[Hint: Compute chord ofconstant of(–2, 0)
and verifyeach alternative w.r.t. the and C.O.C. ]
Q.121158/st.line Straight lines 2x + y= 5 and x  2y= 3 intersect at the pointA. Points B and C are chosen on
these two lines such that AB =AC . Then the equation ofa line BC passing through the point (2, 3) is
(A*) 3x  y  3 = 0 (B*) x + 3y  11 = 0
(C) 3x + y  9 = 0 (D) x  3y + 7 = 0
[Hint: Note that the lines are perpendicular . Find the equation ofthe lines through (2, 3) and parallel to the
bisectors of the given lines, the slopes of the bisectors being 1/3 & 3 ]
Q.122144/circle The centre(s) of the circle(s) passing through the points (0, 0) , (1, 0) and touching the circle
x2 + y2 = 9 is/are :
(A)
3
2
1
2
,





 (B)
1
2
3
2
,





 (C*)
1
2
21 2
, /





 (D*)
1
2
21 2
, /







[Hint: consider familyof 's through(0, 0) and (1, 0)
x(x – 1) + y2 + y = 0
touches x2 + y2 = 9
 common chord = – x + hy + 9 = 0 ....(1)
 perpendicular from(0, 0) on (1) is equal to 3.
2
1
9


= 3  2 = 8   = ± 2
2
circle x (x – 1) + y2 + 2
2 y
 centre 





  2
1
2
,
2
1
]
Q.123172/st.line The x  co-ordinates of the vertices of a square of unit area are the roots of the equation
x2  3x + 2 = 0 and the y  co-ordinates of the vertices are the roots of the equation
y2  3y+ 2 = 0 then the possible vertices of the square is/are :
(A*) (1, 1), (2, 1), (2, 2), (1, 2) (B*) ( 1, 1), ( 2, 1), ( 2, 2), ( 1, 2)
(C) (2, 1), (1, 1), (1, 2), (2, 2) (D) ( 2, 1), ( 1,  1), ( 1, 2), ( 2, 2)
Q.124146/circle Acircle passes through the point 3
7
2
,







 and touches the line pair x2  y2  2x + 1 = 0. The
co-ordinates ofthe centre ofthe circle are :
(A*) (4, 0) (B) (5, 0) (C*) (6, 0) (D) (0, 4)
Q.125180/st.line P(x, y) moves such that the area of the triangle formed by P, Q (a , 2a) and R (a,  2a) is
equalto the area ofthe triangle formed by P, S (a, 2a) & T (2a, 3a). The locus of 'P' is a straight line
givenby:
(A*) 3x  y = a (B*) 5x  3y + a = 0 (C) y = 2ax (D) 2y = ax

Q.110178/circle In the xyplane, the segment with end points (3, 8) and (–5, 2) is the diameter ofthe circle. The
point (k, 10) lies on the circle for
(A) no value ofk (B*) exactlyone integralk
(C) exaclyone nonintegralk (D) two realvalues ofk
[Hint: k = – 1 ]
Q.111199/st.line Let A  (3, 2) and B  (5, 1). ABP is an equilateral triangle is constructed on the side ofAB
remote fromtheoriginthenthe orthocentre oftriangleABPis
(A) 4
1
2
3
3
2
3
 






, (B) 4
1
2
3
3
2
3
 






,
(C) 4
1
6
3
3
2
1
3
3
 






, (D*) 4
1
6
3
3
2
1
3
3
 






,
[Sol. mAB = –1/2
mPM = 2
h =
a 3
2
now a = 5 = AB ; h =
15
2
so point P
x y

cos
=
y  3 2
/
sin
= h
P : x = 4 +
3
2
, y =
3
2
+ 3
so orthocentre/centroid is
 
x y
1 1
3 3
,
= 4 +
3
6
,
3
2
+
3
3
]
Q.112200/st.line The vertexofa right angleofa right angledtriangle lies onthestraight line 2x+y– 10 =0 andthe
two other vertices, at points (2, –3) and (4, 1) then the area oftriangle insq. units is
(A) 10 (B*) 3 (C)
33
5
(D) 11
[Sol. M1M2 = –1
9 2
4
13 2
2




a
a
x
a
a
= –1
117 – 26a – 18a + 4a2 = –(a2 – 6a + 8)
5a2 – 50a + 125 = 0
a = 5
so B is (5, 0)
so area =
2
1
AB × AC =
2
1
2 × 3 2 = 3 ]

