1. QUESTION BANK ON
STRAIGHT LINE
AND
CIRCLE
XI (PQRS & J-Batch)
MATHEMATICS
Time Limit : 6 Sitting Each of 75 Minutes duration approx.

2. Select the correct alternative : (Only one is correct)
Q.1 Ifthe lines x + y + 1 = 0 ; 4x + 3y + 4 = 0 and x + y+  = 0, where 2 + 2 = 2, are concurrent then
(A)  = 1,  = – 1 (B)  = 1,  = ± 1 (C)  = – 1,  = ± 1 (D*)  = ± 1,  = 1
[Sol. Lines are x + y + 1 = 0; 4x + 3y + 4 = 0 and x + y +  = 0 , where 2 + 2 = 2


1
4
3
4
1
1
1
= 0
1 (3 – 4) – 1 (4 – 4) + 1 (4 – 3) = 3 – 4 – 4 + 4 + 4 – 3
= –  + 1 = 0   = 1
  = ± 1 ]
Q.2 The axes are translated so that the new equation ofthe circle x²+y²5x+2y–5 = 0 has no first degree
terms. Thenthenew equationis :
(A) x2 + y2 = 9 (B*) x2 + y2 =
49
4
(C) x2 + y2 =
81
16
(D) none of these
Q.3 Giventhe familyoflines, a(3x+ 4y+ 6) + b(x+y+ 2)=0 . The lineofthe familysituatedat the greatest
distance fromthe point P (2, 3) has equation :
(A*) 4x + 3y + 8 = 0 (B) 5x + 3y + 10 = 0 (C) 15x + 8y + 30 = 0 (D) none
[Hint: point of intersection isA(2, 0) . The required line will be one which passes through (2, 0) and is
perpendicular to theline joining (2, 0) and(2, 3) ortaking (2, 3) ascentre and radius equalto PAdraw
a circle, the required line will be a tangent to the circle at (2, 0) ]
Q.4 The ends ofa quadrant ofa circle have the coordinates (1, 3) and (3, 1) then the centre ofthe such a
circle is
(A*) (1, 1) (B) (2, 2) (C) (2, 6) (D) (4, 4)
Q.5 The straight line, ax+ by= 1 makes withthecurve px2 + 2axy+ qy2 = r a chord whichsubtends a right
angle at theorigin. Then:
(A*) r (a2 + b2) = p + q (B) r (a2 + p2) = q + b
(C) r (b2 + q2) = p + a (D) none
[Hint: Homogeuse equation ofthe curve withline. Coefficient ofx2 + coefficient ofy2=0]
Q.6 The circle described onthe line joining the points (0,1), (a,b) as diameter cuts the xaxis in points
whose abscissae are roots of the equation :
(A) x² + ax + b = 0 (B*) x²  ax+ b = 0 (C) x² + ax b = 0 (D) x²  ax  b = 0
Q.7 Centroid ofthe triangle, the equations of whose sides are 12x2 – 20xy + 7y2 = 0 and 2x – 3y+ 4=0 is
(A) (3, 3) (B*) 





3
8
,
3
8
(C) 





3
8
,
3 (D) 





3
,
3
8
Q.8 The line 2x – y + 1 = 0 is tangent to the circle at the point (2, 5) and the centre of the circles lies on
x – 2y= 4. The radius ofthe circle is
(A*) 5
3 (B) 3
5 (C) 5
2 (D) 2
5

3. [Sol. 2x – y + 1 = 0 is tangent
slope of line OA = –
2
1
equation of OA, (y – 5) = –
2
1
(x – 2)
2y – 10 = – x + 2
x + 2y = 12
 intersection withx – 2y= 4 willgive coordinates ofcentre
solving weget (8, 2)
distance OA = 2
)
5
2
(
)
2
8
( 

  = 9
36  = 45 = 5
3 ]
Q.9 The line x + 3y  2 = 0 bisects the angle between a pair of straight lines of which one has equation
x  7y+ 5 = 0 . The equation ofthe other line is :
(A) 3x + 3y  1 = 0 (B) x  3y + 2 = 0 (C*) 5x + 5y  3 = 0 (D) none
[Hint: L  x – 7y + 5 + (x + 3y – 2) = 0
now equate perpendicular
distance to get ]
Q.10 Given two circles x²+ y² 6x 2y+ 5 = 0 & x² + y² + 6x+ 22y+ 5 = 0. The tangent at (2, 1) to
the first circle :
(A) passes outside the second circle
(B*) touches the second circle
(C) intersects the second circle in 2 real points
(D) passes through the centre of the second circle.
Q.11 A variable rectangle PQRS has its sides parallelto fixed directions. Q & S lie respectivelyon the lines
x = a, x = a & P lies on the xaxis . Then the locus ofR is :
(A*) a straight line (B) a circle (C) a parabola (D) pairofstraight lines
[Hint: Mid point ofQS = Mod point of PS
0 = h + x1 (–a, y2)
 x1 = –h
PQ  y = mx + c
passes through(–h, 0)
c = mh
 y = mx – mh
since Q lies on it
 y1 = ma – mh
now mQR = –
1
m
=
y K
a h
1 

–
1
m
=
m a h K
a h
( )
 

simplifying h + mk = a + (a – h)m2  a st. line]

4. Q.12 To which of the following circles, the line yx+3 = 0 is normalat the point 3
3
2
3
2







, ?
(A) x y
 





  





 
3
3
2
3
2
9
2 2
(B) x y






  





 
3
2
3
2
9
2 2
(C) x² + (y 3)² = 9 (D*) (x 3)² + y² = 9
Q.13 On the portion ofthe straight line, x + 2y= 4 intercepted between the axes, a square is constructed on
the side ofthe line awayfromthe origin. Thenthe point ofintersectionofitsdiagonals has co-ordinates
(A) (2, 3) (B) (3, 2) (C*) (3, 3) (D) (2, 2)
Q.14 The locus of the mid point of a chord of the circle x²+y² = 4 which subtends a right angle at the
origin is
(A) x+ y = 2 (B) x² + y² = 1 (C*) x² + y² = 2 (D) x+ y = 1
Q.15 Given the familyoflines, a(2x + y+ 4) + b(x  2y 3) = 0 .Among the lines ofthe family, the number
oflines situated at a distance of 10 fromthe point M(2, 3) is :
(A) 0 (B*) 1 (C) 2 (D) 
[Hint: The point ofintersection ofthe two lines are (–1, –2)
Distance PM = 10
Hence the required line is one which passes through(–1, –2) and is | to P.M.  B]
Q.16 The equation of the line passing through the points of intersectionof the circles ;
3x² + 3y²  2x+ 12y 9 = 0 & x² + y² + 6x + 2y 15 = 0 is :
(A*) 10x  3y 18 = 0 (B) 5x+ 3y 18 = 0
(C) 5x  3y 18 = 0 (D) 10x+ 3y+ 1 = 0
Q.17 Througha pointAonthe x-axisa straight line isdrawnparallelto y-axis so as to meet thepair ofstraight
lines ax2 + 2hxy + by2 = 0 in B and C. IfAB = BC then
(A) h2 = 4ab (B*) 8h2 = 9ab (C) 9h2 = 8ab (D) 4h2 = ab
[Sol. GivenAB = BC
tan  =
OA
AB
= m1 ; tan  =
OA
AB
2
= m2
1
2
m
m
= 2 ;
1
2
1
2
m
m
m
m


=
1
2
1
2


= 3
 –
b
a
4
b
h
4
b
h
2
2
2

= 3 
b
a
4
b
h
4
2
2
 = 2
2
b
9
h
4

9
8
b
h
4
2
2
 =
b
a
4
 8h2 = 9ab ]

