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Q.B on Straight line and circle [1]

STRAIGHT LINE[STRAIGHT OBJECTIVE TYPE]

Q.1 If the 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

1434111

= 0

1 (3 – 4) – 1 (4 – 4) + 1 (4 – 3) = 3 – 4 – 4 + 4 + 4 – 3= – + 1 = 0 = 1

= ± 1 ]

Q.2 Given the family of lines, a (3x + 4y + 6) + b (x + y + 2) = 0 . The line of the family situated at the greatestdistance from the 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 is A ( 2, 0) . The required line will be one which passes through ( 2, 0) and isperpendicular to the line joining ( 2, 0) and (2, 3) or taking (2, 3) as centre and radius equal to PA drawa circle, the required line will be a tangent to the circle at ( 2, 0) ]

Q.3 A variable rectangle PQRS has its sides parallel to fixed directions. Q and S lie respectively on the linesx = a, x = a and P lies on the x axis. Then the locus of R is(A*) a straight line (B) a circle (C) a parabola (D) pair of straight lines

[Hint: Mid point of QS = Mid point of PS0 = h + x1 x1 = –h hence p (–h, 0)

equation of PQ y = mx + cpasses through (–h, 0)

c = mh y = mx + mhsince Q lies on it

y 1 = ma + mh

now mQR = –1m

= y Ka h1

–1m

=m a h K

a h( )

simplifying h + mk = a + (a + h)m2 a st. line]

Q.4 A rectangular billiard table has vertices at P(0, 0), Q(0, 7), R(10, 7) and S (10, 0). A small billiard ballstarts at M(3, 4) and moves in a straight line to the top of the table, bounces to the right side of the table,then comes to rest at N(7, 1). The y-coordinate of the point where it hits the right side, is(A*) 3.7 (B) 3.8 (C) 3.9 (D) noneUse Image method

Q.5 B and C are points in the xy plane such that A(1, 2) ; B (5, 6) and AC = 3BC. Then(A) ABC is a unique triangle (B) There can be only two such triangles.(C) No such triangle is possible (D*) There can be infinite number of such triangles.

[Hint: Locus of C is a circle 8(x2 + y2) – 88x – 104y + 544 = 0 (D)]


Q.6 A ray of light passing through the point A (1, 2) is reflected at a point B on the x axis and then passesthrough (5, 3). Then the equation of AB is :(A*) 5x + 4y = 13 (B) 5x 4y = 3(C) 4x + 5y = 14 (D) 4x 5y = 6

[Hint : a 1

2 =

53 a

a = 135

]

Q.7 m, n are integer with 0 < n < m. A is the point (m, n) on the cartesian plane. B is the reflection of A in theline y = x. C is the reflection of B in the y-axis, D is the reflection of C in the x-axis and E is the reflectionof D in the y-axis. The area of the pentagon ABCDE is(A) 2m(m + n) (B*) m(m + 3n) (C) m(2m + 3n) (D) 2m(m + 3n)

[Sol. Area of rectangle BCDE = 4mn

Area of ABC = 2)nm(m2

= m2 – mn area of pentagon = 4mn + m2 – mn

= m2 + 3mn Ans. ]

Q.8 The area enclosed by the graphs of | x + y | = 2 and | x | = 1 is(A) 2 (B) 4 (C) 6 (D*) 8

[Hint: figure is a parallelogram

Area =

2·

2312 = 8 Ans. ]

Q.9 If P = (1, 0) ; Q = (1, 0) and R = (2, 0) are three given points, then the locus of the points S satisfyingthe relation, SQ2 + SR2 = 2 SP2 is :(A) a straight line parallel to xaxis (B) a circle passing through the origin(C) a circle with the centre at the origin (D*) a straight line parallel to yaxis .

Q.10 The equation of the pair of bisectors of the 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) 11x + 2y = 0 (C) 38x + 41y = 0 (D) 11x – 2y = 0

[ Hint : (2y x) (y mx) = mx2 xy (2m + 1) + 2y2 = 0 the equation to the pair of bisectors are :

x ym

2 2

2 =

22 1

xym 12x2 7xy 12y2 12

1m2 = 7

)2m(2

or 38m = 41 m = 3841

]

Q.11 Two points A(x1, y1) and B(x2, y2) are chosen on the graph of f (x) = ln x with 0 < x1 < x2. The pointsC and D trisect line segment AB with AC < CB. Through C a horizontal line is drawn to cut the curve atE(x3, y3). If x1 = 1 and x2 = 1000 then the value of x3 equals(A*) 10 (B) 10 (C) (10)2/3 (D) (10)1/3

[Sol. Using section formula


a = 31000·11·2

= 334

b = 31000n·10·2 l

b = 31000nl

....(1)

now line y = b intersects the curve y = ln x b = ln x ....(2)from (1) and (2)

31000nl

= ln x ln (1000)1/3 = ln x

x = (1000)1/3 = 10 Ans. ]

Q.12 If L is the line whose equation is ax + by = c. Let M be the reflection of L through the y-axis, and let Nbe the reflection of L through the x-axis. Which of the following must be true about M and N for allchoices of a, b and c?(A) The x-intercepts of M and N are equal.(B) The y-intercepts of M and N are equal.(C*) The slopes of M and N are equal.(D) The slopes of M and N are reciprocal.