Q.99188/st.line The coordinates of three points A(4, 0) ; B(2, 1) and C(3, 1) determine the vertices of an
equilateraltrapeziumABCD . The coordinates ofthe vertex D are :
(A) (6, 0) (B) ( 3, 0) (C) ( 5, 0) (D*) (9, 0)
[Hint: Equilateralmeanisosceles trapezium]
Q.100168/circle Tangents are drawn from anypoint on the circle x2 + y2 = R2 to the circle x2 + y2 = r2. Ifthe line
joining the pointsofintersectionofthesetangents withthefirst circlealso touchthe second,thenRequals
(A) 2 r (B*) 2r (C)
2
2 3
r

(D)
4
3 5
r

Q.101192/st.line The image ofthe pair oflines represented by ax2 + 2h xy + by2 = 0 bythe line mirror y = 0 is
(A) ax2  2h xy  by2 = 0 (B) bx2  2h xy + ay2 = 0
(C) bx2 + 2h xy + ay2 = 0 (D*) ax2  2h xy + by2 = 0
[Hint : m1  m1 & m2  m2  equation is
(y + m1 x) (y + m2 x) = 0 where m1 + m2 = 
2h
b
& m1 m2 =
a
b
]
Q.102169/circle Pair oftangents are drawn fromevery point on the line 3x + 4y = 12 on the circle x2 + y2 = 4.
Their variable chordofcontact always passes through a fixed point whose co-ordinates are
(A) 





4
3
,
3
4
(B) 





4
3
,
4
3
(C) (1, 1) (D*) 





3
4
,
1
Q.103193/st.line The set ofvalues of 'b' for whichthe originand the point (1, 1) lieonthe same sideofthe straight
line, a2x + a by + 1 = 0  a  R, b > 0 are :
(A) b  (2, 4) (B*) b  (0, 2) (C) b  [0, 2] (D) (2, )
[Hint: a2 + ab + 1 > 0  a  R  D < 0 ]
Q.104170/circle The equation of the circle symmetric to the circle x2 + y2 – 2x – 4y + 4 = 0 about the line
x – y = 3 is
(A*) x2 + y2 – 10x + 4y + 28 = 0 (B) x2 + y2 + 6x + 8 = 0
(C) x2 + y2 – 14x – 2y + 49 = 0 (D) x2 + y2 + 8x + 2y + 16 = 0
[Hint: Circle of the centre is (1, 2) and r = 1
image of (1, 2) in x – y – 3 = 0
1
1
a
2
b




 a + b =3 ....(1)
also
2
2
b
,
2
1
a 

lies on x – y = 3
(a + 1 ) – ( b + 2) = 6  a – b = 7 ....(2)
from (1) and (2) a = 5 and b = – 2
Hence the required circle has the centre (5, – 2) and r = 1
(x – 5)2 + (y + 2)2 = 1  x2 + y2 – 10x + 4y + 28 = 0  (A) ]

Q.91178/st.line Ifin triangleABC, A (1, 10) , circumcentre   
 1
3
2
3
, and orthocentre   
11
3
4
3
, then the
co-ordinates ofmid-point ofside opposite toAis :
(A*) (1, 11/3) (B) (1, 5) (C) (1,  3) (D) (1, 6)
Q.92159/circle Let x & y be the real numbers satisfying the equation x2  4x + y2 + 3 = 0. If the maximum and
minimumvalues ofx2 + y2 are M & mrespectively, then the numericalvalue ofM  mis :
(A) 2 (B*) 8 (C) 15 (D) noneofthese
Q.93181/st.line If the straight lines , ax + amy + 1 = 0 , bx + (m + 1) by + 1 = 0 and cx + (m + 2)cy + 1 = 0,
m 0 are concurrent then a, b, c are in :
(A) A.P. only for m = 1 (B) A.P. for all m
(C) G.P. for all m (D*) H.P. for all m.
[Sol. D =
1
)
2
m
(
c
c
1
)
1
m
(
b
b
1
am
a

