Sự tương giao đường thẳng và đường tròn

GIUPBANQNTAp
•
~Sl1Ji'lf(jNC
CIAO
GIllA DUONG
THANG vA
DUONG
TRON
•
B
ai toan vi su' tuong giao giiia duong
thdng va duong trim thuang xudt hien
trong cdc di thi tuyen sinh vao Dai h9C, Cao
acing va co thi tcng dung di
giai
mot s6
bai
todn dai s6.
I. LiTHUYET
Cho dirong trim (C) co tam 1, ban kinh R
va
dirong thang L'l. D~t h
=
d (I, L'l).
TrU'Ong hQ1>
h
=
R.
L'l
la tiep tuyen cua
dirong trim (C)

M~ T-~~4
I
Hinh 1
N~u tir di~m
M
ngoai (C) ke hai tiep tuyen
MA
va
ME d~n duong trim (C) (h. 1) thi
• L'lMAl
=
L'lMEl, 1M2=R2 +
AM-
=R2 +BM2.
• AB
1
1M tai trung di~m H cua AB.
• SIAMB
=
2S
IAM
=
Al.AM
=
AH.IM.
·:·TrU'Ong hQ1>h < R.
L'l
dt (C) tai hai diem
phan biet
P

va
Q
(h. 2 .
Hinh 2
GQi
H
la trung diem cua doan PQ thi
• Tam giac IPQ can tai 1, SIPQ
= ~
R2 sinPIQ.
NGUYEN TUAN
LAM
(GV THPT Thanh Nhan, TP H6 Chi Minh)
• IH l PQ, R2
=
IH2 + HP2
=
IH2 + PQ2 .
4
• DQ dai doan PQ
IOn
nhelt khi
L'l
di qua tam
1 cua dirong tron (C).
.:. Truong hQ'P
h
>
R.
L'l

va (C) khong co di~m
chung.
• Tir mot diem belt ki tren
L'l
luon ke duoc hai
tiep tuy€n d~n (C). .
• Lely diem
K
n~m tren duong tron (C) thi
h - R ~ d(K,L'l) ~ h + R .
II.
cAc
THi
DV
MINH HQA
.:. Truong hQ1>L'l ti~p xuc
VOl
(C)
* Thi du 1.
Cho duong
tron
(C):
.xl
+
Y
=
2.
n«
phuong trinh tii
p

tuyen
L'l
cua (C) sao
cho L'l ctit cac tia Ox, Oy fdn luot tai A, B va
dien tich tam giac OAB nho nhat.
Lai
giiii. Duong tron (C) co tam trung g6c toa
dQ
0
va ban kinh R =
fi.
Ti~p tuyen
L'l
qua A(a;O) ,B(O;b)(a>O, b>O)
co
P'T
x +y =l<=>bx+ay-ab=O .
a
b
Ta co d( O,L'l)
=
R d,
J
labl
= fi
a
2
+b
2
<=>

ab
=
~2(a2 +b
2
) ~ 2M
=>
ab ~ 4, nen
SOAB
=
!OA.GB
=
!ab ~ 2. Dang thirc xay ra
2 2
khi a=b=2. V~yPT L'l: x+y-2=O.D
*Thi dl}2.
(Ciiu VI.a.! Khbi B 2009)
Cho duong tron (C): (x -
2)2
+ y2
=.±
va cac
5
diarng thang L'l1:X-y=O,L'l2:X-7y=O.
Xac dinh toa d(J tam K va tinh ban kinh cua
duong iron
(C
l
),
biit rang duang iron
(C

l
)
tii
p
xuc vai cac duong thdng L'll ,L'l2 va tam
K thuoc duang iron (C).
lrC:>~~
tic:><: ,~ • : ~~~;;
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Lai
giai.
GQi
K(a;b)E(C)~(a-2)2 +b
2
=~;
(C]) ti~p XUC fl],fl2 ~ la ~bl
=
la ~bl .
"\12
5 2
T
'd' ,
{5(a-2)2 +5b
2
=4
ir
0
ta co

5l
a
-
b
l
=
la-
7b
l·
Gifti
M
nay ta duoc (a;
b)
= ( ~; ~ ).
Ban
kinh dirong trim (C]) la R
=
la ~bl
=
2J2 .0
"\12 5
*Thi
d1}.3.
(ellu
V.a.l
Kh8i B 2006)
Cho duong trim (C): x
2
+y2 -2x-6y+6=0
va diim

M(
-3; 1). G9i TJ va T2 fa cac tiep
diim cua cac tii
p
tuyen ke tit M din (C).
Viit phuong trinh duong thdng TJT
2
.
Liri giai. Duong tron (C) c6 tam 1(1;3) ban
kinh R
=
2 ,IM =
215
>
R nen M
nfun ngoai
(C).
Ta c6 MI;
=
MI;.
=.J MJ2 - R2
=
4
nen T;,T2
thuoc dirong tron
(c,)
c6 tam
M
va ban kinh
R'=4.

