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8/15/2019 JEE Questions Circle
http://slidepdf.com/reader/full/jee-questions-circle 1/15
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http://slidepdf.com/reader/full/jee-questions-circle 5/15
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8/15/2019 JEE Questions Circle
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.
I ~ = R1 11, R
1
(11
1
~ R +R
3
)
(11,+11,+11,)'
The [';iven cHck
;,
2x(.r-a)
4
y(2y- l r l=O
or
xlr-a)+ y(J•-h. 2)=0
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x
2
-
1
J
" ' byl2=0
...(I)
Lot
oftlw cborJ
thmu 'h
(a. h 12)
he
Oi,ccld
at ( lr. n).
I
hen the equation
of
the
chord
having {ir.
01 as
mid poim
T=S
1
+O· v - - ( x
. '
" '
/ 1
'>"+0)- it·
-0-
<1h -0
.
·I
.
I
' o \ hy a ,
11-- •x h=h
-air
' 2; 4 2
...(2)
'Jo\\. (2) \\
l ra."
Li>rough ("·
h
12) if
.,
> ) b b
a
..
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>
1 - a
~ a · .
~ - a 1
.)
2 2
'
'
3
Q b-
h - . - - a / J + - + - · =II
"
...
3)
Accordtng to the eiven c o n d i l i ~ n . 11)
rHU't
IM\C two
dJ>ltnd real
roms
I his i< p o s > J ~ l e if lhe rli<o l iminanl nl"
(31
is greater than 0.
That;,,
i l
" , ~ · · a b ]
" - · l2+H Al
o· _ h' >
- _ ,
4
l
Tile equation ol.thc
Lirde an
the li:>c
joining
the pomt<
3(3
7) and /J 6.
5) '
d ;>meter
; ,
(x-3) (x 6 1 (I ' 7)(l - -5)-11 ...(I)
and the c ~ u a - i o n ni" Ihe lin.• e ' " ' " ~ t."e pointsA 13. 7) "nd
R (0. 5)
is
" 7-5
J -1= - < ' l
1 - 0
2x 13y-27=0
...
(2)
Now tho cqnotwr. oC t;,,.,, J' , , t . c i r ~ k ' possing th·c'.Lg.lJ the
r<>in ol"tntcrsecticn of ( I ) and {2) is
S , ) ,P-0
=>
l.t 3)(>. -6) - (y-7) ( •
-5)
·1 . (2x+Jy-C7)-0
· - 6x -3x 113+y ' -5y-
7)
1
>.<+
) _ , + 3 1 - . " - n J . ~ o
0
s,=x·T,
l- '(2 ' ,-9)+y(:li .-12J
f (5
-27 .)=0 ...
(J)
Again tlte circle. which
Cli S
the member.;
of
Glmil)
drcks
is
s, =x'- y '- ,J.< - 6 y - J n ..
•nd the eq1mtion ofcommon chord to Circles S
1
and S,
s
-s,-o
=>
{x(2/,
?) ' l ' (J) . 12)+(53-271.)}
- ( - 4 x - 6 y - 3
xC i . -9 1 ) r ; J
+(13-27), .. )
2./o,x -5x
+
31-.l
-6y+56
-27),
=I)
=> ( - 5 x - 6 y + 5 6 1 + ) . ( 2 , + 3 y - 2 7 ) ~ 0
w 1ich
rcpre<ents equations ofrvm straight lines pa.,
through the fixed point whose coordinates are obtai
by solving the tvco eqllations
5r 1Gy-56=0
ond2x + 3y-27 =0
Soh-ing for
x
andy, we get
x
= 2 and
y
= 23 I 3.