 = 0 ;
1
c
2
c
1
b
b
1
0
a
= 0
a(b – 2c) +(2bc – bc) = 0
ab – 2ac + bc = 0
b(a + c) = 2ac
b =
c
a
ac
2

 a, b, c are in H.P. ]
Q.94160/circle Aline meets the co-ordinate axes inA&B.Acircle is circumscribedabout the triangle OAB. Ifd1
&d2 are thedistances ofthe tangent to the circle at the originO fromthepointsAandB respectively, the
diameter ofthecircle is :
(A)
2
2
1 2
d d

(B)
d d
1 2
2
2

(C*) d1 + d2 (D)
d d
d d
1 2
1 2

Q.95182/st.line If x1 , y1 are the roots of x2 + 8 x  20 = 0, x2 , y2 are the roots of 4x2 + 32x  57 = 0 and
x3 , y3 are the roots of 9 x2 + 72 x  112 = 0, then the points, (x1 , y1) , (x2 , y2) & (x3 , y3)
(A*) are collinear (B) formanequilateraltriangle
(C) forma right angled isosceles triangle (D) are concyclic
[Sol. Given x1 + y1 = -8  y1 = -8-x1 ; x2 + y2 = -8  y2 = -8 - x2
and x3 + y3 = -8  y3 = -8 -x3
now check
1
x
8
x
1
x
8
x
1
x
8
x
3
3
2
2
1
1






value ofthis determinant comes out to be zero. ]
Q.96164/circle Two concentric circles are such that the smaller divides the largerinto two regions ofequalarea. If
the radius ofthesmaller circle is 2, then the lengthofthe tangent fromanypoint 'P'onthe larger circle
to the smaller circle is :
(A) 1 (B) 2 (C*) 2 (D) none
[Sol.  r1
2
=  r2
2   r1
2
 2 r1
2
= r2
2  r2 = 2 r1
Note P lies on the director circle of radius r1
 L = r1 = 2 cm ]

Q.82148/circle The circle passing through the distinct points (1, t) , (t, 1) & (t, t) for all values of 't ' , passes
throughthepoint :
(A) ( 1,  1) (B) ( 1, 1) (C) (1,  1) (D*) (1, 1)
[Sol. Equation of circle is x2 + y2 + 2gx +2fy + c = 0
(1, t)  1 + t2 + 2g + 2ft + c = 0
(t, t)  t2 + t2 + 2gt + 2ft + c = 0
(t, 1)  1 + t2 + 2gt + 2f + c = 0
subtract 1 + 2g – t2 – 2gt = 0  1 – t2 + 2g(1 – t) = 0  (1 – t)(1 + t + 2g) = 0
 t = 1
 one point (t, t)
 passes through(1, 1) ]
Q.83173/st.line In a triangle ABC, side AB has the equation 2x + 3y = 29 and the sideAC has the equation,
x + 2y = 16 . If the midpoint of BC is (5, 6) then the equation ofBC is :
(A) x  y =  1 (B) 5 x  2 y = 13 (C*) x + y = 11 (D) 3 x  4 y =  9
Q.84149/circle Ifa circle ofconstant radius 3k passes throughthe origin 'O' and meets co-ordinate axes atAand
B thenthe locus ofthecentroid ofthe triangleOAB is
(A*) x2 + y2 = (2k)2 (B) x2 + y2 = (3k)2 (C) x2 + y2 = (4k)2 (D) x2 + y2 = (6k)2
[Sol. Given r = 3k and P is centroid
g2 + f2 = (3k)2 ....(1) because c = 0
 3h1 = a and 3k1 = b also g
2 = a  2g = a and 2f = b
 2g =
2
h
3 1
and 2f =
2
k
3 1
substitute in (1)
4
k
9
4
h
9 2
1
2
1
 = 9k2  x2 + y2 = (2k)2 ]
Q.85174/st.line The circumcentre of the triangle formed bythe lines, xy+ 2x+ 2y+ 4 = 0 and x + y+ 2 = 0 is
(A) ( 2,  2) (B*) ( 1,  1) (C) (0, 0) (D) ( 1,  2)
[Sol. ABC
 isright angle
circumcentre is mid pt. ofBC i.e. (-1,-1)Ans. ]
Q.86153/circle The locus ofthe midpoints ofthe chords ofthe circle x2 + y2  2x  4y  11 = 0 which subtend
600 at the centre is
(A) x2 + y2  4x  2y  7 = 0 (B) x2 + y2 + 4x + 2y  7 = 0
(C*) x2 + y2  2x  4y  7 = 0 (D) x2 + y2 + 2x + 4y + 7 = 0
[Sol. (h, k) are mid points ofchord
4
)
k
2
(
)
h
1
( 2
2