5. Q.18 The number of common tangent(s) to the circles x2 +y2 +2x+8y –23 = 0 and
x2 + y2 – 4x – 10y+ 19 = 0 is
(A) 1 (B) 2 (C*) 3 (D) 4
[Hint: C1 = (–1, – 4) ; C2 = (2, 5) ; r1 = 23
16
1 
 = 10
2 ; r2 = 19
25
4 
 = 10 ;
r1C2 = 81
9 = 10
3 ]
Q.19 A, B and C are points in the xy plane such thatA(1, 2) ; B (5, 6) andAC = 3BC. Then
(A)ABC is a unique triangle (B) Therecan be onlytwo such triangles.
(C) No suchtriangle is possible (D*) Therecanbe infinite number ofsuchtriangles.
Q.20 From the point A(0, 3) on the circle x² +4x+(y3)² = 0 a chord AB is drawn & extended to
a point M suchthat AM = 2AB. The equation of the locus of M is :
(A) x² + 8x+ y² = 0 (B*) x² + 8x+ (y 3)² = 0
(C) (x  3)² + 8x + y² = 0 (D) x² + 8x+ 8y² = 0
Q.21 IfA(1, p2) ;B (0, 1) and C (p, 0) are the coordinates ofthree points then the value of p for which the
area ofthe triangleABC isminimum, is
(A)
3
1
(B) –
3
1
(C)
3
1
or –
3
1
(D*) none
[Sol. A =
1
0
p
1
1
0
1
p
1
2
1
2
=
2
1
[1(1 – 0) + p(p2 – 1)] =
2
1
(p3 – p + 1)
Hence A =
2
1
| p3 – p + 1 |
Now, minimumvalue ofmodulus is zero. SinceA(p) is a cubic it must vanish for some p other than
given as A, B, C  (D) ]
Q.22 The area of the quadrilateral formed by the tangents from the point (4 , 5) to the circle
x²+ y² 4x2y11 = 0 with the pair of radii through the points of contact of the tangents is :
(A) 4 sq.units (B*) 8 sq.units (C) 6 sq.units (D) none
Q.23 The area of triangle formed by the lines x + y – 3 = 0 , x – 3y + 9 = 0 and 3x – 2y + 1= 0
(A)
7
16
sq. units (B*)
7
10
sq. units (C) 4 sq. units (D) 9 sq. units
Q.24 Two circles of radii 4cms & 1cmtouch eachother externallyand  is the angle contained by their
direct common tangents. Then sin =
(A*)
25
24
(B)
25
12
(C)
4
3
(D) none

6. [Hint: sin
2

=
5
3
cos
2

=
5
4
 sin  = 2 ·
5
3
·
5
4
=
25
24
Ans. ]
Q.25 The set of lines ax + by + c = 0, where 3a + 2b + 4c = 0, is concurrent at the point :
(A)
3
4
3
4
,





 (B)
1
2
1
2
,





 (C*)
3
4
1
2
,





 (D) (1, 1)
Q.26 The locus of poles whose polar with respect to x²+y² = a² always passes through (K, 0) is
(A*) Kx  a² = 0 (B) Kx + a² = 0 (C) Ky+ a² = 0 (D) Ky a² = 0
Q.27 The coordinates of the point ofreflection ofthe origin (0, 0) in the line 4x  2y 5 = 0 is :
(A) (1, 2) (B*) (2, 1) (C)
4
5
2
5
, 





 (D) (2, 5)
[Sol. mAA' x mPQ = –1  1
2
·
x
y
1
1

  2y1 = – x1








2
y
,
2
x
R 1
1
lies on 4x – 2y – 5 = 0  2x1 – y1 = 5
)
B
(






Q.28 The locus ofthe mid points ofthe 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 ]

7. Q.29 Arayoflight passing through the point A(1, 2) is reflected at a point B on the xaxis and then passes
through (5, 3). Then the equation ofAB is :
(A*) 5x + 4y = 13 (B) 5x  4y =  3
(C) 4x + 5y = 14 (D) 4x  5y =  6
[Hint :
a  1
2
=
5
3
 a
 a =
13
5
]
Q.30 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.
m1m2 = – 1 ]
Q.31 m, nare integer with 0< n< m.Ais the point (m, n) on the cartesianplane. B is the reflectionofAinthe
line y=x. C is thereflection ofB inthe y-axis, D is thereflectionofC inthe x-axis and E isthe reflection
ofD in the y-axis. The area ofthe pentagonABCDE is
(A) 2m(m+ n) (B*) m(m+ 3n) (C) m(2m+ 3n) (D) 2m(m+ 3n)
[Sol. Area ofrectangle BCDE = 4mn
Area of  ABC =
2
)
n
m
(
m
2 
= m2 – mn
 area ofpentagon = 4mn + m2 – mn
= m2 + 3mn Ans. ]
Q.32 Which one of the following is false ?
The circles x² + y²  6x 6y+ 9 = 0 & x² + y² + 6x+ 6y+ 9 = 0 are such that :
(A) theydo not intersect
(B*) theytouch each other
(C) their exterior common tangents are parallel
(D) their interior common tangents are perpendicular.
Q.33 The lines y  y1 = m (x  x1) ± a 1 2
 m are tangents to the same circle . The radius ofthe circle is
(A) a/2 (B*) a (C) 2a (D) none
[Hint: Two parallellines y= mx + (y1 – mx1) + a 2
m
1 ]
Q.34 The centre of the smallest circle touching the circles x² + y²2y3 = 0 and
x² + y²  8x  18y + 93 = 0 is :
(A) (3 , 2) (B) (4, 4) (C) (2 , 7) (D*) (2, 5)
Q.35 The ends ofthe base ofanisosceles triangle areat (2, 0) and (0, 1) and theequationofone sideis x = 2
thenthe orthocentreofthe triangle is
(A) 





2
3
,
4
3
(B*) 





1
,
4
5
(C) 





1
,
4
3
(D) 





12
7
,
3
4

8. [Sol. 22 + (x1 – 1)2 = x1
2
4 + x1
2 + 1 – 2x1 = x1
2
5 = 2x1 or x1 = 5/2
Equation of (1) from(2, 5/2) to the given base
y – 5/2 = 2 (x – 2)
2y – 5 = 4 (x – 2)
at y = 1 –3/4 = x – 2 or x = 5/4  (B) ]
Q.36 A rhombus is inscribed in the region common to the two circles x2 + y2  4x  12 = 0 and
x2 + y2 + 4x12 = 0 withtwo ofits verticesonthe line joining the centres ofthe circles. The area of the
rhombous is :
(A*) 8 3 sq.units (B) 4 3 sq.units
(C) 16 3 sq.units (D) none
[Hint: circles with centre (2, 0) and (2, 0) each with radius 4
 yaxis is their common chord.
The inscribed rhombus has its diagonals equalto 4 & 4 3
 A =
d d
1 2
2
= 3
8 ]
Q.37 Avariable straight linepasses througha fixed point (a, b) intersecting the coordinates axesatA&B. If
'O' is the origin then the locus ofthe centroid ofthe triangle OAB is :
(A*) bx + ay  3xy = 0 (B) bx + ay  2xy = 0
(C) ax + by  3xy = 0 (D) none
Q.38 The angle between the two tangents fromthe origin to the circle (x7)2 + (y+1)2 = 25 equals
(A)

4
(B)

3
(C*)

2
(D) none
[Sol. The angle between two tangents fromorigin to circle (x7)2 + (y+1)2 = 25
radius = 5
length oftangent = 5
 the figureis square
angle = 90° ]
Q.39 If P = (1, 0); Q = (1, 0) & R = (2, 0) are three given points, thenthe locus ofthe 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.40 The equation of the circle having normal at (3, 3) as the straight line y = x and passing through the
point (2, 2) is :
(A) x² + y²  5x + 5y + 12 = 0 (B) x² + y² + 5x 5y + 12 = 0
(C) x² + y²  5x  5y 12 = 0 (D*) x² + y²  5x 5y + 12 = 0
Q.41 The equation of the base ofanequilateraltriangleABC is x+y= 2and the vertexis(2, 1) .The area
ofthe triangleABC is :
(A)
2
6
(B*)
3
6
(C)
3
8
(D) none

9. Q.42 Inaright triangleABC, right angled atA, onthe legAC asdiameter, a semicircle isdescribed. The chord
joiningAwiththepoint ofintersectionDofthehypotenuseandthesemicircle,thenthelengthACequalsto
(A)
AB AD
AB AD