Tv.o cJrcies touch
cac 1
otht'rcxtcrn"lly lf(
1
c,
=
•;
and i n ~ r n a l l '
i f r
, I '
0
- "· - r
0
and ~ / - • h - 2 y = - 1
(giv
=>
c
1
(2.1) an•
r
1
=I
, ,
and
x - + y - - 1 2 . ~ - 8 1 = - 3 6
(giv
C,(fi·1) and r , = ~
The di>tancc bcm-ccn t h ~ centre< is
2) ' - (4
ThereiC>re. l ~ e c i t t b touch each other
ntorrolly
ond
the p0mt
nt"
" ' " " " ' ~ : •he > "' :' ':• ·he line joi:ling
two
c e n t r ~ s
i n t e n n l l ~ in the ratio ofthctr radii. I ; 4.
lx<i•-1-x2
TI1ercfore.
x
1
=
I -
4 5
h4-4 , ·1
R
I + ~
Agmn
lD
detellllin• the equation<
of common
tange
touching the eire "' m Ji>Linc\ poinlo. we lnow that
tangems pass
thr<H•gh
a
po11n
"·l1ich di,·ides
'he
joining tlle
\Wo
centre' cxtemally
in
the ratio of<heir m
i
e. I :
4.
lx6·--h2 -2 2
x, =
=
. I
-4
-3 3
ence.
and J , ~ l x 4 - 4 x i _
, ' '
\ow,
lel"'
be the slope
~ f t t e
t : u t ~ ; e m
Th''
l i n ~
r ~ " '
through (2
.'
3. io
~ . -G=m{x-213)
,
y m x = m ~ O
3
but for
any
an gent. p =r is >a tstied.
Thcrcfore. -
2
m+(
2
/
31
" ' 1 rc,
=(2.l)andr,
~ l l
}1 +
m
8/15/2019 JEE Questions Circle
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2
~
T1
1e
re
for
o,
o
f
B
Elt
C (
r
c
os
2
13
. rs
in
A
nd
s
lo p
e
o
f A
D
I
S
H
li\
W
-
a
).
Le
t
(x .
y)
be
th
e
co
o
rd
in
a te
s
o
f th
e
p
oin
t D.
E
A
D
i
s
y
=t
a
n
(
)-
u )
(x
<-
r)
( ·
:
•l
op
e
(
fJ -
a
) a
nd
po
in
t
i
s (
-
r 0
))
N
o
w
,
Eq
u
at
io
n
o
f
B
C
; ;
rsm2jl-O
( )
Y-
x
-
r
r
oo
.>
21
}-
r
r
·2
s
in
jl
CO
S I}
y
=
--
x
-
r)
r(
-
2s
in
'
P
l
y
~
2
si
nP
c
os
p
x
-
r)
-
2s
in
2
P
y
=
-O
O
IP
x-
r)
T
o
ob
ta
in
t
he
co
o
rd
in a
te
o
f
D
so
l•
•
(
l)
s
im
u
lt
an
eo
us
ly
8/15/2019 JEE Questions Circle
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=> t a t l
j l - a )
x + r ) ~ -
o t j > ( x
- r )
=:.
l ] l - a
r t o l
a
~
1
=>
-<[tan
{D - a
)H o l
w
r [col
p
-ta
n (j' l- a
)]
~
rsin(
j3-a)
s ~ ]
[' 'p
,;n(
j3-o:)l
x
co ,U
l ,;_-)
_ _ ~ r
sin
j3 -cos (p
- a )
~
[sin (
il u)s
inP+eo
s(j} u
)cosj3]
cos j
3 -a ) ,;n
j3
r rO O
<
j
3 COS
(p • 0:
)
-Sln
jJ s
in
(fl_
-_a )I
'-
s
mjlco>
(j>-a)
x [cos
J> -
a )
cos
j3 + sin
(j
3 -
c )
sin
131
-r
[cos(j3
-a)oo
sj)
sinfls
m
k[cos
(j3 a
Pll=rfc
os(fl u I
fil
l
rw
s(2j3 a
)
x
-
c
osa
P
urring th
is value
in
(2 ), w
e get
-
v
=-cot
jJ' rcos(2
j: l-o:)
- r
L c
oso
J
::::
;, Y- co •
j3 r lcos
(2J>
a
)
co
sa I
""j\
cnsa
.