[Sol. Reflecting a graph over the x-axis results in the line M whoseequation is ax – by = c, while a reflection through the y-axisresults in the line N whose equation is –ax + by = c. Both clearlyhave slope equal to a/b (from, say, the slope-intercept form ofthe equation.) ]

Q.13 The distance between the two parallel lines is 1 unit . A point 'A' is chosen to lie between the lines at adistance 'd' from one of them . Triangle ABC is equilateral with B on one line and C on the other parallelline . The length of the side of the equilateral triangle is

(A) 1dd32 2 (B*)

31dd2

2 (C) 1dd2 2 (D) 1dd2

[ Hint : x cos ( + 3 0º) = d (1) and x sin = 1 d (2)

dividing 3 cot = 11

dd

, squaring equation (2) and putting

the value of cot , x2 = 13

(4d2 4d + 4) x = 2 d d2 13

]

Q.14 Given A(0, 0) and B(x, y) with x (0, 1) and y > 0. Let the slope of the line AB equals m1. Point C lieson the line x = 1 such that the slope of BC equals m2 where 0 < m2 < m1. If the area of the triangle ABCcan 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 = x1yc

; m2 = x1

xmc 1

m2 – m2x = c – m1x(m1 – m2)x = c – m2c = (m1 – m2)x + m2 ....(1)


now area of ABC = 1c11xmx100

21

1 = 21

[cx – m1x] = 21

]xmx)mx)mm[(( 1221

= 21 ]xmxmx)mm[( 12

221 =

21

(m1 – m2)(x – x2) (x > x2 in (0, 1)

Hence, f (x) = 21

(x – x2); f (x)]max = 81

when x = 21

]

Q.15 Consider a parallelogram whose sides are represented by the lines 2x + 3y = 0; 2x + 3y – 5 = 0;3x – 4y = 0 and 3x – 4y = 3. The equation of the diagonal not passing through the origin, is(A) 21x – 11y + 15 = 0 (B) 9x – 11y + 15 = 0(C) 21x – 29y – 15 = 0 (D*) 21x – 11y – 15 = 0

[Sol. Equation of AC(2x + 3y)(3x – 4y) = (3x – 4y – 3)(2x + 3y – 5)

or 5(3x – 4y) + 3(2x + 3y) – 15 = 021x – 11y – 15 = 0 Ans.

equation of BD is(2x + 3y)(3x – 4y – 3) = (3x – 4y)(2x + 3y – 5)5(3x – 4y) – 3(2x + 3y) = 09x – 29y = 0 ]

Q.16 What is the y-intercept of the line that is parallel to y = 3x, and which bisects the area of a rectangle withcorners at (0, 0), (4, 0), (4, 2) and (0, 2)?(A) (0, – 7) (B) (0, – 6) (C*) (0, – 5) (D) (0, – 4)

[Sol. Area of OADE = Area of ABCDsince Area of ADD' = Area of A'DAso area of rectangle AA'OE = Area of rectangle D'DCBBD' = OA4 – h2 = h1h1 + h2 = 4 ....(1)

mAD = 3 12 hh

02

= 3 h2 – h1 = 32

....(2)

from (1) and (2) 2h1 = 4 – 32

= 310

h1 = 35

Equation of line AD is y – 0 = 3

35x

put x = 0, get y = – 5(0, – 5) Ans. ]

Q.17 Given A (1, 1) and AB is any line through it cutting the x-axis in B. If AC is perpendicular to AB andmeets the y-axis in C, then the equation of locus of midpoint P of BC 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 = )1x(m1

2h = 1 m1

2k = 1 + m1

____________ locus is x + y = 1

____________ ]

Q.17 Given xa

yb

= 1 and ax + by = 1 are two variable lines, 'a' and 'b' being the parameters connected by

the relation a2 + b2 = ab. The locus of the point of intersection has 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 of intersection then

01xyyx1hkkh

1)ba

ab(hkkhmultiply

1ab

ba1kbah

abbagiven1bk

ah

22

22

22__________________

22

Note that the locus is not physically viable ]Q.18 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 of its 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 of circumcentre 22 mll

, 22mm

l

h = 22 mll

....(1)

and k = 22 mm

l

....(2)

also l2 + m2 = 1square and add (1) and (2) and then subtract (1) and (2)

h2 + k2 = 222

22

)m(m

ll

= 222 )m(1l

and h2 – k2 = 222

22

)m(m

ll

= 22 m1l

hence h2 + k2 = (h2 – k2)2

locus is x2 + y2 = (x2 – y2)2 ]


Q.19 Area of the triangle formed by the line x + y = 3 and the angle bisectors of the line pairx2 – y2 + 4y – 4 = 0 is(A*) 1/2 (B) 1(C) 3/2 (D) 2

[Sol. x2 – (y2 – 4y + 4) = 0 x2 – (y – 2)2 = 0 (x + y – 2)(x – y + 2) = 0

Area = 21·1

= 21

Ans. ]

Q.20 If the straight lines , ax + amy + 1 = 0 , b x + (m + 1) b y + 1 = 0 and cx + (m + 2)cy + 1 = 0, m 0are 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)2m(cc1)1m(bb1ama

= 0 ;

1c2c1bb10a

= 0

a(b – 2c) +(2bc – bc) = 0 ab – 2ac + bc = 0 b(a + c) = 2ac

b = caac2

a, b, c are in H.P. ]

Q.21 A is a point on either of two lines y + 3 x= 2 at a distance of 43 units from their point of

intersection. The co-ordinates of the foot of perpendicular from A on the bisector of the angle betweenthem are

(A)

23

2, (B*) (0, 0) (C) 23

2,

(D) (0, 4)