2 2 (B)
AB AD
AB AD


(C) AB AD
 (D* )
AB AD
AB AD


2 2
[Sol. l · x = y 2
2
x

l
l2 x2 = y2(l2 + x2)
l2(x2 – y2) = x2y2
l =
2
2
y
x
xy

=
2
2
AD
AB
AD
·
AB

Ans. ]
Q.43 The equationofthe pair ofbisectorsofthe angles betweentwo straight lines is,
12x2  7xy  12y2 = 0. Ifthe equationof one line is 2y  x = 0 then the equation ofthe 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.44 If the circle C1 : x2 + y2 = 16 intersects another circle C2 of radius 5 in such a manner that the
commonchordis ofmaximumlengthand has a slopeequalto 3/4, thenthe co-ordinates of the centreof
C2 are :
(A)  






9
5
12
5
, (B*) 






9
5
12
5
,  (C)  






12
5
9
5
, (D) 






12
5
9
5
, 
[Hint: Equation ofcommon chord is y =
3
4
x  3x– 4y= 0
familyof x2 + y2 – 4 +  (3x – 4y) = 0
equate radius ofthis as 5.Alternately...............]
Q.45 Area ofthe rhombus bounded bythe four lines, ax ± by ± c = 0 is :
(A)
c
ab
2
2
(B*)
2 2
c
ab
(C)
4 2
c
ab
(D)
ab
c
4 2
Q.46 Two lines p1x + q1y + r1 = 0 & p2x + q2y + r2 = 0 are conjugate lines w.r.t. the circle x² + y² = a² if
(A) p1p2 + q1q2 = r1r2 (B) p1p2 + q1q2 + r1r2 = 0
(C*) a²(p1p2 + q1q2) = r1r2 (D) p1p2 + q1q2 = a² r1r2
Q.47 Area ofthe quadrilateral formed by the lines x + y = 2 is :
(A*) 8 (B) 6 (C) 4 (D) none
Sol. givenlines are
x + y = 2
-x + y = 2
x - y = 2
-x -y = 2
ABCD is a square asAC = BD and BD
AC 
also AB = BC = CD = AD now are ABCD = 4 area  AOB
= units
.
sq
8
2
2
2
1
4 


 ]

10. Q.48 If the two circles (x1)² + (y3)² = r² & x²+y²8x+2y+ 8 = 0 intersect in two distinct points then
(A*) 2 < r < 8 (B) r < 2 (C) r = 2 (4) r > 2
Q.49 Let thealgebraic sumofthe perpendicular distances fromthe points (3, 0), (0, 3) & (2, 2) to a variable
straight line be zero, then the line passes through a fixed point whose co-ordinates are :
(A) (3, 2) (B) (2, 3) (C)
3
5
3
5
,





 (D*)
5
3
5
3
,






Q.50 If a 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.51 Consider a quadratic equation in Z with parameters x and yas
Z2 – xZ + (x – y)2 = 0
The parameters x and y are the co-ordinates of a variable point P w.r.t. an orthonormal co-ordinate
systemina plane. Ifthequadratic equation has equalroots then the locus ofP is
(A) a circle
(B) a line pair through the originofco-ordinates with slope 1/2 and 2/3
(C) a line pair through the originofco-ordinates with slope 3/2 and 2
(D*) a line pair through the originofco-ordinates with slope 3/2 and 1/2
[Sol. D = 0
x2 = 4(x – y)2
x = 2(x – y) or x = –2(x – y)
x = 2y or 3x = 2y  line pair withslope 3/2 and 1/2  D ]
Q.52 Consider the circle S  x2 + y2 – 4x – 4y+ 4 = 0. Ifanother circle ofradius 'r' less than the radius ofthe
circle S is drawn, touching the circle S, and the coordinate axes, then the value of'r' is
(A) 3 – 2
2 (B) 4 – 2
2 (C) 7 – 2
4 (D*) 6 – 2
4
[Hint: 2(2 – r)2 = (2 + r)2 ]
Q.53 Vertices of a parallelogramABCD are A(3, 1), B(13, 6), C(13, 21) and D(3, 16). If a line passing
throughtheorigindivides the parallelograminto two congruent partsthenthe slope ofthe line is
(A)
12
11
(B*)
8
11
(C)
8
25
(D)
8
13
[Hint: as shown
m =
13
x
21
=
3
1
x 
63 – 3x = 13x + 13
16x = 50
x =
8
25
; Hence m =
3
1
·
1
8
25






 =
24
33
=
8
11
Ans. ]

11. Q.54 The distance between the chords ofcontact 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 ofthe pair oftangent from(0, 0) and (g, f) are
gx + fy + c = 0 ....(1) and gx + fy +
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.55 Two mutually perpendicular straight lines through the origin froman isosceles triangle with the line
2x + y = 5. Thenthe area ofthe triangle is :
(A*) 5 (B) 3 (C) 5/2 (D) 1
[Hint: Area ofright isosceles triangle = p2
Area ofequilateraltriangle interms of
altitude =
h2
3
]
Q.56 The locus of the centers of the circles which cut the circles x2 + y2 + 4x  6y + 9 = 0 and
x2 +y2 5x+ 4y2 = 0 orthogonally is :
(A) 9x+ 10y 7 = 0 (B) x  y+ 2 = 0
(C*) 9x  10y+ 11 = 0 (D) 9x+ 10y+ 7 = 0
[Hint: Locus ofthe centre ofthe cutting S1 = 0 and S2 = 0 orthogonallyis the radical axis between S1 = 0
and S2 = 0 ]
[Hint: Let out circle be x2 + y2 + 2gx + 2fy + c = 0
conditions 2 (– g) (–2) + 2( – f ) (3) = c + 9
and 2 (– g) (5/2) + 2( – f ) (–2) = c – 2
 ag – 10 f = 11
 locus of centre 9x – 10y + 11 = 0 ]
Q.57 Distance between the two lines represented bythe line pair,
x2  4xy + 4y2 + x  2y  6 = 0 is :
(A)
1
5
(B*) 5 (C) 2 5 (D) none
[Hint : Line pair represents two parallel lines, x  2y + 3 = 0 & x  2y  2 = 0 ]
Q.58 The locus of the center of the circles such that the point (2 , 3) is the mid point of the chord
5x+ 2y = 16 is :
(A*) 2x 5y + 11 = 0 (B) 2x+ 5y 11 = 0
(C) 2x+ 5y+ 11 = 0 (D) none
[Hint : Slope ofthe given line =  5/2
 


5
2
3
2
.
f
g
=  1  15 + 5f = 4 + 2g
 locus is 2x  5y + 11 = 0 ]

12. Q.59 The distance betweenthe two parallellines is 1 unit .Apoint 'A' is chosento lie between the lines at a
distance 'd'fromone ofthem.TriangleABC is equilateralwith B ononeline and C onthe other parallel
line . The lengthofthe side ofthe equilateraltriangle is
(A) 1
d
d
3
2 2

 (B*)
3
1
d
d
2
2


(C) 1
d
d
2 2

 (D) 1
d
d2


[Hint: x cos( + 3 0º) = d  (1) & x sin  = 1  d  (2)
dividing 3 cot  =
1
1


d
d
, squaring equation(2) & putting
the value of cot , x2 =
1
3
(4d2 4d + 4)  x = 2
d d
2
1
3
 
]
Q.60 The locus of the mid points ofthe chords ofthecircle x²+y²+4x6y12 = 0 whichsubtend anangle
of

3
radians at 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.61 GivenA(0, 0) and B(x, y) with x (0, 1) and y> 0. Let the slope ofthe lineAB equals m1. Point C lies
on the line x = 1 suchthat the slope ofBC equals m2 where 0 < m2 < m1. Ifthe area ofthe triangleABC
can be expressed as (m1 – m2) f (x), then the largest possible value of f (x) is
(A) 1 (B) 1/2 (C) 1/4 (D*) 1/8
[Sol. Let the coordinates of C be (1, c)
m2 =
x
1
y
c