~ y=
- rco
s
p
l-2;in
2j3
; -a ,
;n a -2
1-t-a
l
sinp
co
sa
i
J
=>
,
_
_
s [ J [ 2 s
n ~ - s i n
Pll
·
.ini3
c
osa
~
- 2 r c o s
~ - > i n
(a
- jl)
i
co
' a
l h
ere fore.
coo rdin
ate ' of
D .m :
[
r en
• (
2jl
- )
_
2':_
cos_Jl
si
n
(«
-I. _)_ 'J
cos
a
cos a
T
hus, coo
mm:rtes
of
E
'
1
r e ~ > : ~
- ~ _ : < _
_ o o _ > _ a . _
r
co
s jhin
(a
--
j
lJ)
2cosa
cos
a
a
j
'
u
'
c o s ~ · c . o
s ( ~ a)
~ s i n (
· a )
r·
-
.
CO>
0: 0:
Smce .
.
J
t" - d, we ge
t
d
-
r'
r os pe
ns (IJ -a )+
] ] '
+
c'
-"'" D_<in
m
_
cos
u
L
cosa
=
.....>:;...- f
cos'
P
cos' (p
a ) + c
os
a
CO<-
<L
I '
I ' I
'
.
'
- , - · c o
cos
~ - O , -
S l n (
1
•
c
os· a
'
='
'
o
s· p •CO
O u +2
co '
~ t o >
"
Tilere
fare. ar
ea
of
the
circle
=JU'
'
os·
j;
1
c
os ·" 1
2cosCl
co;Bcoo
(j) "
The give
n circle
is,
ax
+2hxy
+i>_,
=I
Let
the po
in t P
~ i n g
(1)
bo (x
,J
1
),
le l 0
inclina
tion
of
l
ine throu
gh
P
w
hich
iH
ters<:<h
th
e
curve
>11 Q
01
1d R
.
Th
en equat
ion
of
line
til rough
I
r-
x 1 = y
- Yt =>"
co
sf l ,[r.
f.l
X= . r
1
+rcosf ,y=y,
+rsinO
fo r po
m tQ
and
R,
above
point m
ust lie o
n ( ll
~
a(x
1
+ rc
osf. l)
2
+2h
(x
1
~ r c - o s
I J ( y
, ' ' · '
- l>(y
1
'
" i1H
-=;,
(a
co
s'
f
.l
-
2h
sin f.lco
s 9
+ h'in
'
,
. 21
1
e
n< B. , m,
<m l
l
•
11<
·
1
11
h1·
- (axi
2h1, )",
+ lo
v; 1-
n
ll
is ~ u a d
a t i c in r
,
givi
ng mo
v
al"os
r JS
'{
""
:.
R ~ - -
.. x1+ 2
h 1y1
r h
,2 _
a cos
' e , 2hs
in e
c<>>
O
+
b,;,' o
He
ro, +
:-Ju, y,
+
b)\
- I
-,t 0.
a:; (.,
1
•
J.
)l l
, ,, ,
011 ( I)
Als
o,
a cos
'
0
+
2
h>in fl COl
fl + hoin
2
2 h s i
l c o s €
(b
,i
Jsin'O
=a + l (2hc
n ,fl
+ (
b -
)si
n
€1)
=
a +oin
fl. ;
.'4h
1
-
h
- a )
c o s 0
"
si'lL'c
b- 1
w
here.tan
fJ-
_
_
"
'
=a
+
• ~ l b ~ - - o - l
.1in e,in
€1 , 1
w
llich
wi
ll
inJq
;c nJcnl
ura. i
l'
4h
1 ~ - a J
~ o
:. Equati
on (1
i
1e
duces to
'
., '
y =
which
is
tl
circl
e.