= cos 30°
= (1 – h)2 + (2 – k)2 =
2
4
2
3









h2 + 1 – 2h + k2 + 4 – 4k = 12
hence locus is x2 + y2 – 2x – 4y – 7 = 0 ]

p = 4 3
2 2
 = 7 & c3M =
7
4
 CM = 7 
4
7
=
3 7
4
 A =
3 7
4
.
3
2
.
1
2
=
9 7
16
]
Q.73155/st.line Let the co-ordinates of the two points A& B be (1, 2) and (7, 5) respectively. The line AB is
rotated through 45º in anticlockwise directionabout thepoint oftrisectionofAB whichis nearerto B.
The equationofthe line innew positionis :
(A) 2x  y  6 = 0 (B) x  y  1 = 0
(C*) 3x  y  11 = 0 (D) none of these
[Hint : point oftrisectionis (5, 4) .After rotation the slopeofthe line
tan 1 =
2
2
tan
45
tan
1
tan
45
tan






=
2
1
2
1
1
1


= 3  C ]
Q.74132/circle Apair oftangents are drawnto a unit circle with centre at the origin and these tangents intersect at
A enclosing anangle of60°. The area enclosed bythese tangents and the arc ofthe circle is
(A)
3
2
–
6

(B*) 3 –
3

(C)
3

–
6
3
(D) 




 

6
1
3
[Hint: r = 1 ; L = 3
quad = 3
sector =
2
1
· 1 ·
3
2
=
3

shaded region = 3 –
3

Ans. ]
Q.75161/st.line The true set ofreal values of for which the point P with co-ordinate (, 2) does not lie inside
the triangle formed by the lines, x  y = 0 ; x + y  2 = 0 & x + 3 = 0 is :
(A) (, 2] (B) [0, ] (C) [ 2, 0] (D*) (,  2]  [0, ]
[Hint : Point lies on the parabola y = x2 . Solve y = x2 with the line x + y = 2 ]
Q.76138/circle If the line x cos + y sin = 2 is the equation of a transverse common tangent to the circles
x2 + y2 = 4 and x2 + y2  6 3 x  6y + 20 = 0, then the value of  is :
(A) 5/6 (B) 2/3 (C) /3 (D*) /6
[Sol. C1C2 = r1 + r2
C1 = (0, 0) ; C2 = ( 3
3 , 3) & r1 = 2, r2 = 4
 circle toucheachother
T.C.T = 3 x + y – 4 = 0
comparing with x cos + y sin = 2
 =
6

]

Q.64124/circle If two chords, each bisected by the xaxis canbe drawn to the circle,
2 (x2 + y2)  2ax  by = 0 (a  0 , b  0) from the point (a, b/2) then :
(A) a2 > 8b2 (B) b2 > 2a2
(C*) a2 > 2b2 (D) a2 = 2b2
[Hint: (h, b/2) must lie onthe given circle
 2h2  2ah + b2 = 0
For two distinct values of origin h D > 0  a2 > 2b2  C ]
Q.65146/st.line Ifthe line y= mx bisects the angle between the lines ax2 + 2h xy + by2 = 0 then m is a root of
the quadraticequation:
(A*) hx2 + (a  b) x  h = 0 (B) x2 + h(a  b) x  1 = 0
(C) (a  b) x2 + hx  (a  b) = 0 (D) (a  b) x2  hx  (a  b) = 0
[Hint : Equation of the angle bisector
x y
a b
2 2