[Sol. Draw figure

2x3y for x > 0

2x3y for x < 0 ]

Q.22 P is a point inside the triangle ABC. Lines are drawn through P, parallel to the sides of the triangle. Thethree resulting triangles with the vertex at P have areas 4, 9 and 49 sq. units. The area of the triangle ABCis(A) 32 (B) 12 (C) 24 (D*) 144

[Sol.49

= 2

2

aq

or aq7

....(1)

|||ly ap2

....(2) and a

r3

....(3)

(1) + (2) + (3) gives 12

= arqp

= 1 = 144 ]


Q.23 Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is2x + y = 3, then the equation representing the pair of lines PQ and PR is(A) 3x2 3y2 + 8xy + 20x + 10y + 25 = 0 (B*) 3x2 3y2 + 8xy 20x 10y + 25 = 0(C) 3x2 3y2 + 8xy + 10x + 15y + 20 = 0 (D) 3x2 3y2 8xy 10x 15y 20 = 0

Q.24 An equilateral triangle has each of its sides of length 6 cm . If (x1, y1) ; (x2, y2) and (x3, y3) are its verticesthen the value of the determinant,

2

33

22

11

1yx1yx1yx

is equal to :

(A) 192 (B) 243 (C) 486 (D*) 972

[Hint: 21

1yx1yx1yx

33

22

11 = A D = 4A2 D = 4

2

36·43

= 972 Ans. ]

Q.25 A graph is defined in polar co-ordinates as r() = cos + 21

. The smallest x-coordinates of any point on

the graph is

(A*) – 161

(B) – 81

(C) – 41

(D) – 21

[Sol. x-coordinate = r cos

= (cos + 21

) cos

= cos2 + 2cos

= (cos + 41

)2 – 161

Hence smallest is – 161

Ans. ]

Paragraph for Question Nos. 26 to 28Consider a general equation of degree 2, as

x2 – 10xy + 12y2 + 5x – 16y – 3 = 0Q.26 The value of '' for which the line pair represents a pair of straight lines is

(A) 1 (B*) 2 (C) 3/2 (D) 3

Q.27 For the value of obtained in above question, if L1 = 0 and L2 = 0 are the lines denoted by the givenline pair then the product of the abscissa and ordinate of their point of intersection is(A) 18 (B) 28 (C*) 35 (D) 25

Q.28 If is the acute angle between L1 = 0 and L2 = 0 then lies in the interval(A) (45°, 60°) (B) (30°, 45°) (C) (15°, 30°) (D*) (0, 15°)

[Sol.

(i) a = ; h = – 5; b = 12; g = 25

; f = – 8, c = – 3

(12)(–3) + 2(–8)

25

(– 5) – (64) –

425

· 12 + 3 · 25 = 0

– 36 + 200 – 64 – 75 + 75 = 0 100 = 200 = 2 Ans.


(ii) 2x2 – 10xy + 12y2 + 5x – 16y – 3 = 0consider the homogeneous part2x2 – 10xy + 12y2

2x2 – 6xy – 4xy + 12y2 or 2x(x – 3y) – 4y(x – 3y) or (x – 3y)(x – 2y) 2x2 – 10xy + 12y2 + 5x – 16y – 3 (2x – 6y + A)(x – 2y + B)solving A = – 1; B = 3hence lines are 2x – 6y – 1 = 0 and x – 2y + 3 = 0

solving intersection point

27,10

product = 35 Ans.

(iii) tan = ba

abh2 2

=

1424252

= 71

(0, 15°) Ans. ]

Paragraph for Question Nos. 29 to 31Consider a family of lines (4a + 3)x – (a + 1)y – (2a + 1) = 0 where a R

Q.29 The locus of the foot of the perpendicular from the origin on each member of this family, is(A) (2x – 1)2 + 4(y + 1)2 = 5 (B) (2x – 1)2 + (y + 1)2 = 5(C) (2x + 1)2 + 4(y – 1)2 = 5 (D*) (2x – 1)2 + 4(y – 1)2 = 5

Q.30 A member of this family with positive gradient making an angle of 4 with the line 3x – 4y = 2, is(A*) 7x – y – 5 = 0 (B) 4x – 3y + 2 = 0 (C) x + 7y = 15 (D) 5x – 3y – 4 = 0

Q.31 Minimum area of the triangle which a member of this family with negative gradient can make with thepositive semi axes, is(A) 8 (B) 6 (C*) 4 (D) 2

[Sol. Given (4a + 3)x – (a + 1)y – (2a + 1) = 0(3x – y – 1) + a(4x – y – 2) = 0family of lines passes through the fixed point P which is the intersection of

3x – y = 1 and 4x – y = 2 solving P(1, 2), now

(i) 1h2k·

hk

= – 1

locus is x(x – 1) + y(y – 2) = 0x2 + y2 – 2y – x = 0

45)1y(

21x 2

2

(2x – 1)2 + 4(y – 1)2 = 5 Ans.(ii) We have y – 2 = m(x – 1) ....(1)

this makes an angle of /4 with 3x – 4y = 2 with slope 3/4

4m31)43(m

= 1; m343m4

= ± 1 4m – 3 = 4 + 4m (with + ve sign) m = 7

with – ve sign 4m – 3 = – 4 – 3m 7m = – 1 m = – 71

(rejected)

hence the line isy – 2 = 7(x – 1)7x – y – 5 = 0 Ans.