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·
Equat
ion' ot
;my
circle
C w it h
ce m
re
(
0,
,1 2) is gi
ven by
where
r ->
0
o
'
(x
01
· , (
y-,'2
1- ~
•
{
,·
,,
- y -
~ , ~
Y -
= r
-- 1 )
Let(x
1
, y
1
),
(x,,
Y z)
,(-< ,.
y
3
)be
thrcc
d isti
nctra
tio na
l
po
ints o
n l
). Sin
re a s tr
a igh t
line p
<m l.ll
elto x-
axis
mee
t;
a
oirol
e in a
t most
t w o
poin ts
, h e r
,1), y
2
or
Jl -
J',
Pu
tting
in
(1
).
we
get
S
uhtrac
iing
(2) fro
m (3 )
. we
obta in
,-
,f2q
,=O
wh
ere
q
=y
-y
Sub
tr,1cti
ng (2)
lm m
(4),
we ob
tain
p
, --
4, =0
whe1
e
p , = r i + y ~ - x , - ; 1 ,
q,-- > -F i
.( 2)
,(3)
- .. ( 4
)
N.,w,
11 , p
,,
q
1
, q ,a
re ration,
_]
num
bcro AI>
<>
eith
er
q
1
z0
orq
2
'>'0. If
q
1
lq
1
a
nd ifq
2
ct 0,
th
en ,h
=
p
,
I q
2
.
In an
y ca s
e .f
i
is
a r
ation
al num
be r.
This
i ; a c
ontra
dictio
n.
19 .
Le
t the p
oint P be
(2
r cos
e. 2r
>-in f
l)
W e h
ave
O
=
r,O
P
=2r
A•
A O
Af is
a r ig h t
a ngle
d tr ia
ng le ,
•
\
cos¢>
= l/2
=;.
$ = 7 t/
3
:. C o
ordin
ate. o
fA are
{r
co
s (G
- 1t I
3 .
rs i
n
(0
- 1t I J)
)
an
d th a
t of
B
are[
rcos
(
0 -rn
/ 3), rsin
(9 -r 1ti
3)]
tf p,
q is th
e cen
troid
of 1 'AB
,
the
n p = ~
[r e
o> 9
1
3) -r rro
s
(0 -r
1
t.
3) -> 2rc
rn;9
]
[
r
{co
s
(6
-
1
rl 3
+ cr
(6 + 1 /3
)}+
lr·o
os
0]
[[
·
-··
···
•
-'-•
-']
i rz
co s
~
3_uJ
>
_32
3
+
2rco
s e]
~
1
[r
{2
cos£1
cos 7
t/3} +
2rc<
Js6]
•
'
[r
·c
o s 9
rco
se
•
an d
q
=_
_
[n in
(o
- ::<
)., n
.i n (
e +
\2 r
s i
J
3
J)
= [r {>
in
l
;
s i n [ e
~
}
+ 2n
- [I,
..
:···:
:;
rl _sm
2
-
0
- -1
1
j
0.-
-\
.l
1
[r {2
>ine
co sn
l3 )+
2n in
6]
•
=
_ _ 1,. {s in
9} +2
cs in
O
J= >
·sine
l
p,q )=(roos6,nin9)
lies on
r
w h
ich is
cal lc
dC,_
20
.
'
-
c,
~ _ /
f
b :
:
V
~
Fr
om t1
gure i
t is cl
ear th
at tria
ngle
0/.
S
i
a rig
·ith
right
angle
m L.
Abu
UL=l
an d
S
=2
si<
l (L
L S O
) = =>
L
LSO
=
300
in
ce SA
1
=
.SA
2
, t1 SA
1
A
, is an
eq u
ilatera
l tri
T
he c
ircle w
ith c
entr<
at
C
is a
circl
e insc
r
,
\
S
A
1
A,_
The
refor
e, cen
tre
C, is c
entro
ld o
This.