=
xy
h
; Now put y = mx ]
Q.66125/circle Tangents aredrawnto a unit circle withcentre at the originfromeachpoint onthe line 2x + y= 4.
Then the equation to the locus ofthe middle point ofthe chord ofcontact is
(A) 2 (x2 + y2) = x + y (B) 2 (x2 + y2) = x + 2y
(C*) 4 (x2 + y2) = 2x + y (D) none
[Sol. (x1, y1) lies on 2x + y = 4
 2x1 + y1 = 4 ....(1)
chord ofcontact w.r.t. (x1, y1)
xx1 + yy1 = 1
also equation ofchord whose mid point is (h, k)
h2 + k2 = hx + ky
 2
2
1
1
k
h
1
k
y
h
x


  x1 = 2
2
k
h
h

; y1 = 2
2
k
h
k

substitute in(1)
2 · 2
2
k
h
h

+ 2
2
k
h
k

= 4
locus = 4(x2 + y2) = 2x + y ]
Q.67149/st.line An equilateraltriangle has each of its sides of length 6 cm. If(x1, y1) ; (x2, y2) & (x3, y3) are its
vertices thenthevalue ofthe determinant,
2
3
3
2
2
1
1
1
y
x
1
y
x
1
y
x
is equal to :
(A) 192 (B) 243 (C) 486 (D*) 972
Q.68126/circle Two circles whose radiiare equalto 4 and 8 intersect at right angles. The lengthoftheir common
chord is
(A*)
16
5
(B) 8 (C) 4 6 (D)
8 5
5

Q.54116/circle The chord ofcontact of the tangents drawn from a point on the circle, x2 + y2 = a2 to the circle
x2 + y2 = b2 touches the circle x2 + y2 = c2 then a, b, c are in :
(A) A.P. (B*) G.P. (C) H.P. (D) A.G.P.
[Hint: equation ofchordofcontactAB
x a cos + y a sin = b2 .....(1)
this is tangent to the circle x2 + y2 = c2
 perpendicular from(0, 0) on (1) is equalto c



 2
2
2
2
2
sin
b
cos
a
b
= c  b2 = ac ]
Q.55106/st.line Alight beamemanating fromthe pointA(3, 10) reflects fromthe line 2x+ y- 6 = 0and thenpasses
throughthe point B(5, 6) . The equationofthe incident and reflected beams are respectively :
(A*) 4 x  3 y + 18 = 0 & y = 6 (B) x  2 y + 8 = 0 & x = 5
(C) x + 2 y  8 = 0 & y = 6 (D) none of these
[Hint : slope ofAB = 2 = slope ofgiven line
 PAB is isosceles . Equation ofline MPis
x  2y + 12 = 0 . Solving it with 2x + y  6 = 0
point P(0, 6) . Equation ofAP is 4x  3y + 18 = 0
and equation of BP is y  6 = 0 ] [ M.E. with Q. No. 91 ]
Q.56118/circle If the two circles, x2 + y2 + 2 g1x + 2 f1y = 0 & x2 + y2 + 2 g2x + 2 f2y = 0 touch each then:
(A) f1 g1 = f2 g2 (B*)
f
g
1
1
=
f
g
2
2
(C) f1 f2 = g1 g2 (D) none
[Hint: circles can touchonlyat (0, 0)
 (–g, –f1); (0, 0) and (–g2, –f2) collinear 
f
g
f
g
1
1
2
2
 ]
Q.57109/st.line If P 
1
x
p
p
,







 ; Q =
1
x
q
q
,







 ; R =
1
x
r
r
,





 where xk  0, denotes the kth term of an H.P.
.
for k  N, then :
(A) Ar. ( PQR) =
p q r
p q q r r p
2 2 2
2 2 2
2
( ) ( ) ( )
    
(B)  PQR is a right angled triangle
(C*) the points P, Q, R are collinear
(D) none
[Hint :
1
xp
= a + (p  1) d ;
1
xq
= a + (q  1) d ;
1
xr
= a + (r  1) d 
1
r
d
)
1
r
(
a
1
q
d
)
1
q
(
a
1
p
d
)
1
p
(
a






= 0]
Q.58119/circle Tangents are drawn to the circle x2 + y2 = 1 at the points where it is met by the circles,
x2 + y2  ( + 6)x + (8  2) y  3 = 0 .  being the variable . The locus ofthe point ofintersection of
these tangentsis :
(A*) 2x y+ 10 = 0 (B) x + 2y 10 = 0 (C) x  2y+ 10 = 0 (D) 2x + y 10 = 0
[Hint: compare chord ofcontact ofthe pair oftangents from(x1, y1) to thecircle x2 + y2 = 1 withthe common
chord betweenthe two circles and eliminate  ]
[Sol. Locus ofpoint ofintersection oftangents
chord of contact of (x1, y1) w.r.t. x2 + y2 = 1 is xx1 + yy1 = 1 (AB) .....(1)