; m2 =
x
1
x
m
c 1


m2 – m2x = c – m1x
(m1 – m2)x = c – m2
c = (m1 – m2)x + m2 ....(1)
now area of ABC =
1
c
1
1
x
m
x
1
0
0
2
1
1 =
2
1
[cx – m1x] =
2
1
]
x
m
x
)
m
x
)
m
m
[(( 1
2
2
1 


=
2
1
]
x
m
x
m
x
)
m
m
[( 1
2
2
2
1 

 =
2
1
(m1 – m2)(x – x2) (x > x2 in (0, 1)
Hence, f (x) =
2
1
(x – x2); f (x)]max =
8
1
when x =
2
1
]
Q.62 If two chords of the circle x2 + y2  ax  by = 0, drawn from the point (a, b) is divided by the
xaxis in the ratio 2 : 1 then:
(A*) a2 > 3 b2 (B) a2 < 3 b2 (C) a2 > 4 b2 (D) a2 < 4 b2
[Hint: (a, b) lie on the circumference ofcircle. Get theother end ofchordpassing through (a, b) withsection
formula & substitute in circle . Put D > 0 ]

13. Q.63 P lies on the line y = x and Q lies on y = 2x. The equation for the locus of the mid point of PQ, if
| PQ | = 4, is
(A) 25x2 + 36xy + 13y2 = 4 (B*) 25x2 – 36xy + 13y2 = 4
(C) 25x2 – 36xy – 13y2 = 4 (D) 25x2 + 36xy – 13y2 = 4
Q.64 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.65 Ifthe vertices Pand 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, then the locus ofthe centroid ofthe 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.66 The angle at which the circles (x – 1)2 + y2 = 10 and x2 + (y – 2)2 = 5 intersect is
(A)
6

(B*)
4

(C)
3

(D)
2

[Hint: cos =
5
·
10
·
2
5
5
10 

=
2
1
  =
4

]
Q.67 The coordinates of the points A, B, C are ( 4, 0) , (0, 2) & ( 3, 2) respectively . The point of
intersectionofthe line whichbisectsthe angle CAB internallyand the line joiningCto themiddlepoint of
AB is :
(A) 






7
3
4
3
, (B) 






5
2
13
2
, (C)
7
3
10
3
, 





 (D*) 






5
2
3
2
,
[Hint: sides of the  ABC are 3, 5 and 2 5 , now proceed ]

14. Q.68 Two congruentcircleswithcentres at(2, 3)and (5,6) whichintersect at right angles has radiusequalto
(A) 2
2 (B*) 3 (C) 4 (D) none
[Hint: 2r2 = 32 + 32 = 18
r2 = 9
r = 3 ]
Q.69 Three lines x + 2y + 3 = 0 ; x + 2y– 7 = 0 and 2x – y – 4 = 0 formthe three sides oftwo squares. The
equationto the fourth side ofeach square is
(A) 2x – y + 14 = 0 & 2x – y + 6 = 0 (B) 2x – y + 14 = 0 & 2x – y – 6 = 0
(C) 2x – y – 14 = 0 & 2x – y – 6 = 0 (D*) 2x – y – 14 = 0 & 2x – y + 6 = 0
[Sol. d =
5
10
3rd side is parallel to the line 2x – y – 4 = 0
Hence line is 2x – y +  = 0
now
5
4


=
5
10
 + 4 = ± 10
 = 6 or  = – 14  (B) ]
Q.70 Acircle ofradius unityis centred at origin. Two particles start moving at the same time fromthe point
(1, 0) and move around the circle in opposite direction. One ofthe particle moves counterclockwise
with constant speed v and the other moves clockwise with constant speed 3v.After leaving (1, 0), the
two particlesmeet first at apoint P, and continue untiltheymeet next at point Q. Thecoordinatesofthe
point Q are
(A) (1, 0) (B) (0, 1) (C) (0, –1) (D*) (–1, 0)
[Sol. The particlewhich moves clockwise is moving three times
as fast asthe particle moving anticlockwise
This tells usthat inthe time that theclockwise particle travels (3/4)th of
the wayaround the circlethe anticlockwise particle willtravel(1/4)th of
the wayaround the circle and so the 2nd particle willmeet at p (0, 1).
Using thesame logic theywillmeet at Q (–1, 0) when theymeet the 2nd time]
Q.71 The pointsA(a, 0), B(0, b), C(c, 0) & D(0, d) are such that ac = bd & a, b, c, d are allnonzero. The
thepoints :
(A) forma parallelogram (B) do not lie on a circle
(C) forma trapezium (D*) are concyclic
[Hint : a/d = b/c  cyclic ]
Q.72 The value of 'c' for which the set, {(x, y)x2 + y2 + 2x  1}  {(x, y)x  y+ c  0} contains only
one point incommon is :
(A) (, 1]  [3, ) (B) {1, 3}
(C) {3} (D*) { 1 }
[Hint: x2 + y2 + 2x – 1 = 0; centre (–1, 0) and rad. = 2
line x – y + c = 0
 

1
2
2
c ; |c – 1| = 2; c – 1 = ± 2
 c = 3 or –1 ]

15. Q.73 GivenA  (1, 1) andAB is anyline throughit cutting the x-axis inB. IfAC is perpendiculartoAB and
meets the y-axis in C, thenthe equation of locus ofmid- point P ofBC is
(A*) x + y = 1 (B) x + y = 2 (C) x + y = 2xy (D) 2x + 2y = 1
[Hint: y – 1 = m (x – 1)
y – 1 = )
1
x
(
m
1


2h = 1
m
1

2k = 1 +
m
1
____________
locus is x + y = 1
____________ ]
Q.74 Acircle is inscribed into a rhombousABCD with one angle 60º. The distance from the centre of the
circle to the nearest vertex is equalto 1 . IfP is anypoint ofthe circle, then
PA PB PC PD
2 2 2 2
   is equal to :
(A) 12 (B*) 11 (C) 9 (D) none
[Hint: OA = 1
r = OA cos300 =
3
2
equation of is x2 + y2 = 3/4
PA2 + PB2 + PC2 + PD2
= 2
1
x + (y1 – 1)2 + (x1 + 3 )2 + 2
1
y
= 2
1
x + (y1 + 1)2 + (x1 – 3 )2 + 2
1
y
 8
y
4
x
4 2
1
2
1 
 = 8
)
y
x
(
4 2
1
2
1 
 = 8
4
3
·
4  = 11
1 ]
Q.75 Thenumber ofpossiblestraight lines, passing through(2, 3)andforminga trianglewithcoordinate 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. ]

16. Q.76 P is a point (a, b) in the first quadrant. If the two circles which pass through P and touch both the
co-ordinate axes cut at right angles, then:
(A) a2  6ab + b2 = 0 (B) a2 + 2ab  b2 = 0
(C*) a2  4ab + b2 = 0 (D) a2  8ab + b2 = 0
[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.77 In a triangleABC , ifA(2, – 1) and 7x – 10y+ 1 = 0 and 3x – 2y + 5 = 0 are equations ofan altitude
and anangle bisector respectivelydrawnfromB, then equationofBC is
(A) x + y + 1 = 0 (B*) 5x + y + 17 = 0 (C) 4x + 9y + 30 = 0 (D) x – 5y – 7 = 0
[Sol. BD and BE are intresect at B
Coordinates of B are (–3, –2)
mAB = 1/5
tan =
10
3
1
5
1
2
3


=
2
m
3
1
m
2
3


 1 =
m
3
2
m
2
3


or + 1 =
m
3
2
m
2
3


 m = 1/5 (rejected) or – 5
equation of BC = y + 2 = – 5 (x + 3)  5x + y + 17 = 0
Alternatively:Take image of(2, –1) inthe line BD to get a point on BC]
Q.78 The range of values of 'a' such that the angle  between the pair of tangents drawn from the 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.79 Distance of the point (2, 5) from the line 3x + y + 4 = 0 measured parallel to the line
3x  4y + 8 = 0 is
(A) 15/2 (B) 9/2 (C*) 5 (D) None
Q.80 Three concentric circles ofwhich the biggest is x2 + y2 = 1, have their radiiinA.P. Ifthe line y = x + 1
cuts allthe circles inrealand distinct points. Theintervalinwhichthecommon difference oftheA.P. will
lieis
(A) 0
1
4
,