(iii) Again y – 2 = m(x – 1)

x = 0; y = 2 – m; y = 0, x = 1 – m2

2A = (2 – m)

m21 (m < 0)

2A = 2 – m – m4

+ 2 = 4 +

m4m

let – m = M (M > 0)

2A = 4 + M + M4

= 4M2M4

2

=

2

M2M8

area is minimum if M = 2 m = – 22A|min = 8 A|min = 4 Ans. ]

[MULTIPLE OBJECTIVE TYPE]

Q.32 If xc

yd

= 1 is a line through the intersection of xa

yb

= 1 and xb

ya

= 1 and the lengths of the

perpendiculars drawn from the origin to these lines are equal in lengths then :

(A*) 1 12 2a b = 1 1

2 2c d (B) 1 1

2 2a b = 1 1

2 2c d

(C*) 1 1a b =

1 1c d (D) none

[Hint : Use the condition of concurrency for three lines ]

Q.33 A and B are two fixed points whose co-ordinates are (3, 2) and (5, 4) respectively . The co-ordinates ofa point P if ABP is an equilateral triangle, 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.34 Straight lines 2x + y = 5 and x 2y = 3 intersect at the point A . Points B and C are chosen on thesetwo lines such that AB = AC . Then the equation of a 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 of the lines through (2, 3) and parallel to thebisectors of the given lines, the slopes of the bisectors being 1/3 and 3 ]

Q.35 The straight lines x + y = 0, 3x + y 4 = 0 and x + 3y 4 = 0 form a triangle which is(A*) isosceles (B) right angled (C*) obtuse angled (D) equilateral


CIRCLE[STRAIGHT OBJECTIVE TYPE]

Q.1 The line 2x – y + 1 = 0 is tangent to the circle at the point (2, 5) and the centre of the circles lies onx – 2y = 4. The radius of the circle is(A*) 53 (B) 35 (C) 52 (D) 25

[Sol. 2x – y + 1 = 0 is tangent

slope of line OA = – 21

equation of OA, (y – 5) = – 21

(x – 2)

2y – 10 = – x + 2x + 2y = 12

intersection with x – 2y = 4 will give coordinates of centresolving we get (8, 2)

distance OA = 2)52()28( = 936 = 45 = 53 ]

Q.2 If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect in two distinct pointsP and Q then the line 5x + by – a = 0 passes through P and Q for(A) exactly one value of a (B*) no value of a(C) infinitely many values of a (D) exactly two values of a

[Sol. S1 – S2 = 5ax + (c – d)y + a + 1 = 0 and 5x + by – a = 0 must represent the same line

a1a

bdc

1a

ab = c – d and + a2 + a + 1 = 0a is imaginary onlyso no value of a (B) ]

Q.3 Four unit circles pass through the origin and have their centres on the coordinate axes. The area of thequadrilateral whose vertices are the points of intersection (in pairs) of the circles, is(A) 1 sq. unit (B) 22 sq. units(C*) 4 sq. units (D) can not be uniquely determined, insufficient data

[Sol. Ar. (square of sides 2) = 4 sq. units

]


Q.4 To which of the following circles, the line y x + 3 = 0 is normal at the point

23,

233 ?

(A) 92

3y2

33x22

(B) 9

23y

23x

22

(C) x2 + (y 3)2 = 9 (D*) (x 3)2 + y2 = 9[Hint: Line must pass through the centre of the circle ]

Q.5 Three concentric circles of which the biggest is x2 + y2 = 1, have their radii in A.P. If the line y = x + 1cuts all the circles in real and distinct points. The interval in which the common difference of the A.P. willlie is

(A)

41,0 (B)

221,0 (C*)

4

22,0 (D) none

[Sol. Equation of circle arex2 + y2 = 1; x2 + y2 = (1 – d)2 ; x2 + y2 = (1 – 2d)2

solve any of circle with line y = x + 1e.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

22]

Q.6 The circle with equation x2 + y2 = 1 intersects the line y = 7x + 5 at two distinct points A and B. Let Cbe the point at which the positive x-axis intersects the circle. The angle ACB is

(A) tan–1

34

(B) tan–1

43

(C*) tan–1(1) (D) tan–1

23

[Sol. Solving y = 7x + 5 and the circle x2 + y2 = 1 [13th, 07-10-2008]

A

54,

53 and B

53,

54

mOA = 34

; mOB = 43

Hence AOD = 90° ACB = 4

= tan–1(1) Ans. ]

Q.7 The number of common tangents of the circles (x + 2)² + (y 2)² = 49 and(x 2)² + (y + 1)² = 4 is(A) 0 (B*) 1 (C) 2 (D) 3

Q.8 In the xy-plane, the length of the shortest path from (0, 0) to (12, 16) that does not go inside the circle (x– 6)2 + (y – 8)2 = 25 is

(A) 310 (B) 510 (C*) 310 + 3

5(D) 10 + 5

[Sol. Let O = (0, 0), P = (6, 8) and Q = (12, 16).As shown in the figure the shortest route consists of tangentOT, minor arc TR and tangent RQ.Since OP = 10, PT = 5, and OTP = 90°,


it follows that OPT = 60° and OT = 35 .

By similar reasoning, QPR = 60° and QR = 35 .Because O, P and Q are collinear (why?),

RPT = 60°, so arc TR is of length 35

.