[Hint: equation of the two circles be (x  r)2 + (y  r)2 = r2
i.e. x2 + y2  2rx  2ry+ r2 = 0 where r = r1 & r2. Condition of orthogonality gives
2 r1r2 + 2 r1r2 = r1
2 + r2
2  4 r1r2 = r1
2 + r2
2. Circle passes through (a, b)
 a2 + b2  2ra  2rb + r2 = 0 i.e. r2  2r (a + b) + a2 + b2 = 0
r1 + r2 = 2 (a + b) and r1 r2 = a2 + b2 ]
Q.4785/st.line Ifthe verticesPand Q ofa triangle PQR are given by(2, 5)and (4, –11) respectively, and the point
R moves along the line N: 9x+ 7y+ 4= 0, thenthelocus ofthe centroidofthe triangle PQR is a straight
line parallelto
(A) PQ (B) QR (C) RP (D*) N
Sol. R (x, y) lies on 9x + 7y + 4 =0
 





 

7
a
9
4
,
a
R , centroid of  PQR = (h, k)
3
a
6
3
a
4
2
h







 

 ....(1)
3
7
a
9
46
3
7
)
a
9
4
(
11
5
k







 ....(2)
from(1) &(2) we get
equatingx 0
46
54
21
27
9
)
46
21
(
6
3 







 k
h
k
h
or locus is 9x + 7y - 8/3 = 0
this line is  to N ]
Q.48111/circle The range ofvalues of 'a' suchthat the angle  between the pair oftangents drawnfromthe point
(a, 0) to the circle x2 + y2 = 1 satisfies

2
<  <  is :
(A) (1, 2) (B)  
1 2
, (C)  
 
2 1
, (D*)  
 
2 1
,   
1 2
,
[Hint: Use the concept ofdirector circle ]
Q.4989/st.line The pointsA(a, 0), B(0, b), C(c, 0) & D(0, d) are suchthat ac = bd & a, b, c, d are allnonzero.
The the points :
(A) forma parallelogram (B) do not lie on a circle
(C) forma trapezium (D*) are concyclic
[Hint : a/d = b/c  cyclic ]
Q.50112/circle If(,) is a point on the circle whose centre is on the x -axis and which touches the line x + y= 0
at (2, –2), thenthe greatest value ofis
(A) 4 – 2 (B) 6 (C*) 4 + 2 2 (D) 4 + 2

Q.3566/st.line Astick oflength10 unitsrests against the floor&a wallofa room. Ifthe stick begins to slide onthe
floor thenthelocus ofits middlepoint is :
(A) x2 + y2 = 2.5 (B*) x2 + y2 = 25 (C) x2 + y2 = 100 (D) none
[Sol. given (a2 + b2) = 100 ---(1)
p (h, k) be the middle pt then
h = a/2, k = b/2
 a = 2h b = 2k
from(1) 4h2 + 4k2 = 100
h2 + k2 =25
locus x2 + y2 = 25 ]
Q.3680/circle The locus of the mid points ofthe chords ofthe circle x²+y²+4x6y12 = 0 which subtend
an angle of

3
radiansat its circumference is :
(A) (x 2)² + (y + 3)² = 6.25 (B*) (x + 2)² + (y3)² = 6.25
(C) (x + 2)² + (y 3)² = 18.75 (D) (x + 2)² + (y + 3)² = 18.75
[Hint : radius = 5  p = 5 cos 60º = 2.5
 locus is (h + 2)2 + (k  3)2 = 6.25 ]
Q.3772/st.line Througha givenpoint P(a, b)astraight line is drawnto meet the axesat Q &R.Ifthe parallelogram
OQSR is completed then the equationofthe locus ofS is
(given 'O' isthe origin) :
(A*)
a
x
+
b
y
= 1 (B)
a
y
+
b
x
= 1 (C)
a
x
+
b
y
= 2 (D)
a
y
+
b
x
= 2
Q.3892/circle The points (x1, y1) , (x2, y2), (x1, y2) & (x2, y1) are always :
(A) collinear (B*) concyclic
(C) vertices of a square (D) vertices ofa rhombus
[Hint : All the points lie on the circle (x  x1) (x  x2) + (y  y1) (y  y2) = 0 ]
Q.39202/st.line The numberofpossible straight lines, passing through(2, 3) and forming atriangle withcoordinate
axes, whose area is 12 sq. units , is
(A) one (B)two (C*) three (D)four
[Sol equation ofanyline through (2, 3) is y -3 = m(x - 2)
y = mx - 2m + 3
with the help ofthe fig. area of  OAB =  12
ie. 12
)
2
3
(
3
2
2
1