 (B) 0
1
2 2
,





 (C*) 0
2 2
4
,








 (D) none

17. [Sol. Radius of circle are r1, r2 and 1
line y = x + 1
perpendicular from(0, 0) on line y = x + 1
=
2
1
now r1 >
2
1
 r1 = 1 – 2d 
2
r
1 1

= d
 d =
2
2
1
2 
Aliter :Equationofcircle are
x2 + y2 = 1
x2 + y2 = (1 – d)2
x2 + y2 = (1 – 2d)2
 solve any ofcircle with line y= x + 1
e.g. x2 + y2 = (1 – d)2  2x2 + 2x + 2d – d2 = 0 cuts the circle in real and distinct point hence  > 0
 2d2 – 4d + 1 > 0  d =
4
2
2 
]
Q.81 The co-ordinatesofthe vertices P, Q, R &S ofsquare PQRS inscribedinthe triangleABCwithvertices
A (0, 0), B  (3, 0) &C  (2, 1)given that two ofits vertices P, Qare onthe sideAB are respectively
(A)
1
4
0
3
8
0
3
8
1
8
1
4
1
8
, , , , , & ,























 (B)
1
2
0
3
4
0
3
4
1
4
1
2
1
4
, , , , , & ,
























(C) (1, 0) ,
3
2
0
3
3
1
2
1
1
2
, , , & ,

















 (D*)
3
2
0
9
4
0
9
4
3
4
3
2
3
4
, , , , , & ,
























Q.82 Atangent at a point on the circle x2 + y2 = a2 intersects a concentric circle Cat two points P and Q. The
tangents to the circle X at P and Q meet at a point onthe circle x2 + y2 = b2 then the equation ofcircle
is
(A*) x2 + y2 = ab (B) x2 + y2 = (a – b)2
(C) x2 + y2 = (a + b)2 (D) x2 + y2 = a2 + b2
[Hint: C.O.C. of the point Aw.r.t. x2 + y2 = r2 is
xbcos + ybsin = r2 .............(1)
This must be a tangent to the circle x2 + y2 = a2
r
b a
2
2 2 2 2
cos sin
 

= a  r2 = ab ]
Q.83 AB is the diameter of a semicircle k, C is an arbitrary point on the
semicircle (other thanAor B) and S is the centre ofthe circle inscribed
into triangleABC, thenmeasure of
(A) angleASB changes as C moves on k.
(B) angleASB is the samefor allpositions ofC but it cannot be determined without knowing the radius.
(C*) angleASB = 135° for all C.
(D) angleASB = 150° for all C.

18. Q.84 Tangents are drawn to the circle x2 + y2 = 1 at the points where it is met bythe 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
[Sol. Locus ofpoint ofintersection oftangents
chord of contact of (x1, y1) w.r.t. x2 + y2 = 1 is xx1 + yy1 = 1 (AB) .....(1)
AB is also common chord betweentwo circles
 – 1 + ( + 6)x – (8 – 2)y + 3 = 0
 ( + 6)x – (8 – 2)y + 2 = 0 .....(2)
comparing (1) and (2) we get
2
1
8
2
y
6
x1 






eliminate   2x – y + 10 = 0 ]
Q.85 Given
x
a
y
b
 = 1 and ax + by = 1 are two variable lines, 'a' and 'b' being the parameters connected by
the relation a2 + b2 = ab. The locus ofthe point ofintersectionhas the equation
(A*) x2 + y2 + xy  1 = 0 (B) x2 + y2 – xy + 1 = 0
(C) x2 + y2 + xy + 1 = 0 (D) x2 + y2 – xy – 1 = 0
[Sol. Let (h, k) be point ofintersection then
0
1
xy
y
x
1
hk
k
h
1
)
b
a
a
b
(
hk
k
h
multiply
1
a
b
b
a
1
kb
ah
ab
b
a
given
1
b
k
a
h
2
2
2
2
2
2
________
__________
2
2




















Note that the locus is not physically viable ]
Q.86 B &C are fixed points having coordinates (3, 0) and (3, 0) respectively. Ifthe verticalangle BAC is
90º, then the locus ofthe centroid ofthe ABC has the equation :
(A*) x2 + y2 = 1 (B) x2 + y2 = 2 (C) 9 (x2 + y2) = 1 (D) 9 (x2 + y2) = 4
[Hint: Let A(a, b) & G(h. k) Now A, G, O are collinear
 h =
20
3
.  a
 a = 3 h & similarly b = 3 k.
Now (a, b) lies on the circle x2 + y2 = 9  A ]
Q.87 The set ofvalues of 'b' for which theorigin and the point (1, 1) lie onthe same side ofthe 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 ]

19. Q.88 If a
a
,
1





 , b
b
,
1





 , c
c
,
1





 & d
d
,
1





 are four distinct points on a circle of radius 4 units then,
abcd is equalto
(A) 4 (B) 1/4 (C*) 1 (D) 16
[Sol. Let us assume that circle : x2 + y2 = 16
points are ofform 





t
1
,
t  t2 + 2
t
1
= 16 should satisfy
 t4 – 16t2 + 1 = 0
 product of roots = 1 ]
Q.89 Triangle formed by the lines x + y = 0 , x – y = 0 and lx + my = 1. If l and m vary subject to the
condition l 2 + m2 = 1 then the locus ofits circumcentre is
(A*) (x2 – y2)2 = x2 + y2 (B) (x2 + y2)2 = (x2 – y2)
(C) (x2 + y2) = 4x2 y2 (D) (x2 – y2)2 = (x2 + y2)2
[Sol. Coordinates ofcircumcentre 2
2
m

l
l
, 2
2
m
m
l

h = 2
2
m

l
l
....(1) and l2 + m2 = 1
and k = 2
2
m
m


l
....(2)
square and add (1) and (2)
h2 + k2 = 2
2
2
2
2
)
m
(
m


l
l
= 2
2
2
)
m
(
1

l
 2
2
2
)
m
(
1

l
= (h2 – k2)2
locus is x2 + y2 = (x2 – y2)2 ]
Q.90 Tangents aredrawn to a unit circle withcentre at the origin fromeachpoint 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 ]

20. Q.91 The coordinates ofthree points A(4, 0) ; B(2, 1)and C(3, 1)determine the vertices ofanequilateral
trapeziumABCD. The coordinates ofthe vertex D are :
(A) (6, 0) (B) ( 3, 0) (C) ( 5, 0) (D*) (9, 0)
[Hint: Equilateralmeanisosceles trapezium]
Q.92 ABCD isa square ofunit area.Acircle istangent to two sidesofABCD and passes through exactlyone
ofits vertices. The radius ofthe circle is
(A*) 2
2  (B) 1
2  (C)
2
1
(D)
2
1
[Hint: 2(1 – r)2 = r2
2 (1 – r) = r
r  
1
2  = 2
r =
1
2
2

=  
1
2
2  = 2
2  ]
Q.93 A parallelogramhas3 ofits vertices as (1, 2), (3, 8) and (4, 1).
The sumofallpossible x-coordinates for the 4th vertex is
(A) 11 (B*) 8
(C) 7 (D) 6
[Hint: Referdiagram ]
Q.94 Apair oftangents are drawn to a unit circle with centre at the origin and these tangents intersect atA
enclosing an angle 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
area ofquadrilateral= 3
sector =
2
1
· 1 ·
3
2
=
3

shaded region = 3 –
3

Ans. ]
Q.95 The image ofthe pair oflines represented by ax2 + 2h xy + by2 = 0 by the 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
]

21. Q.96 A straight line with slope 2 and y-intercept 5 touches the circle, x2 + y2 + 16x + 12y+ c = 0 at a point
Q. Then the coordinates ofQ are
(A) (–6, 11) (B) (–9, –13) (C) (–10, – 15) (D*) (–6, –7)
[Hint: y1 = 2x1 + 5 and 8
x
)
6
y
(
1
1