Hence the length of the shortest route is 2( 35 ) + 35

Ans. ]

Q.9 Triangle ABC is right angled at A. The circle with centre A and radius AB cuts BC and AC internally atD and E respectively. If BD = 20 and DC = 16 then the length AC equals(A) 216 (B*) 266 (C) 30 (D) 32

[Sol. b2 + r2 = (36)2 ....(1)Also CD · CB = CE · CX (using power of the point C)

16 · 36 = (b – r)(b + r) b2 – r2 = 16 · 36from (1) and (2)

2b2 = 36(36 + 16) = 36 · 52b2 = 36 · 26 b = 266 Ans. ]

Q.10 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 of line be y = x + cy – x = c ....(1)

perpendicular from (0, 0) on (1) is 2c

= 2c

In AON,2

2

2c2

= AN

and in CPM, 2c232 = CM

perpendicular from (5, 7) on line y – x = c = 2c2

Given AN = CM = 2c4

2 = 9 –

2)c2( 2

c = – 23

equation of line y = x – 23

of 2x – 2y – 3 = 0 ]

Q.11 If x = 3 is the chord of contact of the circle x2 y2 = 81, then the equation of the corresponding pair oftangents, is(A) x2 8y2 + 54x + 729 = 0 (B*) x2 8y2 54x + 729 = 0(C) x2 8y2 54x 729 = 0 (D) x2 8y2 = 729


[Sol. Equation of tangent to the circle at A 26,3 [11th, 23-12-2007]

3x + y26 = 81 ....(1)

and at B 26,3

3x – y26 = 81 ....(2)hence equation of the line pair

(x + y22 – 27)(x – y22 – 27) = 0

x2 – 8y2 – 27(x + y22 ) – 27(x – y22 ) + 729 = 0x2 – 8y2 – 54x + 729 = 0 Ans.]

Q.12 Let C1 and C2 are circles defined by x2 + y2 – 20x + 64 = 0 and x2 + y2 + 30x + 144 = 0.The length of the shortest line segment PQ that is tangent to C1 at P and to C2 at Q is(A) 15 (B) 18 (C*) 20 (D) 24

[Sol. Centres are (10, 0) and (– 15, 0)r1 = 6; r2 = 9d = 25r1 + r2 < d circles are separated

PQ = l = 221

2 )rr(d

= 225625 = 20 Ans. ]Q.13 If the straight line y = mx is outside the circle x2 + y2 – 20y + 90 = 0, then

(A) m > 3 (B) m < 3 (C) | m | > 3 (D*) | m | < 3[Sol. Centre (0,10), radius 10 .

Distance of (0,10) from y = mx is greater than 10 i.e. 101m

102

< 3 | m | < 3]

Q.14 The centre of the smallest circle touching the circles x2 + y2 – 2y 3 = 0 andx2 + y2 8x 18y + 93 = 0 is(A) (3 , 2) (B) (4 , 4) (C) (2 , 7) (D*) (2 , 5)

Q.15 Suppose that two circles C1 and C2 in a plane have no points in common. Then(A) there is no line tangent to both C1 and C2.(B) there are exactly four lines tangent to both C1 and C2.(C) there are no lines tangent to both C1 and C2 or there are exactly two lines tangent to both C1 and C2.(D*) there are no lines tangent to both C1 and C2 or there are exactly four lines tangent to both C1 and C2.

[Hint: or ]

Q.16 A variable line moves in such way that the product of the perpendiculars from (a, 0) and (0, 0) is equalto k2. The locus of the feet of the perpendicular from (0, 0) upon the variable line is a circle, the squareof whose radius is (Given: | a | < 2 | k |)

(A*) 22

k4a

(B) 4

ka 22 (C) a2 +

4k2

(D) 2

ka 22


Sol. Let the equation of the variable line is [13th, 10-08-2008, P-1]xx1 + yy1 – (x1

2 + y12) = 0

p1p2 = 21

21

21

21

yx

yx

2

121

21

211

yx

)yx(ax

= k2

i.e. |x12 + y1

2 – ax1| = k2

locus is x2 + y2 –ax ± k2 = 0

r2 = 4

a2

– (± k2)(0,0) (a,0)

p1 p2

(x , y )1 1 Variable line xm= – y11

+ve sign r2 = 22

k4a

(not possible as r2 becomes – ve)

–ve sign r2 = 4

a2

+ k2 Ans. ]

Q.17 The shortest distance from the line 3x + 4y = 25 to the circle x2 + y2 = 6x – 8y is equal to(A*) 7/5 (B) 9/5 (C) 11/5 (D) 32/5

[Sol. Centre: (3 , – 4) and r = 5perpendicular distance from (3, – 4) on

3x + 4y – 25 = 0 is

p = 525169

= 532

d = 532

– 5 = 57

Ans. ]

Q.18 If the circle C1 : x2 + y2 = 16 intersects another circle C2 of radius 5 in such a manner that thecommon chord is of maximum length and has a slope equal to 3/4, then the co-ordinates of the centre ofC2 are

(A)

95

125

, (B*)

95

125

, (C)

125

95

, (D)

125

95

,

[Hint : Equation of common chord is y = 34

x 3x – 4y = 0

family of circle x2 + y2 – 4 + (3x – 4y) = 0equate radius of this circle as 5. Alternately ...............]

Q.19 The points (x1, y1) , (x2, y2) , (x1, y2) and (x2, y1) are always :(A) collinear (B*) concyclic(C) vertices of a square (D) vertices of a rhombus

[Hint: All the points lie on the circle (x x1) (x x2) + (y y1) (y y2) = 0Note that points form a rectangle]

Q.20 Let C be the circle of radius unity centred at the origin. If two positive numbers x1 and x2 are such thatthe line passing through (x1, – 1) and (x2, 1) is tangent to C then(A*) x1x2 = 1 (B) x1x2 = – 1 (C) x1 + x2 = 1 (D) 4x1x2 = 1


[Sol. Let the equation tangent in parametric form isx cos + y sin = 1

put (x1, – 1)x1 cos – sin = 1x1 cos = 1 + sin ....(1)

now put (x2, 1)x2 cos = 1 – sin ....(2)

(1) × (2) x1x2 cos2 = 1 – sin2 = cos2x1 x2 = 1 Ans.

