi 89 Trong mt phng vi h ta Oxy, cho ng trn (C): x2+ y2+ 4x + 6y +5 = 0 v hai ng thng 1: 2y 6 = 0, 2: x + y = 0. Tm im A thuc 1v im B thuc (C) sao cho A v B i vi xng nhau qua 2.Bi 90 Trong mt phng vi h ta Oxy, cho im M(5;1) v ng trn (C): x2+ y24x + 6y 3 = 0. Vit

phng trnh tip tuyn ca (C) sao cho khong cch t M n tip tuyn ln nht.Bi 91 (KD - 07)Trong mt phng vi h to Oxy, cho ng trn (C) :

( x - 1 )2

+ ( y + 2 )2

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

Bi 92 (DBKA - 07)Trong mt phng vi h to Oxy ,cho ng trn (C) : x2+y2= 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.

Bi 93 (DBKB - 07)Cho ng trn (C) : x2+ y2-8x +6y +21 = 0 v ng thng d : x + y -1 = 0.

Xc nh to cc nh ca hnh vung ABCD ngoi tip (C) ,bit A thuc d.

Bi 94.(DBKB - 07)Cho ng trn C: x2+y2-2x+4y+2 = 0. vit phng trnh ng trn (C') tm M(5;1) ,bit (C

ct (C) ti cc im A,B sao cho AB = 3

Bi 95(KB - 06) (KB - 06) Trong mt phng vi h to Oxy ,Cho ng trn (C) : x2+y2-2x -6y +6 = 0 v i

M(-3;1).Gi T1v T2l cc tip im ca cc tip tuyn k t M n (C) .Vit phng trnh ng thng T1T2.


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Bi 96(KD- 06) Trong mt phng vi h ta Oxy, cho ng trn (C): x2+ y2- 2x - 2y + 1 = 0 v ng thnd: x-y+3=0. Tm ta im M nm trn d sao cho ng trn tm M, c bn knh gp i bn knh ng trn (Ctip xc ngoi vi ng trn (C).

Bi 97(DBKD- 06) Trong mt phng vi h to Oxy ,cho ng thng d: x -y +1- 2 = 0 v imA(-1;1).Vit phng trnh ng trn (C) i qua A,gc to O v tip xc vi ng thng d.

Bi 98(DBKA- 05)Trong mt phng vi h to Oxy cho ng trn (C1):x2+y2-12x-4y+36 = 0. Vit phng

trnh ng trn (C2) tip xc vi hai trc to Ox ,Oy ,ng thi tip xc vi ng trn (C1).

Bi 99(DBKA- 05)Trong mt phng vi h to Oxy cho ng trn (C) : x2+y2-4x-6y -12 = 0. Gi I l

tm v R l bn knh ca (C) .Tm to im M thuc ng thng d: 2x -y +3 = 0 sao cho MI = 2 R.

Bi 100(DBKD- 05)Trong mt phng vi h ta Oxy cho hai ng trn :

(C1): x2+y2= 9 v (C2) : x

2+y2-2x -2y -23 =0.Vit phng trnh trc ng phng d ca hai ng trn (C1) v (C2).Tm to im K thuc d sao cho khongcch t K n tm (C1) bng 5.

Bi 101 (DB -KB-03)Trong mt phng vi h to Oxy cho ng thng d: x -7y +10 = 0.Vit phng trnh

ng trn c tm thuc ng thng : 2x + y = 0 v tip xc vi ng thng d ti im A(4;2).

Bi 102(CT -KD-03)Trong mt phng vi h ta cac vung gc Oxy cho ng trn :

(C): (x-1)2+ (y-2)2= 4 v ng thng d: x - y1 = 0.Vit ph-ng trnh -ng trn (C) i xng vi -ng trn (C) qua -ng thng d. Tmta cc giao im ca (C

v (C).

Bi 103(DB -KA-02)Trong mt phng vi h to 0xy cho ng thng d: x-y+1=0 v ng trn

(C) :x2+y2+2x- 4y = 0. Tm to im M truc ng thng d m qua ta k c hai ng thng tip xc vi

(C) ti A v B sao cho gc AMB =600.

Bi 104(DB -KB-02)Trong mt phng vi h to Oxy cho hai ng trn

(C1) : x2+y2-4y -5 = 0 v (C2) : x

2+y2-6x +8y +16 = 0.Vit ph ng trnh cc tip tuyn chung ca hai ng trn (C1) v (C2).