× 2 = – 1
 x1 = – 6 and y1 = – 7 ]
Q.97 The acute anglebetweentwo straight lines passing throughthe point M(6, 8) andthe pointsinwhich
the line segment 2x + y + 10 = 0 enclosed between the co-ordinate axes is divided in the ratio
1 : 2: 2 in the directionfromthe point ofits intersectionwiththe xaxis to the point ofintersectionwith
the yaxis is :
(A) /3 (B*) /4 (C) /6 (D) /12
[Hint: the co-ordinate of P and Q are (4, 2) and (2, 6) respectively ]
Q.98 Avariable circle cuts each of the circles x2 + y2  2x = 0 & x2 + y2  4x  5 = 0 orthogonally. The
variable circle passes throughtwo fixed points whose coordinates are :
(A)
 








5 3
2
0
, (B*)
 








5 3 5
2
0
, (C)
 








5 5 3
2
0
, (D)
 








5 5
2
0
,
[Hint: c =  5 ; g = 5/2  B ]
Q.99 If in triangle ABC , 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)
[Hint: O, G, C are collinear. Get G ]
Q.100 The radicalcentre ofthree circlestaken in pairs describedonthe sides ofa triangleABC asdiametres is
the:
(A) centroid ofthe ABC (B) incentre ofthe ABC
(C) circumcentre o the ABC (D*) orthocentre ofthe ABC
[Hint: observe bymaking a figure or proceed analyticallybytaking (xr, yr).
r = 1, 2, 3 as the coordinates ofthe vertices ]
Q.101 The line x + y = p meets the axis of x & y at A& B respectively .AtriangleAPQ is inscribed in the
triangle OAB, O being the origin, withright angle at Q . Pand Q lie respectivelyon OB andAB. Ifthe
area of the triangleAPQ is 3/8th ofthe area ofthe triangle OAB, then
AQ
BQ
is equal to :
(A) 2 (B) 2/3 (C) 1/3 (D*) 3
[Sol.
8
3
AOB
AQP



or
8
3
p
2
1
)
1
(
p
2
2
2




B
A
O
Q
P

1
(0,p)







 1
,
1 


P
P








1
)
1
(
,
0

 P
(p, 0)

3
1
,
3



22. 3
1
or
3
BQ
AQ

3
1
is rejected because this gives negative coordinater ofP and it is gives that P lies on OB. ]
Q.102 Two circles are drawn through the points (1, 0) and (2, 1) to touch the axis of y. Theyintersect at an
angle:
(A*) cot–1
3
4
(B) cos 1
4
5
(C)

2
(D) tan1 1
[Hint: m1 =  ; m2 =  . The two circles are
x2 + y2  2x + 2y + 1 = 0 and x2 + y2  10x  6y + 9 = 0 ]
Q.103 In a triangle ABC, side AB has the equation 2 x + 3 y = 29 and the side AC 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.104 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 externally
equation ofcommon tangent is, 3 x + y – 4 = 0 ....(1)
comparing it with x cos + ysin = 2
 =
6

]
Q.105 ABC is anisosceles triangle . Ifthe co-ordinates ofthe base are (1, 3) and (2, 7), then co-ordinates
ofvertexAcan be :
(A)  
 1
2
5
, (B)  
 1
8
5
, (C)  
5
6
5
,  (D*)  
7 1
8
,
[Hint : Vertex willsatisfythe equation ofthe perpendicular bisectors except themiddle point ]
Q.106 Acircleis drawnwithy-axisas a tangent andits centre at thepoint whichis thereflectionof (3, 4) inthe
line y= x. The equation ofthe circle is
(A) x2 + y2 – 6x – 8y + 16 = 0 (B) x2 + y2 – 8x – 6y + 16 = 0
(C*) x2 + y2 – 8x – 6y + 9 = 0 (D) x2 + y2 – 6x – 8y + 9 = 0

23. Q.107 A is a point on either of two lines y + 3 x= 2 at a distance of
4
3
units from their point of
intersection. Theco-ordinates ofthe foot ofperpendicular fromAonthe bisector ofthe angle between
themare
(A) 






2
3
2
, (B*) (0, 0) (C)
2
3
2
,





 (D) (0, 4)
[Sol. Drawfigure
2
x
3
y 
 for x > 0
2
x
3
y 

 for x < 0 ]
Q.108 A circle ofconstant radius 'a' passes through origin 'O' and cuts the axes ofcoordinates inpoints P
and Q, thenthe equation ofthe locus ofthe foot ofperpendicular fromO to PQ is :
(A) (x2 + y2) 1 1
2 2
x y



 

 = 4 a2 (B) (x2 + y2)2 1 1
2 2
x y



 

 = a2
(C*) (x2 + y2)2 1 1
2 2
x y



 

 = 4 a2 (D) (x2 + y2) 1 1
2 2
x y



 

 = a2
[Sol. Equationofline PQ is
y – k = –
k
h
(x – h)  hx + ky = h2 + k2 ; also 2a = 2
1
2
1 y
x   2
1
2
1 y
x  = 4a2
 points Q 






 
0
,
h
k
h 2
2
and P 






 
k
k
h
,
0
2
2
(x2 + y2)2








 2
2
y
1
x
1
= 4a2 ]
Q.109 Three straight linesaredrawnthroughapoint Plying intheinterior ofthe ABCandparallelto its sides.
The areas ofthe three resulting triangles with P as the vertexare s1, s2 and s3. The area ofthe triangle
in terms of s1, s2 and s3 is :
(A) s s s s s s
1 2 2 3 3 1
  (B) s s s
1 2 3
3
(C*)  
s s s
1 2 3
2
  (D) none
[Hint:
s
s
1
=
p
a
2
2 
s
s
1
=
p
a
similarly
s
s
2
=
q
a
and
s
s
3
=
r
a

s s s
s
1 2 3
 
=
p q r
a
 
= 1
 s =  
s s s
1 2 3
2
  ]

24. Q.110 The circle passing throughthe distinct points (1, t), (t, 1) & (t, t) for allvalues of 't' , passes through
the point :
(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.111 The sides of a ABC are 2x  y + 5 = 0 ; x + y  5 = 0 and x  2y  5 = 0 . Sum of the tangents
ofits interiorangles is :
(A) 6 (B*) 27/4 (C) 9 (D) none
[Hint : tanA + tan B + tan C = tan A tan B tan C =
3
4
. 3 . 3 =
27
4
]
Q.112 If a circle of constant radius 3k passes through the origin 'O' and meets co-ordinate axes atAand B
then the locus ofthe centroid ofthe triangle OAB 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.113 Chords ofthe curve 4x2 + y2  x +4y= 0 which subtend a right angle at the originpassthrough a fixed
point whose co-ordinates are :
(A*)
1
5
4
5
, 





 (B) 






1
5
4
5
, (C)
1
5
4
5
,





 (D)  






1
5
4
5
,
Q.114 Let x &ybe the realnumbers satisfyingthe equationx2  4x+ y2 +3 = 0. Ifthe maximumand minimum
values ofx2 + y2 are M & mrespectively, then the numericalvalue of M  mis :
(A) 2 (B*) 8 (C) 15 (D) noneofthese
Q.115 Ifthe straight lines joining the originand the points ofintersectionofthe curve
5x2 + 12xy  6y2 + 4x  2y + 3 = 0 and x + ky  1 = 0
are equallyinclined to the co-ordinate axes then the value of k :
(A) is equalto 1 (B*) is equal to 1
(C) is equal to 2 (D) does not exist inthe set ofrealnumbers .