Q.21 Tangents are drawn from any point on the circle x2 + y2 = R2 to the circle x2 + y2 = r2. If the line joiningthe points of intersection of these tangents with the first circle also touch the second, then R equals

(A) 2 r (B*) 2r (C) 2

2 3r

(D)

43 5

r

[HInt: ]

Q.22 The locus of the middle points of the system of chords of the circle x² + y² = 16 which are parallel to theline 2y = 4x + 5 is(A) x = 2y (B*) x + 2y = 0 (C) y + 2x = 0 (4) y = 2x

Q.23 The distance between the chords of contact of tangents to the circle x2+ y2 + 2gx + 2fy + c = 0 from theorigin and the point (g, f) is

(A) g f2 2 (B) g f c2 2

2 (C*) g f c

g f

2 2

2 22

(D)

g f c

g f

2 2

2 22

[Sol. Equation of chord or contact of the pair of tangent from (0, 0) and (g, f) are

gx + fy + c = 0 ....(1) and gx + fy + 2

cfg 22 = 0 ....(2)

These lines are parallel hence distance = 22

22

fg2

cfgc

]

Q.24 The locus of the center of the circles such that the point (2 , 3) is the mid point of the chord5x + 2y = 16 is(A*) 2x 5y + 11 = 0 (B) 2x + 5y 11 = 0(C) 2x + 5y + 11 = 0 (D) none

[ Hint : Slope of the given line = 5/2

52

32

. fg

= 1 15 + 5f = 4 + 2g

locus is 2x 5y + 11 = 0 ]


Q.25 The points A (a , 0) , B (0 , b) , C (c , 0) and D (0 , d) are such that ac = bd and a, b, c, d are allnon-zero. Then the points(A) form a parallelogram (B) do not lie on a circle(C) form a trapezium (D*) are concyclic

[Sol. If OA · OC = OB · OD (Power of point)then points are concylic a · c = bd (true) ]

Q.26 The locus of the centers of the circles which cut the circles x2 + y2 + 4x 6y + 9 = 0 andx2 + 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: Let out circle be x2 + y2 + 2gx + 2fy + c = 0conditions 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.27 If the angle between the tangents drawn from P to the circle x2 + y2 + 4x – 6y + 9 sin2 + 13 cos2 =0is 2, then the locus of P is(A) x2 + y2 + 4x – 6y + 14 = 0 (B) x2 + y2 + 4x – 6y – 9 = 0(C) x2 + y2 + 4x – 6y – 4 = 0 (D*) x2 + y2 + 4x – 6y + 9 = 0

[Sol. C(–2, 3); R2 = 4 + 9 – 9 sin2 – 13 cos2 = 4 sin2R = 2 sin

2 sin

(–2,3)C

P(x,y)

yNow use sin = CP

sin2 CP = 2

Hence locus is (x + 2)2 + (y – 3)2 = 4 ]Q.28 The locus of the mid points of the chords of the circle x² + y² + 4x 6y 12 = 0 which subtend an angle

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.29 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 = 5·10·25510

= 21

= 4

]


Q.30 A circle of radius unity is centred at origin. Two particles start moving at the same time from the point(1, 0) and move around the circle in opposite direction. One of the particle moves counterclockwisewith constant speed v and the other moves clockwise with constant speed 3v. After leaving (1, 0), thetwo particles meet first at a point P, and continue until they meet next at point Q. The coordinates of thepoint Q are(A) (1, 0) (B) (0, 1) (C) (0, –1) (D*) (–1, 0)

[Sol. The particle which moves clockwise is moving three timesas fast as the particle moving anticlockwiseThis tells us that in the time that the clockwise particle travels (3/4)th ofthe way around the circle the anticlockwise particle will travel (1/4)th ofthe way around the circle and so the 2nd particle will meet at p (0, 1).Using the same logic they will meet at Q (–1, 0) when they meet the 2nd time]

Q.31 The value of 'c' for which the set, { (x, y)x2 + y2 + 2x 1} {(x, y)x y + c 0} contains onlyone point in common is :(A) (, 1] [3, ) (B) {1, 3}(C) {3} (D*) { 1 }

[Hint: x2 + y2 + 2x – 1 = 0; centre (–1, 0) and rad. = 2line x – y + c = 0

12

2c ; |c – 1| = 2; c – 1 = ± 2

c = 3 or –1 ]Q.32 Locus of the middle points of a system of parallel chords with slope 2, of the circle

x2 + y2 – 4x – 2y – 4 = 0, has the equation(A*) x + 2y – 4 = 0 (B) x – 2y = 0 (C) 2x – y – 3 = 0 (D) 2x + y – 5 = 0

[Hint: Locus will be a line with slope – 1/2and passing through the centre (2, 1) of the circle

y – 1 = – 21

(x – 2)

2y – 2 = – x + 2 x + 2y – 4 = 0 Ans. ]Q.33 Two circles are drawn through the points (1, 0) and (2, 1) to touch the axis of y. They intersect at an

angle

(A*) cot–1 34

(B) cos 1 45

(C) 2

(D) tan1 1

[Hint : The two circles arex2 + y2 2x + 2y + 1 = 0 and x2 + y2 10x 6y + 9 = 0 (use family of circles) ]