C
1
di>id
e ' SM
in
the r
atio
2 : I
.
co
ordin
.-tes
of ( '
1
a
re (-4/3
,0)
a
nd
-C M
- l
/3
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-. its equ
atLon io {x +
4 I
3)
1
<
i = (I
I 3
2
(1)
T h
e othc-r circle to
uche s the eq uila
tcrai triangle S
B
1
B,
e
xternally_ h.s
radiusr is give n
by ~ . w h
e r e
B
1
B,
=a. ut i\ = (a)(S.V)
= ~ a
' 2
2
Olld
s -a=-a -
a=a /2
2
l'hliS
,
r
=3
=oo
Coordinate
> ofC
2
are (4. 0)
s -a
·_
equation
of
i r d ~
with cen tre
m
c,
(x -4 )
+ y
=
3
2
-- .(2)
E q
ua t ion\ of
c om m
on tangents to c
ircle (I) and
mcle
C
.,
x • - 1 and
y=-T
-
2
- { x+2 )
J;
[T
1
nd
T,
1
Equation
of
com
m o n t
angent;
to
circlo
(2)
aud c
ircle
C
"
x= l and y - ± ~
( x , : : )
J2
7
1
andT,]
Two tangents co
m m o n
to
(l) an d
(2) are 1
1
and
l2 at
0
To lind the rema
inh>g
mo
traJls
-erse tangent>
to 1
)
and
(2),
we
t
lnd a potnl I whi
ch divides t l1e jo
int ofC
1
C
2
in
r a t 1 o ~
:r ,=
l 3 : 3 · : ~
Therefore,
coordinates
ofl are (-
4
I
5,
OJ
Equation
of an)' line
t h r o u I i
y
=
m
(x
+ 4 I 5). It il
l
touch
d ) if
~ - ~ ~ ~ ~ ~ 1
6 m
2
=25(1+m
2
t
3
9m
2
=25
'
= i
The
refore, these tan
gent< are
5 '
4)
v =
\ G < i l +
5
2 F.quati<>n
of
any tang
ent to
drclex
0
+
y
=r is
xcmO
+ys inO
-r
Suppose (
1)
i
tangent to
+
25
y
I
00
'
'
: _ ~ L -
1 a t ( x
,,2
25 4
1hen (1)
and
::::::c + m =1 ;ne identical
?_•
- _ ~ = ± =-
c
osO sinO
r
25co<9
4 s i n ~
=
>
x,
=
·Yt
~
'
'
The l
ine( ) meet the co ord
inates axes
in
A (r sec
e. 0) and
(0..- cosec 0). Let (h.
k) be 'nid
<>fA B.
k=rco;ece
'
h
erefore.
'
h= - - and 2
k =-'
>in e
As
(x
1
w
e
ge
l
cos 0
'
,•--
and
'
.l l =
2k
' '
j
lie'
on th
e ellipse
':_+ L
= 1
-
2
5 4
-(625 ~ I . . (
' ) ~ )
254
h i4k
~ ~ + - 2 _ , ,
4
h'
k
2
5k +4h =4h
k
There
fore, required lu
cuo is
4x
2
+25 y = 4x' J
• ' 0
2
.
2;(" + y
- ] J . } -
---
Lx -hr -J
:y+
v
=0
_;, h
x-y ) -y x -y
)= O
=>
(2x-y)(_,-y)
=0
---,
y
=2x
y=x
art the equations
o f strai[ h
l
pass ing thro
ugh origin.
N
ow, let the ang
le bet.,.,ccn
t ~ c
lin
es be
28
and th
y makes angl
e of 45" witll x-a
xis
Thcrcf<"
" '-"" ( 45° +
2 9)
= 2 (slope
of
he
line v -
~
~
tan45"-
tan29 _ =
2
1
tan
45'xtan2fl
l+tan20
a
n 29
(1
+tan
2fl)
-(1
-t•rllaJ
H
(l+tan20)+(1
ton29)
2 + 1)
1 tan 2fl
2
2
-
-
tan2fl=-
;
'
2
'
-
;
2 tan e
-
---
.