PT dirong tron (C') la r+y+6x-2y-6=0.
Tir d6 (C) ciit
(C,)
tai hai di~m
T;
,T2 .
Xet he PT {X2 +y2 - 2x - 6y +6
=
0
. x
2
+y2 +6x - 2y - 6
=
0
~ 2x+ y-3
=
0 (1). Do T;,T2 la giao diem
cua (C) va
(c,)
nen toa dQ cua cac diem
T;
,T2 thoa man d~ng thirc (1).
Do d6 PT duong thang T;T
2
la 2x+y-3=0.0
*Thi
d1} 4. Cho
duirng trim
(Q:(x_l)2 +(y_2)2=4
va diim

A
(2; 1).
DU'Cmg thdng
d thay dJi di
qua A ciit (C) tai hai diim T] va
h
Hai tiep
tuyen cua (C) tai hai diim T] va T2 cdt nhau
tai diim M Tim quy tich cua diim M
Liri giai: Duong tron (C) c6 tam 1(1; 2) ban
kinh R = 2, lA
=
J2 < R nen di~m A (y trong
duong tron (C), suy ra dirong thang d qua A
luon ciit (C) tai hai di~m
T]
va
T
2
.
Gift su M (xo;Yo), PT dirong tron (C') tam
M va ban kinh R'
=
MJj
=
.JMJ2 _R2 la
x
2
+y2 - 2xox- 2yoY +2xo +4yo -1
=

O.
Tuong tir Till du 3, ta tim diroc PT T;T2:
(2xo - 2)x+ (2yo -4)y - 2xo -4yo +
2
=
O.
Do duong thang
T;
T2di qua
A
(2; 1) nen
(2xo -2).2+(2Yo-4).1-2xo -4Yo+2=0
~ Xo - Yo - 3
=
0 (2)
Di~m M c6 toa dQ thoa man (2), suy ra quy
tich di~m Mia dirong thang x - y - 3
=
0
.0
*Thi
d1}.5.
Cho duong trim (C): x
2
+y2
=
4
va duong thiing
fl :
x +y -

6
=
O.
Gia su Mfa
diim thuoc duong thang
fl.
a) Chung minh rang tit M luon ke duac hai
tiip tuyen MJj , MT2 din (C) (T;
va
T2 fa hai
tiip diim)
va
duong thdng T;T
2
luon di qua
mot diim
cd
dinh.
b) Tim toa d6 diim M di doan thdng T;T
2
co
~A
dai b ~
815
ao ai
ang
. 5
c) Tim vi tri diim M di dien tich tu giac
OT;MT2 dat gia tri nho nhdt.
Liri gidi: a) Duong tron (C) c6 tam 0(0; 0) va

R
=
2, ME
fl,
d(O,fl)
=
3J2 > R nen Mn~m
ngoai, duOn~ tr~n. Vi vay
ill
M
luon ke diroc
hai tiep tuyen den duong tron (C).
Do
Met:
nen
M(t,6-tj, Me =MI2 =.JMJ2 _R2 .
Tuong tu Thi du 3, ta c6
T;,
T2 I~ giao diem
cua dirong tron
(C)
va dirong tron
(C')
(C'
c6
tam M
va
ban kinh R'
=
MJj ).