 
m
m
m
B
A
O
(0,0)
.(2,3)
(0, 3–2m)
taking + sign me get (2m+3)2 = 0
this gives one value ofm= -3/2
takingnegativesignwe get
4m2 - 36m + 9 = 0 (D > 0)
quadratic inmgives 2 values ofm
 3 st. lines are possible. ]
Q.4073/st.line Two mutuallyperpendicular straight lines throughthe origin froman isosceles trianglewiththe line
2x + y= 5 . Thenthe area ofthe triangle is :
(A*) 5 (B) 3 (C) 5/2 (D) 1

Q.2367/circle The equation of the circle having the lines y2 2y+4x2xy= 0 as its normals & passing through
the point (2, 1) is :
(A*) x2 + y2  2x 4y + 3 = 0 (B) x2 + y2  2x + 4y 5 = 0
(C) x2 + y2 + 2x + 4y 13 = 0 (D) none
[Sol. normals are (y– 2)(y– 2x) = 0 should pass throughcentre (a, b)
 2a = b and b = 2  a = 1 and b = 2
 equation x2 + y2 – 2x – 4y + c = 0 ]
Q.2448/st.line If P = (1, 0) ; Q = (1, 0) & R = (2, 0) are three given points, then the locus of the points S
satisfying the relation, SQ2 + SR2 = 2SP2 is :
(A) a straight line parallelto xaxis (B) a circle passing throughthe origin
(C) a circle with the centre at the origin (D*) a straight line parallelto yaxis .
Q.2570/circle Ifa circle passes through the point (a, b) & cuts the circle x²+ y² = K² orthogonally, then the
equation of the locus of its centre is :
(A*) 2ax + 2by (a² + b² + K²) = 0 (B) 2ax + 2by (a²  b² + K²) = 0
(C) x² + y²  3ax  4by+ (a² + b²  K²) = 0 (D) x² + y²  2ax  3by+ (a²  b²  K²) = 0
Q.2650/st.line The coordinates ofthe orthocentre ofthe triangle bounded bythe lines, 4x7y+10 = 0;x + y=5
and 7x + 4y = 15 is :
(A) (2, 1) (B) ( 1, 2) (C*) (1, 2) (D) (1, 2)
Q.2774/circle The distance between the chords of contact of tangents to the circle ; x2+y2 +2gx+2fy+ c=0
from the origin & the point (g, f) is :
(A) g f
2 2
 (B)
g f c
2 2
2
 
(C*)
g f c
g f
2 2
2 2
2
 

(D)
g f c
g f
2 2
2 2
2
 

[Sol. Equation ofchord or contact are gx + fy + c = 0 ....(1)
& 2gx + 2fy +
2
c
f
g 2
2


= 0 ....(2)
These lines are parallel
hence distance =
2
2
2
2
f
g
2
c
f
g
c




]
Q.2852/st.line The equation ofthe pair ofbisectors ofthe angles between two straight lines is,
12x2  7xy  12y2 = 0 . If the equation of one line is 2y  x = 0 then the equation of the other line is :
(A*) 41x  38y = 0 (B) 38x  41y = 0 (C) 38x + 41y = 0 (D) 41x + 38y = 0
[Hint : (2y  x) (y  mx) = mx2  xy (2m + 1) + 2y2 = 0  the equation to the pair of bisectors are :
x y
m
2 2
2


=


2
2 1
xy
m
 12x2  7xy  12y2  m = 41/38 ]
Q.2976/circle The points A(a, 0) , B (0 , b), C (c, 0) & D(0 , d) are such that ac = bd & a, b, c, d are all
non-zero. Thenthepoints :
(A) forma parallelogram (B) do not lie on a circle
(C) forma trapezium (D*) are concyclic

Q.1233/st.line IfA& B are the points (3, 4) and (2, 1), then the coordinates ofthe point C onAB produced
such that AC = 2BC are :
(A) (2, 4) (B) (3, 7) (C*) (7, 2) (D) 