25. [Hint: Homogenise and put co-efficient of xy= 0 ]
Q.116 A line meets the co-ordinate axes in A & B. A circle is circumscribed about the triangle OAB. If
d1 &d2 are the distances ofthetangent to the circleat the originOfromthe pointsAand B respectively,
the diameter ofthe circle is :
(A)
2
d
d
2 2
1 
(B)
2
d
2
d 2
1 
(C*) d1 + d2 (D)
2
1
2
1
d
d
d
d

[Sol. Let the circle be x2 + y2 + 2gx + 2fy = 0
Tangent at the origin is gx + fy= 0
d1 =
2
2
2
f
g
g
2

d2 = 2
2
2
f
g
f
2

 d1d2 = 2
2
f
g
2  = diameter of the circle]
Q.117 A pair ofperpendicular straight lines is drawn through the origin forming with the line 2x + 3y= 6 an
isosceles triangle right angled at the origin. The equationto the line pairis :
(A*) 5x2  24xy  5y2 = 0 (B) 5x2  26xy  5y2 = 0
(C) 5x2 + 24xy  5y2 = 0 (D) 5x2 + 26xy  5y2 = 0
[Sol. 2x + 3y =6
tan 45o =
)
3
2
(
1
)
3
2
(




m
m
get two value ofm m1 = -5, m2 = 1/5 ]
Q.118 The equation of a line inclined at an angle

4
to the axis X, such that the two circles
x2 + y2 = 4, x2 + y2 – 10x – 14y + 65 = 0 intercept equal lengths on it, is
(A*) 2x – 2y – 3 = 0 (B) 2x – 2y + 3 = 0 (C) x – y + 6 = 0 (D) x – y – 6 = 0
[Sol. Let equation ofline be y = x + c
y – x = c ....(1)
perpendicular from (0, 0) on (1) is
2
c

=
2
c
In AON,
2
2
2
c
2 





 = AN
and in CPM,
2
c
2
32

 = CM
perpendicular from(5, 7) on line y– x = c =
2
c
2 

26. Given AN = CM =
2
c
4
2
 = 9 –
2
)
c
2
( 2

 c = –
2
3
 equation of line y = x –
2
3
of 2x – 2y – 3 = 0 ]
Q.119 If the line y = mx bisects the angle between the lines ax2 + 2h xy + by2 = 0 then m is a root of the
quadratic equation:
(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.120 Tangents are drawn fromanypoint on the circle x2 + y2 = R2 to the circle x2 + y2 = r2. Ifthe line joining
the points ofintersection ofthese tangentswith the first circlealso touch the second, thenR equals
(A) 2 r (B*) 2r (C)
2
2 3
r

(D)
4
3 5
r

[HInt: ]
Q.121 An equilateraltriangle has each ofits sides oflength 6cm. If(x1, y1); (x2, y2) &(x3, y3) are its vertices
thenthe valueofthe 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
[Hint:
2
1
1
y
x
1
y
x
1
y
x
3
3
2
2
1
1
= A  D = 4A2  D = 4
2
36
·
4
3








= 972 Ans. ]
Q.122 A variable circle C has the equation
x2 + y2 – 2(t2 – 3t + 1)x – 2(t2 + 2t)y + t = 0, where t is a parameter.
Ifthe power ofpoint P(a,b) w.r.t. the circle C is constant then the ordered pair (a, b) is
(A) 






10
1
,
10
1
(B*) 






10
1
,
10
1
(C) 





10
1
,
10
1
(D) 







10
1
,
10
1
[Sol. Power : P = a2 + b2 – 2(t2 – 3t + 1)a – 2(t2 + 2t)b + t [12 & 13th test (29-10-2005)]
= – (2a + 2b)t2 + (6a – 4b + 1)t + a2 + b2 – 2a
This poweris independent ofthe parameter t
ifandonlyif 2a + 2b = 0  a = – b

27. and 6a – 4b + 1 = 0
 a = –
10
1
and b =
10
1
]
Q.123 Points A& B are inthe first quadrant ; point 'O' is the origin . Ifthe slope ofOAis 1, slope ofOB is 7
and OA= OB, then the slope ofAB is :
(A)  1/5 (B)  1/4 (C)  1/3 (D*)  1/2
[Sol. tan = 7 ; OA = OB = r sin =
2
5
7
cos =
2
5
1
now mAB =
2
1

]
Q.124 Let C be a circle with two diameters intersecting at anangle of30 degrees.Acircle S istangent to both
the diameters and to C, and has radius unity. The largest radius ofC is
(A*) 1 + 2
6  (B) 1 + 2
6  (C) 2
6  – 1 (D) noneofthese
[Hint: cosec15° =
1
x
x = cosec15°
R = x + 1 = 1 + cosec 15°
= 1 +
1
3
2
2

= 1 +
2
6
4

= 1 + 2
6  ]
Q.125 The co-ordinates ofa point Pon the line 2x  y+ 5 = 0 suchthat PA PB is maximum where A
is (4,  2) and B is (2, 4) will be :
(A) (11, 27) (B*) ( 11, 17) (C) ( 11, 17) (D) (0, 5)
[Hint : PA PB will be maximumif P, A & B willbe collinear . Hence
2 a  b + 5 = 0 and the determinant
1
4
2
1
2
4
1
b
a

 = 0 ]
Q.126 A straight line l1 withequation x – 2y+ 10 = 0 meets the circle with equation x2 + y2 = 100 at B in the
first quadrant.Aline through B, perpendicular to l1 cuts the y-axis at P(0, t). The value of't' is
(A) 12 (B) 15 (C*) 20 (D) 25
[Sol. slope of l1 =
2
1
slope of l2 = – 2
equation of l2
y = – 2(x – 10)  y + 2x = 20
Hence t = 20 Ans.]
Q.127 A variable circle C has the equation
x2 + y2 – 2(t2 – 3t + 1)x – 2(t2 + 2t)y + t = 0, where t is a parameter.
The locus ofthe centre ofthe circle is
(A*) a parabola (B) an ellipse (C) a hyperbola (D) pairofstraight lines

28. [Sol. centre is x = t2 – 3t + 1 ....(1)
y = t2 + 2t ....(2)
eliminating t,weget
x = t2 + 2t – 5t + 1 = y – 5t + 1
t =
5
1
x
y 

Substituting thevalue of t in(2)
y =
2
5
1
x
y





 

+ 2 




 

5
1
x
y
25y = (y – x + 1)2 + 10(y – x + 1)
25y = y2 + x2 + 1 – 2xy – 2x + 2y + 10y – 10x + 10
x2 + y2 – 2xy – 12x – 13y + 11 = 0
which is a parabola
as   0 and h2 = ab ]
Q.128 Let aand brepresent thelengthofa right triangle's legs. Ifdis the diameter
of a circle inscribed into the triangle, and D is the diameter of a circle
superscribed onthe triangle, then d + D equals
(A*) a + b (B) 2(a + b)
(C)
2
1
(a + b) (D) 2
2
b
a 
[Sol. AB = 
 b
a2
hence D = 2
2
a
b  ....(1)
Now
2
d
=
s

=
s
2
ab

2
d
= 2
2
b
a
b
a
ab



or d = 2
2
b
a
b
a
ab
2



....(2)
from(1) and (2)
d + D =
2
2
2
2
2
2
b
a
b
a
ab
2
b
a
)
b
a
(
b
a









 



= 2
2
2
2
2
b
a
b
a
b
a
)
b
a
(
)
b
a
(







 (A) ]
Select the correct alternatives : (More than one are correct)
Q.129 The area of triangleABC is 20 cm2. The coordinates of vertexAare (5, 0) and B are (3, 0). The
vertex C lies on the line, x  y = 2 . The coordinates ofC are
(A) (5, 3) (B*) ( 3,  5) (C) ( 5,  7) (D*) (7, 5)

29. Q.130 Apoint (x1, y1) is outside the circle, x2 + y2 + 2gx+ 2fg + c = 0 with centre at the origin andAP,AQ are
tangents to thecircle. Then:
(A*) area of the quadriletralformed by the pair oftangents and the corresponding radii through the
points ofcontact is   
g f c x y gx fy c
2 2
1
2
1
2
1 1
2 2
     
(B*) equation ofthe circle circumscribing the APQ is, x2 + y2 + x(g – x1) + y(f y1) – (gx1 + fy1)= 0
(C) least radius ofa circle passing through the points 'A' & the originis, ( ) ( )
x g y f
1
2
1
2
  
(D*) the  between the two tangent is,
  
1
2
2 2 2
1
2 2
1
2
1
2
1 1
1
2
1
2
sin
( ) ( )