Tir d6 ta c6 PT cua duong thang T;T
2
la
tx+(6-t)y-4=0.
Tir day suy ra dirong thang T;T
2
luon di qua
d
·;:. ;.
dinh
H(2 2)
rem
co.
3;3 .
b) Tac6
SOTlM72
=l T;T
2
.GM=0T;.MJj
2
= R JM02 - R2. Suy ra
1r<:>~r.
tic:><: ~
~~~ji
&
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4J5
O'M

=
2.J M02 - 4
=>
OM
=
2J5
5
¢::>
t
2
+
(6-
t)2
=
20
¢::>
t
=
4 hoac
t
=
2 .
V~y co hai diem
M,
(4;2) va M2 (2;4).
c) Ta co
S01\MI2
=
OTj.MY;
=

R JM02 - R2
= 2.J M02 - 4. Tir do
S01\MI2
nho nhat khi
dQ dai MO nho nhat. Khi do diem M la hinh
chien cua 0 len ~, suy ra M(3;3).0
*Thi dl}
6. (Ciiu V.a.2 Kh6i D 2007)
Cho duong trim (C): (x_l)2 +(y+2)2 =9
va
dutmg thdng
t:.:
3x -
4
y + m
=
O.
Tim m
di tren
t:.
co duy nhdt mot diim P ma tir do
co thi ke duac hai tiep tuyen PA, PB d~n
duong trim (C) (A, B la cac tiep diim) sao
cho tam giac PAB d~u.
Loi
gidi.
Duong trim (C) co tam I
(1;-2)
va
ban kinh R

=
3. Tam giac PAB deu nen IP
=
21A
=
2R
=
6 nen P thuoc duong trim
(c,)
tam
I
ban kinh R'
=
6.
Tren ~ co duy nhat di€m
P
thoa man yeu du
bai toan
¢::>
d(J,~)
=
6
=>
m
=
19;m
=
-41.0
·:·TfU'Ong hQP ~ dt (C) tlili hai di~m
phan bi~t

*Thi dl}7.
Cho
iIuOng tMng ~:1n(+y-2m-l=O
ill duong trim
(C) :
(x_l)2 +(y - 2)2
=
4.
a) Tim m di ~ cat (C) tai hai diim phdn
biet A
va
B sao cho d(J dai doan thang AB
ngan nhdt.
b) Tim quy tich trung diem H cua doan
thang AB khi duong thang ~ thay d6i.
Lai giai.
a) Duong trim (C) co tam I
(1;2)
va
ban kinh R = 2. Ta thfty ~ luon di qua di€m
M(2;1) va 1M
= J2
< R nen M
a
trong (C),
suy ra ~ .luon c~t (C) tai hai diem phan biet
A,B.
G9i
H
Ia trung diem cua doan AB va

d(J,~)
=
IH. DQ dai doan AB ngan nhat khi
IH dai nhat. Do IH l AB va M
E
AB nen
IH ~ 1M
=>
IH
max
=
1M. Khi ~ .L 1M , suy ra
m
=
-1.
V~y PT ~: x - y
-1
=
0.
b) Di€m H la trung di€m cua doan AB nen
IH .L HM. Do do H nam tren duong trim
(e')
co dirong kinh la 1M.
PT dirong trim
(e'):
(x-~J
+~-~J
=~.O
*Thi du 8.
Xet hai

s6
thuc x,
y
thoa
man
x
2
+4x +y2 -
5
=
O.
Tim gia tri Ian
nhdt,
va
gia tri nho nhdt cua biiu thiec T
=
3x + 4y.
Lai giai.
Tren mat phang toa dQ Oxy, Ifty
di€m M(X,Y)E~:3x+4y-T=0.
Hai s6 x,y thoa man x
2
+ 4x + y2 - 5
=
0
nen M(X,Y)E(C):X2+4x+y2-5=0 la dirong
tron
tam
1(-2;0) va
ban kinh R

=
3.
MIa diem chung cua
(e)
va
t:.
¢::>d(l,~) ~R
¢::>
16
+
TI ~
15
¢::>
-21 ~
T <
9. D~ng thirc xay
ra khi ~ Ia tiep tuyen cua (C).
Nhtr v~y max(T)
=
9 khi
{
X2 +4x+ y2
-5
=
0
¢::>
{x
=-~
3x +4y - 9
=

0 12
y=-
5
rnin(T)
=
-21 khi
{
X2 +4x+ y2
-5 ~0
¢::>
{x
=
_1:
3x+4y+21=0 y=-g.D
5
*Thi dl}9.
(Ciiu VI.h.1 Kh6iA 2009)
Cho Quang tron
(C):
X2+y2+4x+4y+6=;0
va