1
2
5
2
,
Q.1352/circle The locus ofthe mid points of the chords ofthe circle x2 + y2 axby= 0 which subtend a right
angle at
a
2
b
2
,





 is :
(A) ax + by = 0 (B) ax + by = a2 + b2
(C*) x2 + y2  ax  by+
8
b
a 2
2

= 0 (D) x2 + y2  ax  by 
8
b
a 2
2

= 0
[Sol. r =
4
b
4
a 2
2
 =
2
b
a 2
2

sin 45° =
2
b
a
2
b
k
2
a
h
2
2
2
2

































 2
2
2
2
b
a
4
)
b
k
2
(
4
)
a
h
2
(
4
2
1
simplify to get locus x2 + y2 – ax – by –
8
b
a 2
2

= 0 ]
Q.1434/st.line The base BCofa triangleABC is bisected at the point (p, q) and the equation to the sideAB &AC
are px + qy = 1 & qx + py = 1 . The equation ofthe median throughAis :
(A) (p  2q) x + (q  2p) y + 1 = 0
(B) (p + q) (x + y)  2 = 0
(C*) (2pq  1) (px + qy  1) = (p2 + q2  1) (qx + py  1)
(D) none
[Sol. EquationthroughAis
(px + qy -1) +  (qx + py - 1) = 0 .....(1)
ifthis representAB then, point D (p, q) satisfies (1)
 [pp + qq – 1] + [pq + pq – 1] = 0
1
q
p
)
1
pq
2
(
2
2





 ]
Q.1555/circle From (3, 4) chords are drawn to the circle x² + y²4x = 0 . The locus of the mid points of
the chords is :
(A*) x² + y²  5x  4y + 6 = 0 (B) x² + y² + 5x 4y + 6 = 0
(C) x² + y²  5x + 4y + 6 = 0 (D) x² + y²  5x 4y 6 = 0
[Hint: Arc ofthe with OP as diameter
intercepted bythe givencircle.
]

Question bank on Circle & Straight line
There are 115 questions in this question bank.
Select the correct alternative : (Only one is correct)
Q.14/circle Coordinates ofthe centre ofthe circle which bisects the circumferences ofthe circles
x2 + y2 = 1 ; x2 + y2 + 2x – 3 = 0 and x2 + y2 + 2y – 3 = 0 is
(A) (–1, –1) (B) (3, 3) (C) (2, 2) (D*) (– 2, – 2)
Q.26/st.line One sideofa square is inclined at an acute angle  withthe positive x-axis, and one ofits extremities
is at theorigin. Ifthe remaining three vertices ofthe square lie above the x-axis and the side ofa square
is 4, thenthe equation ofthe diagonalofthe square whichis not passing throughthe origin is
(A) (cos  + sin ) x + (cos  – sin ) y = 4 (B) (cos  + sin ) x – (cos  – sin ) y = 4
(C*) (cos  – sin ) x + (cos  + sin ) y = 4 (D) (cos  – sin ) x – (cos  + sin ) y = 4 cos 2
[Sol.  slope of lineAC =










sin
4
cos
4
cos
4
sin
4
 equationofline =





sin
cos
sin
sin
y – sin  =





sin
cos
sin
sin
(x – 4cos )
y cos  + y sin  – 4 sin  · cos  = x sin  – x cos 
– 4 sin2  – sin  cos  + 4 cos2
x (cos  – sin ) + y (cos  + sin ) = 4  (D) ]
Q.310/circle The line 2x – y+ 1 = 0 is tangent to the circle at the point (2, 5) and the centre ofthe circleslies on
x – 2y= 4. The radius ofthe circle is
(A*) 5
3 (B) 3
5 (C) 5
2 (D) 2
5
[Sol. 2x – y + 1 = 0 is tangent
slope of line OA = –
2
1
equation (y – 5) = –
2
1
(x – 2)
2y – 10 = – x + 2
x + 2y = 12
 intersectionwithx – 2y= ywillgive coordinatesofcentre
x + 2y = 12
– x – 2y = 4
+
———————
4y = 8  y = 2
x – 4 = 4  x = 8
C = (8, 2)
distance OA = 2
)
5
2
(
)
2
8
( 

  = 9
36  = 45 = 5
3 ]