~ (2
t
an e 1
·3=1-ta
n
1
-tan'O
'
•n
9 + 6 t a n 0 - 1= 0
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2
44
O
CA
wne=J ·.OA= -
O
A
tan
O
=
JO
Ji
O
J
(
- 3
O
J
(3
o
· -
- =
J
v
+ -
.1
J
(1 0
-
9 )
le t
th
e gi
ven
ci
rcl
es C
1
an
d
c,
hav
e
ce
ntr
.,.
o
an
d
0
2
a
nd
ra
dii
,. ,
a
nJ
r re
sp
oc
liv
dj.
L
et
rh e
'a r
i b
lc
cir
cle
C
to
uc
h i
ng
C
1
i
ntc
ma
ll}
, C
,
c'
tc r
na
ll y
h
a'e
a
r&
liu>
ra
nd
cen
tre
at 0
.
N
ow
,
\
·-·
--<
..
.
0
0
2
~ r
r
,
an
d
1
r ·
· r
1
-
00
~
r
1 r
2
wh
1ch
i
gr
ea t
.,. th
an
0
1
0
2
a
s
0
1
0 , <
r
1
r
0
(
C
, lie
s
in ,
.dc
C
1
)
=
> iO
<U
< o
f()
is
an
el
hp>
e
W
iih
f
oc
i 0
1
a
nd
0
,.
l
te r
na
te
su
lu
tiu
n
:
l.e
t
cq
uat
iOl
l.> o
f
C
1
b
e
x <·
y
an
d
oi
c
,
be
.
.
n
-
a l
·
+
(y
·
b
r
le
i
cem
re
C
b
t:
(h
. k
) an
d r
adi
uo
r
th
en
by
tJ,c
g
ive
n
\i
(h
a
l
2
+(
k-
b
/
=r
+r
, a
nd
,lh
2
.
k
2
- ~
r
SS
ER
TIO
N N
D
RE
S
ON
L.
S
in c
e th
e
ar
e p
erp
en
dic
ula
r.
S
o.
l
<>e
u'
of p
er p
end
ic
ula
r
tru
1g
em
s
to
cir
elc
T
CH
T
HE
CO
LU
MN
l .
(A )
W
he
n
tw
o
cir
cle
s
arc
i
nte
rse
cti
ng
: t
hey
h
a,
·e
a
co
m1
non
n
orm
al
an
d
C
<Jm
mo
n tan
ge
n l
2
5.
rep
re
sen
t.
an
e
llip
><e
wh
ose
li
,ci
ar
e a
t
(0
. 0)
.
I fle
eq
ua
tio
n o
f
cir
cle
ha
vin
g
t
on
ge
nt
2
c +
3
(
I, - I
=>
( r
· l l
+
(y
+
l)
w
htc
h ;,
Ol1
1w
go
nal
(0
1l:
e c
ird
e h
avi
ng
en
d
mm
et
'
(0,
-
I)
an
d (
-
2,
3
)
=
>
.
r (x
+
2 )
(y
o
i) (
J ·
·3
)=
0
=>
2i
--·
·J
=
>
Jn
. .f
rom
e<
J< '-
Ii<>
n (
J)
eq
ua
ti o
n of
c
irc
le .
.
>:
2 y - 1 0 x - 5 y I ~
O
t
he
ci
rc l
es
w i
th
ra
dii
3. 4
an
d 5
t
ou
ch
ext
ern
al]
}·
and
P
is
the
po
in
t o
f i
nte
«e
ci i
on
ot
=>
1
ts
in ccntr c
ofi' .C 1
C
,c :.
T
hu
s d i
,ta
nc
e
of
p
oin
t
P f
rom
t
he
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