Q.34 A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the equation2x – y – 2 = 0. Then the equation of the circle is(A) x2 + y2 – 4y + 2 = 0 (B) x2 + y2 – 4y + 1 = 0(C*) x2 + y2 – 2x – 1 = 0 (D) x2 + y2 – 2x + 1 = 0

[Sol. Diameter of the circle is given as 2x – y – 2 = 0 ....(1)

slope of PN: 2413

= 1


equation of normal through PN isy – 1 = (x – 2)x – y – 1 = 0 ....(2)

solving (1) and (2) centre is (1, 0)hence equation of the circle is

(x – 1)2 + y2 = (2 – 1)2 + 1x2 + y2 – 2x – 1 = 0 Ans. ]

Q.35 A rhombus is inscribed in the region common to the two circles x2 + y2 4x 12 = 0 andx2 + y2 + 4x 12 = 0 with two of its vertices on the line joining the centres of the circles. The area of therhombous 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 equal to 4 and 4 3

A = d d1 2

2 = 38 ]

Q.36 A circle is drawn touching the xaxis and centre at the point which is the reflection of(a, b) in the line y x = 0. The equation of the circle is(A) x2 + y2 2bx 2ay + a2 = 0 (B*) x2 + y2 2bx 2ay + b2 = 0(C) x2 + y2 2ax 2by + b2 = 0 (D) x2 + y2 2ax 2by + a2 = 0

[Sol. 11ahbk

k – b = h – a or a – b = h – k

also 2ha

and 2kb

lies on y-axis 2ha

= 2kb

a – b = k – h

and touches the x-axis cg2 2 = 0 centre (b, a)

g2 = c = b2

equation x2 + y2 – 2hx – 2ay + b2 = 0 ]Q.37 A(1, 0) and B(0, 1) and two fixed points on the circle x2 + y2 = 1. C is a variable point on this circle. As

C moves, the locus of the orthocentre of the triangle ABC is(A*) x2 + y2 – 2x – 2y + 1 = 0 (B) x2 + y2 – x – y = 0(C) x2 + y2 = 4 (D) x2 + y2 + 2x – 2y + 1 = 0

[Sol. Let C (cos , sin ); H(h, k) is the orthocentre of the ABC

h = 1 + cos k = 1 + sin O A(1, 0)

B(0, 1)

C(cos

, sin

)

(x – 1)2 + (y – 1)2 = 1x2 + y2 – 2x – 2y + 1 = 0 ]


Q.38 A line meets the co-ordinate axes in A and B. A circle is circumscribed about the triangle OAB. If d1 andd2 are the distances of the tangent to the circle at the origin O from the points A and B respectively, thediameter of the circle is :

(A) 2

dd2 21 (B) 2

d2d 21 (C*) d1 + d2 (D) 21

21dd

dd

[Sol. Let the circle be x2 + y2 + 2gx + 2fy = 0Tangent at the origin is gx + fy = 0

d1 = 22

2

fg

g2

d2 = 22

2

fg

f2

diameterfg2dd 2221

]

Q.39 In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chordjoining A with the point of intersection D of the hypotenuse and the semicircle, then the length AC equals to

(A) AB AD

AB AD

2 2 (B) AB ADAB AD

(C) AB AD (D*) AB AD

AB AD

2 2

[Sol. l · x = y 22 xl where l = AC; x = AB, y = ADl2 x2 = y2(l2 + x2)l2(x2 – y2) = x2y2

l = 22 yx

xy

=

22 ADAB

AD·AB

Ans. ]

Q.40 A straight line l1 with equation x – 2y + 10 = 0 meets the circle with equation x2 + y2 = 100 at B in thefirst quadrant. A line 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 = 21

slope of l2 = – 2equation of l2

y = – 2(x – 10) y + 2x = 20Hence t = 20 Ans.]

Q.41 B and C are fixed points having coordinates (3, 0) and ( 3, 0) respectively . If the vertical angle BACis 90º, then the locus of the centroid of the 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) and G (h. k) Now A, G, O are collinear

h = 2 0

3. a

a = 3 h and similarly b = 3 k.

Now (a, b) lies on the circle x2 + y2 = 9 A ]


Paragraph for Question Nos. 42 to 44Let C be a circle of radius r with centre at O. Let P be a point outside C and D be a point on C. A linethrough P intersects C at Q and R, S is the midpoint of QR.

Q.42 For different choices of line through P, the curve on which S lies, is(A) a straight line (B) an arc of circle with P as centre(C) an arc of circle with PS as diameter (D*) an arc of circle with OP as diameter

Q.43 Let P is situated at a distance 'd' from centre O, then which of the following does not equal the product(PQ) (PR)?(A) d2 – r2 (B) PT2, where T is a point on C and PT is tangent to C(C) (PS)2 – (QS)(RS) (D*) (PS)2

Q.44 Let XYZ be an equilateral triangle inscribed in C. If , , denote the distances of D from vertices X,Y, Z respectively, the value of product ( + – ) ( + – ) ( + – ), is

(A*) 0 (B) 8

(C) 6

3333 (D) None of these

[Sol.(i) Locus of S is a part of circle with OP as diameter passing inside the circle 'C'

MP

RQ

N

S(h,k)