duong thang ~: x + my - 2m +
3
=
O.
G9i I
la
tam duong tron (C). Tim m d~ ~ dt (C)
tai hai di~m ph an biet A, B sao cho dien tich
tam giac lAB Ian nhdt.

Lili
giiii.
Duong tron
(C)
co tam
1(-2;-2) va
1 R2
R
=
J2.
Ta co
SlAB
=-IA.IB.sinAIB~- =1.
2 2
loAN HOC.• • •.
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Yay
SlAB
IOn nh~t khi IA1JB=>d(I,tl) = ~ =1
. ~2
1-2-2m-2m+31
8
¢::>
=1=>m=O
hoac m= D
Jl

+nr- .
15
*Thi du 10.
(Cau
VI.h.1 KhaiD20JI)
Cho diim A(O;I) va
duang iron
(C):
X2 +y2 - 2x +4y -
5
=
O.
n«
phuong trinh
duong thdng
tl
eftt (C) tai hai diim M va N
saG cho tam giac AMN vuong can tai A.
Loi
giai. Duong trim (C) co tam 1(1;-2)
ve
R
=
J1O;
IA
=
(0;-2). Ta co IM= IN
va
AM
=

AN
=>
AI
1
MN nen PT
tl:
y
=
m .
G9i hai giao diem M (Xl;m), N (X2;m).
Khi do Xl va X2 la nghiem cua PT
X2 - 2x +
m?
+4m - 5
=
0 (3)
De PT (3) co hai nghiem Xl;X2 phan biet thi
m
2
+4m-6<0 (4)
AM
1
AN
¢::>
AM.AN
=
0
¢::>
(Xl -1)( X2-1) +m
2

=
0
¢::>
XlX2- (XI + X2) + 1+
m?
=
0
Ap dung dinh li Viete d6i voi PT (3), SUY ra
2m
2
+ 4m - 6 = 0
¢::>
m = - 3 hoac m = 1,
thoa man (4).
V~yphuongtrinh
tl:
y
=
1 hoac
tl:
y
=
-3.0
*Thi du 11.
(Cau
VI.a.1 Khai
A
2010)
Cho hai duong thdng d,: 13.x + y
=

0 va
d
2
:13.x - y
=
O. G9i (1) fa duang
tron
ti~p
xuc vai d
l
tai A, eftt d
2
tai hai diem B,
C
saG
cho tam giac ABC vuong tai B.
n«
phuong
trinh cua (1), bi~t rdng tam giac ABC co dien
, h b ~
13
'd';: A
i
h ' h ~~
d.
tic
ang
2
va iem co oan
ao

uong.
un gidi. Ta co A
E
d,
=>
A(
a ;-aJ3)
(a> 0).
Tir AC.l
a.
=>
AC: x-13y-4a
=
0;
C
=
d
2
(\
AC
=>
c(
-2a; - 2J3a);
BA.l d
2
=>BA:x+13y+2a=0;
B~d, nBA=>
+~;_
a~}
Ta co

S
ABC
=!
BA.BC
=
13
¢::>
13a.3a
=
13
2 2
=>a= ~ =>A(~;-I} C(~;-2}
Duong tron
(1)
co tam 1(-
2~
;-%)
(Ila
trung diem cua A C) va ban kinh R
=
IA
=
1.
Tir do PT cua duong tron (1) la
(x+
2~)'
+(Y+H
~LD
.:. Truimg hQ1>
tl

khong
dt
(C)
*Thi du 12.
Cho duong trim
(C):r+y2-2x
-2y
-7
=
0
va duong thdng tl:3x+4y+13=0.
Tim
gia
tri
tan
nhdt va gia tri nho nhdt cua
khoang each tir
mot aiim
M tren (C)
a~n
duong thdng
tl.
Loi
giiii.
Duong tron (C) co tam 1(1; 1)va
ban kinh R
=
3; d(I,tl)=4>R nen dirong
thang
tl

khong
d.t
duong tron (C).
Ta vi~t PT tiep tuyen cua (C) va song song
voi
tl.
Co hai tiep tuyen la
tll
:3x+4y+8=0
voi tiep diem
M,
( 4 ;-7)
va ~ :3x+4y- 22=0
55'
r ,~
d'~
M(1417)
VO'I tiep tern 2
5;5 .
Khi d6 d(tl,tll)=I,d(tl,tl2)=7. Voi diem
M
thuoc (C) suy ra
l=d(tl,tll):S;d(M,tl):S;d(tl,tl2)=7.
Nhu vay
, (14 17)
max d(M,
tl)
=
7
khi M

=-
M2
5;5 .
,
,(~~)
mmd(M,tl) =1 khi M
=-
MJ
5;5
.0
ToAN HOC •••••.•••••~,.••••.•.•••••••••••