.Bi 104 (DB -KD-02)Trong mt phng vi h to Oxy cho hai ng trn

(C1) : x2+y2-10x =0 , (C2) : x

2+y2+4x -2y -20 = 0.1.(DB -KD-02)Vit ph ng trnh ng trn i qua cc giao im ca (C1) ,(C2) v c tm nm trn ng thng x +6y -6 = 0.2.(DB -KD-02)Vit phng trnh tip tuyn chung ca cc ng trn (C1),(C2).

Bi 105(CT -KA-09) Trong khng gian vi htoOxyz, cho mt phng P : 2x 2y z 4 0 v mt c

2 2 2S : x y z 2x 4y 6z 11 0 . Chng minh rng mt phng (P) ct mt cu (S) theo mt ng trn.

Bi 106 CT -KA-09) Trong mt phng vi h to Oxy, cho ng trn 2 2C : x y 4x 4y 6 0

ng thng : x my 2m 3 0 , vi m l tham sthc. Gi I l tm ca ng trn (C). Tm m ct

ti hai im phn bit A v B sao cho din tch tam gic IAB ln nht.

Bi 107 (CT -KB-09) Trong mt phng vi htoOxy, cho ng trn (C) : 2 2 4

(x 2) y

5

v hai ng

thng 1: xy = 0, 2: x7y = 0. Xc nh totm K v tnh bn knh ca ng trn (C1); bit ng trn

(C1) tip xc vi cc ng thng 1, 2v tm K thuc ng trn (C)

Bi 108(CT -KD-09) Trong mt phng vi hta Oxy, cho ng trn (C) : (x1)2+ y2= 1. Gi I l tm

(C). Xc nh ta im M thuc (C) sao cho IMO = 300.

Bi 109 (CT -KA-10) Trong mt phng ta Oxy, cho hai ng thng d1: 3 0 x y v d2: 3 0x y . G

(T) l ng trn tip xc vi d1ti A, ct d2ti 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

3

2v im A c honh dng.


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Bi 110 (CT -KA-11) Trong mt phng ta Oxy, cho ng thng : x+ y + 2 = 0 v ng trn (C): x2 + y2

2y = 0. Gi I l tm ca (C), M l im thuc . Qua M kcc ng tip tuyn MA, MB n (C) ( A, B l cc t

im). Tm ta im M, bit tgic MAIB c din tch bng 10.

Bi 111(CT -KB-11) Trong mt phng ta Oxy, cho tam gic ABC c nh B1

;12

. ng trn ni tip t

gic ABC tip xc vi cc cnh BC, CA, AB tng ng ti cc im D, E, F. Cho D (3; 1) v ng thngc phng trnh y 3 = 0. Tm ta nh A, bit A c tung dng.

Bi 112(CT -KD-11) Trong mt phng ta Oxy, cho im A(1; 0) v ng trn (C) : x2+ y22x + 4y

0. Vit phng trnh ng thng ct (C) ti im M v N sao cho tam gic AMN vung cn ti A.

Bi 113Trong mt phng to Oxy, cho tam gic ABC c nh A(3;-7), trc tm l H(3;-1), tm ng trn ngtip l I(-2;0). Xc nh to nh C, bit C c honh dng.

Bi 114Cho phng trnh ng trn (Cm) : x2+y2-4mx -2(m+1)y = 1.

a)Tm qu tch tm cc ng trn (Cm)b)Chng minh rng qu tch tip xc vi parabol (p) : y2=2x.

Bi 115Trong mt phng toOxy cho ng trn (T): 2 2 2 5 0x y x y v ng thng (d):

3x+4y-5=0. Chng minh rng (d) ct (T) ti hai im phn bit B, C. Tm trn (T) im A c honh sao cho tam gic ABC c bn knh ng trn ni tip r=1.Bi 116 Trong mt phng toOxy, cho tam gic ABC vung ti A, B(-5;0), C(7;0), bn knh ng

trn ni tip tam gic l 2

13 6r . Tm totm I ca vng trn ni tip tam gic bit im I c tun

dng.

Bi 117 Trong mt phng vi hta Oxy, cho hai ng trn 13yx:)C( 22

v

25y)6x(:)'C(22 . Gi A l mt giao im ca )C( v )'C( vi 0

Ay . Vit phng trnh ng

thng (d)i qua A v ct )'C(),C( theo hai dy cung c di bng nhau (hai dy cung ny khc nhau).

Bi 114. Trong mt phng vi h to Oxy cho ng trn (C) : x2+y2-4x-6y -12 = 0.

Gi I l tm v R l bn knh ca (C) .Tm to im M thuc ng thng

d: 2x -y +3 = 0 sao cho MI = 2 R. ...........///...........

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