     
  










g f c x y gx fy c
x g y f
Q.131 Let u  ax + by + a b
3 = 0 v  bx  ay + b a
3 = 0 a, b  R be two straight lines. The equation of
the bisectors of the angle formed by k1u  k2v = 0 & k1u + k2v = 0 for non zero real k1 & k2 are:
(A*) u = 0 (B) k2u + k1v = 0 (C) k2u  k1v = 0 (D*) v = 0
[Hint: Note that the lines are perpendicular.
Assume the co-ordinate axes to be directed along u = 0 & v= 0.
Now the lines k1 u  k2 v = 0 & k1 u + k2 v = 0 are equally inclined with u v axes.
Hence the bisectors are u = 0 & v = 0. i.e.
2
2
2
1
2
1
k
k
u
k
u
k


= ±
2
2
2
1
2
1
k
k
u
k
u
k


]
Q.132
x x
 1
cos
=
y y
 1
sin
= r, represents :
(A*) equation of a straight line, if is constant & r is variable
(B*) equation of a circle, if r is constant &  is a variable
(C*) a straight line passing through a fixed point & having a known slope
(D*) a circle witha known centre &a given radius.
Q.133 Allthe points lying inside the triangle formed bythe points (1, 3), (5, 6) & (1, 2) satisfy
(A*) 3x + 2y  0 (B*) 2x + y + 1  0 (C) 2x + 3y  12  0 (D*)  2x + 11  0
Q.134 The equations ofthe tangents drawn fromthe originto the circle, x² + y²  2rx  2hy+ h² = 0 are
(A*) x = 0 (B) y = 0
(C*) (h²  r²) x  2rhy = 0 (D) (h²  r²)x + 2rhy = 0
Q.135 The co-ordinates of the fourth vertex of the parallelogram where three of its vertices are ( 3, 4);
(0,  4) & (5, 2) can be :
(A*) (8, 6) (B*) (2, 10) (C*) ( 8,  2) (D) none
Q.136 The equation of a circle with centre (4, 3) and touching the circle x2 + y2 = 1 is :
(A) x2 + y2  8x  6y  9 = 0 (B) x2 + y2  8x  6y + 11 = 0
(C*) x2 + y2  8x  6y  11 = 0 (D*) x2 + y2  8x  6y + 9 = 0
[Hint: Drawfigure]

30. Q.137 Two vertices of the ABC are at the pointsA(1, 1) and B(4, 5) and the third vertex lines on the
straight line y= 5(x  3) . Ifthe area ofthe  is19/2 then the possible coordinates ofthe vertex C are:
(A*) (5, 10) (B*) (3, 0) (C) (2,  5) (D) (5, 4)
Q.138 A circle passesthrough the points (1, 1) , (0, 6) and (5, 5) . The point(s) on this circle, the tangent(s)
at whichis/are parallelto thestraight line joining theorigin to its centreis/are :
(A) (1, 5) (B*) (5, 1) (C) ( 5,  1) (D*) ( 1, 5)
[Hint : Note that  is right angled at (0, 6) . Centre ofthe circle is (2, 3) . Slope ofthe line joining originto the
centre is 3/2. Take parametric equationofa line through(2, 3) with
tan  = 
2
3
as
x  2
cos
=
y  3
sin
= ± r where r = 13 .
Get the coordinates onthe circle ]
Q.139 Line
x
a
y
b
 = 1 cuts the coordinate axes at A(a, 0) & B (0, b) & the line
x
a
y
b



=  1 at
A (a, 0) & B(0, b). If the points A, B, A, B are concyclic then the orthocentre ofthe triangle
ABA is:
(A) (0, 0) (B*) (0, b') (C*) 0 ,
aa
b






 (D) 0 ,
'
bb
a






Q.140 Point M moved along thecircle (x 4)2 + (y 8)2 = 20 . Then it broke awayfromit and moving along
a tangent to the circle, cuts the xaxis at the point (2, 0) . The coordinates ofthe point onthe circle
at which the moving point broke awaycan be :
(A) 






3
5
46
5
, (B*) 






2
5
44
5
, (C*) (6, 4) (D) (3, 5)
[Hint: Compute chord ofconstant of(–2, 0)
and verifyeach alternative w.r.t. the and C.O.C. ]
Q.141 If one vertex of an equilateral triangle of side 'a' lies at the origin and the other lies on the line
x  3 y = 0 then the co-ordinates ofthe third vertex are :
(A*) (0, a) (B*)
3
2 2
a a
, 







 (C*) (0,  a) (D*) 








3
2 2
a a
,
[Hint : make a figureand interpret ]
Q.142 The circles x2 + y2 + 2x + 4y  20 = 0 & x2 + y2 + 6x  8y + 10 = 0
(A*) aresuchthat the numberofcommontangents onthemis 2
(B) arenot orthogonal
(C*) are suchthat the length oftheir commontangent is 5(12/5)1/4
(D*) are such that the lengthoftheir common chord is 5
3
2
.
[Hint: r1 = 5 ; r2 = 15 ; d = 40  ABCD ]

31. Q.143 Two straight lines u = 0 and v= 0 passes through the origing forming an angle of
tan1 (7/9) witheachother . Iftheratio ofthe slopesof u = 0 and v = 0 is 9/2 thentheir equations are:
(A*) y = 3x & 3y = 2x (B*) 2y = 3x & 3y = x
(C*) y + 3x = 0 & 3y + 2x = 0 (D*) 2y + 3x = 0 & 3y + x = 0
[Hint:
m1
9
=
m2
2
= k  m1 = 9k ; m2 = 2k &
m m
m m
1 2
1 2
1


= ± 7/9 ]
Q.144 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
Alternatively: OP = 3/2
2
4
1

 =
2
3
;
2
4
1

 =
4
9
 2  l = ± 2 ]
Q.145 Given two straight lines x  y 7 = 0 and x  y+ 3 = 0. Equationofa line whichdivides the distance
between themin the ratio 3 : 2 can be :
(A*) x  y  1 = 0 (B*) x  y  3 = 0 (C) y = x (D) x  y + 1 = 0
Q.146 The circles x2 + y2  2x  4y + 1 = 0 and x2 + y2 + 4x + 4y  1 = 0
(A) touchinternally
(B*) touch externally
(C*) have 3x + 4y  1 = 0 as the common tangent at the point ofcontact.
(D) have 3x + 4y+ 1 = 0 as the common tangent at the point of contact.
Q.147 Three vertices of a triangle are A(4, 3) ; B(1,  1) and C(7, k) . Value(s) of k for which centroid,
orthocentre, incentre and circumcentre ofthe ABC lie on the same straight line is/are :
(A) 7 (B*) 1 (C*)  19/8 (D) none
[Hint: For K = 7, the points are collinear ]

32. Q.148 Aand B are two fixed points whose co-ordinates are(3, 2) and (5, 4) respectively. The co-ordinates of
a point PifABPis anequilateraltriangle, is/are :
(A*)  
4 3 3 3
 
, (B*)  
4 3 3 3
 
,
(C)  
3 3 4 3
 
, (D)  
3 3 4 3
 
,
[Hint : use parametric ]
Q.149 Which ofthe following lines have the intercepts ofequallengths on the circle, x2 + y2  2x + 4y= 0?
(A*) 3x  y = 0 (B*) x + 3y = 0 (C*) x + 3y + 10 = 0 (D*) 3x  y  10 = 0
[Hint: Chords equidistance fromthe centre are equal ]
Q.150 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 . Thenthe equation ofa line BC passing throughthe 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.151 Equationofa straight line passingthroughthe point (2, 3) and inclined at anangle of
arctan
1
2
with the line y+ 2x = 5 is :
(A) y = 3 (B*) x = 2 (C*) 3x + 4y  18 = 0(D) 4x + 3y  17 = 0
[Hint : m =  3/4 or  ]
Q.152 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.153 Consider the equation y y1 =m(x x1). If m& x1 are fixedand different lines are drawnfor different
values of y1, then
(A) the lines willpass througha fixed point (B*) there willbe a set ofparallellines
(C*) allthe lines intersect the line x= x1 (D) allthe lines willbe parallel to the line y= x1.