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Nh~n xet.
Hai diem MJ va M2 la giao diem
cua (C) va dirong thang d qua tam I, vuong
goc voi
f:: ;
maxd(M,f:: )
=
d(I,f:: )+ R,
mind(M,f:: )
=
d(I,/),.)-R.
*Thi d1}13.Cho
duong trim
(Q:r+j-a-t4y+l=<l
va

duang thdng thay d6i
f:: :
1m+y-2m-2=0.
Tim m d~ khodng each
nho
nh6t tir di~m M th
QC
(C) din duang thdng
f::
dat
gia
tri
fern
nh6t.
Liri
gidi:
(h. 3)
N
Hinh 3
Duong trim (C) co tam
1(1;-2)
va
R
=
2.
GQi h la khoang each nho nhat ill M dSn
f::
Ta th~y
f::
luon di qua di~m N(2;2)

a
ngoai
duong tron (C).
• Truong h9P
f::
cat
(C)
thi h
=
O.
• Truong h9P
f::
khong clit (C)
GQi H la chan dirong vuong goc ha ill I xuong
f:: ,
A la giao diem cua doan IH va (C) thi
h=HA.
R5 rang la h dat gia tri Ian nhat khi H
==
N .
Khi do, h = IN - R =
JU -
2
va
luc nay
f::
1.
IN re do suy ra m
= ~
.0

4
*Thi d1}14. Xet cac
s6
thuc a, b, c, d thoa
man a2+b2-2a+2b-23=0
va
3c-4d=-23.
Tim gia tri nho nh6t cua bi~u thuc
T= (a-c)2 +(b-dr
Lai
giiii. Xet diem
A
(a;
b)
thuoc duong tron
(C): x2+y2-2x+2y-23=0 co tam
1(1;-1)
va R = 5; di~m B (c ;d) thuoc duong thang
f:: :3x-4y+23=0 thi T= (a-c/ +(b-d)2 =AIJ2.
Ta co d(I,f:: )
=
6 > R nen
f::
khong c~t (C).
T dat gia tri nho nh~t khi doan AB ngan nhat.
Theo Thi
du
l
Zi
ta co minAB,=d(I,f:: )-R=l.

Khi do B la hinh chieu cua I len
f:: ,
A leigiao
diem cua doan thang IB va duong tron (C).
-13 19
A
Tir
do
a
=
-2;
b
=
3,' c
= -'
d
=
Yay
5' 5 .
minT
=
1.0
BAIT~P
1. Cho duong tron (C): X2+ y2 - 6x -10
=
O.
Duong thang d qua diem A dt (C) tai hai
di~mM, N.
a) ViSt phuong trinh duong th~ngd trong cac
tnrong hQ'Pdoan

MN
nho nhat, Ian nhat.
b) Tim quy tich trung diem
H
cua doan
MN.
A
,{1m+3
y
+m+3=0
2. Cho h~ phirong tnnh X2+
y2 _
2x-15
=
0
(m la tham so).
a) Chung minh h~ da cho co hai nghiem
phan biet,
b) GQi (Xl; Yl )va (X2; Y2) la hai nghiem cua
he. Tim gia tri Ian nh~t, gia tri nho nh~t cua
bi~u tlnrc
F
=
(Xl - X2)2 +(Yl - Y2)2.
3. Cho duong thang
f:: :
X
+
Y
+ 2

=
0 va
duong tron (C): X2
-ta
y2 - 4x - 2
Y
=
O. GQi Ila
tam cua (C), M la d~~m thuoc dirong thang
f::
Qua Mke cac tiep tuyen MA va ME dSn (C) (A,
B la cac tiSp diem). Tim toa dQ diem M, biSt
rang ill giac MAlB co dien tich bang 10.
4.
Cho duong tron (C):X2
+y2
-8x+6y+21=0
va dirong thang d:
X
+Y -1
=
0 . Xac dinh toa
dQ cac dinh hinh vuong ABCD ngoai tiep
duong tron (C), biSt rang diem A nam tren d.
5. Cho dirong tron(C):r+y-4x-6y-12=O.
Tim toa dQ diem M thuoc duong thang :
d: 2x - Y + 3
=
0 sao cho MI = 2R, trong do I
la tam va RIa ban kinh cua duong tron (C).

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