O

C

(ii) (PR) (PQ) = PT2 = (PN)(PM) = (d – r) (d + r) = d2 – r2

= (PS – SR) (PS + SQ) = PS2 – SQ2 ( SQ = SR)= PS2 – (SQ)(SR)

(PQ) (PR) (PS)2

(iii) Using Ptolemy's theorem,(YD) (XZ) = (XY)(ZD) + (YZ) (XD)

= XZ (ZD + XD) { (XY = YZ = ZX) } = + (A)

X

D

ZY

60°60°

30° 30°

Alternatively:

21

= )YD(2)XY()YD( 222

from XYD 21

= )YD(2)YZ()YD( 222

from YZD

(YD) = 2 + (YD)2 – (XY)2 ....(1)(YD) = 2 + (YD)2 – (YZ)2 ....(2)(1)-(2) ( – ) = 2 – 2

= + ]


Paragraph for question nos. 45 to 47Consider 3 circles

S1 : x2 + y2 + 2x – 3 = 0S2 : x2 + y2 – 1 = 0S3 : x2 + y2 + 2y – 3 = 0

Q.45 The radius of the circle which bisect the circumferences of the circles S1 = 0 ; S2 = 0 ; S3 = 0 is(A) 2 (B) 22 (C*) 3 (D) 10

Q.46 If the circle S = 0 is orthogonal to S1 = 0 ; S2 = 0 and S3 = 0 and has its centre at (a, b) and radius equalsto 'r' then the value of (a + b + r) equals(A) 0 (B) 1 (C) 2 (D*) 3

Q.47 The radius of the circle touching S1 = 0 and S2 = 0 at (1, 0) and passing through (3, 2) is(A) 1 (B) 12 (C*) 2 (D) 22

[Sol. • (1,0)O(–1,0)

S1

S2

(i)

(–1,0)

2r

(a,b) (0,–1)2

(0,0) 1

r

S1 S = 03

S3

r2 = a2 + b2 + 1 = (a + 1)2 + b2 + 4 and (a + 1)2 + b2 + 4 = a2 + (b + 1)2 + 4 2a + 4 = 0 2a = 2b

a = – 2 b = – 2r2 = 9 r = 3 Ans.

(ii) S1 – S2 = 0 x = 1S2 – S3 y = 1 Radical centre = (1, 1)

radius LT = 1S = 1 equation of circle is (x – 1)2 + (y – 1)2 = 1 radius = 1 and a = 1 ; b = 1 a + b + r = 3 Ans.


(iii)

(3,2)

(1,0)

x = 1

family of circles touches the line x – 1 = 0 at (1, 0) is (x – 1)2 + (y – 0)2 + (x – 1) = 0passing through (3, 2) 4 + 4 + 2 = 0 = – 4 x2 + y2 – 6x + 5 = 0

radius 59 = 2 Ans. ]Paragraph for Question Nos. 48 to 50

Consider a line pair ax2 + 3xy – 2y2 – 5x + 5y + c = 0 representing perpendicular lines intersecting eachother at C and forming a triangle ABC with the x-axis.

Q.48 If x1 and x2 are intercepts on the x-axis and y1 and y2 are the intercepts on the y-axis then the sum(x1 + x2 + y1 + y2) is equal to(A) 6 (B*) 5 (C) 4 (D) 3

Q.49 Distance between the orthocentre and circumcentre of the triangle ABC is(A) 2 (B) 3 (C*) 7/4 (D) 9/4

Q.50 If the circle x2 + y2 – 4y + k = 0 is orthogonal with the circumcircle of the triangle ABC then 'k' equals(A) 1/2 (B) 1 (C) 2 (D*) 3/2

[Sol.(i) As lines are perpendicular

a – 2 = 0 a = 2 (coefficient of x2 + coefficient of y2 = 0)using = 0 c = – 3 (D abc + 2fgh – af2 – bg2 – ch2)hence the two lines are

x + 2y – 3 = 0 and 2x – y + 1 = 0

1y;23y

21x;3x

21

21

interceptsy

interceptsx x1 + x2 + y1 + y2 = 5 Ans.

(ii) (CM)2 = 259

51

45 2

= 25

920

425 2

= 259

400441

= 400784441

= 4001225

= 1649

CM = 47

Ans.

(iii) Circumcircle of ABC

0y)3x(21x 2

(2x + 1)(x – 3) + 2y2 = 0

2(x2 + y2) – 5x – 3 = 0 x2 + y2 – 25

x – 23

= 0 ....(1)

given x2 + y2 – 4y + k = 0 which is orthogonal to (1) using the condition of orthogonality

we get, 0 + 0 = k – 23

k = 23

Ans. ]


Paragraph for question nos. 51 to 53Consider a circle x2 + y2 = 4 and a point P(4, 2). denotes the angle enclosed by the tangents from P onthe circle and A, B are the points of contact of the tangents from P on the circle.

Q.51 The value of lies in the interval(A) (0, 15°) (B) (15°, 30°) (C) 30°, 45°) (D*) (45°, 60°)

Q.52 The intercept made by a tangent on the x-axis is(A) 9/4 (B*) 10/4 (C) 11/4 (D) 12/4

Q.53 Locus of the middle points of the portion of the tangent to the circle terminated by the coordinate axes is(A*) x–2 + y–2 = 1–2 (B) x–2 + y–2 = 2–2 (C) x–2 + y–2 = 3–2 (D) x–2 – y–2 = 4–2

[Sol. Tangenty – 2 = m(x – 4)mx – y + (2 – 4m) = 0

p = 2m1

m42

= 2

(1 – 2m)2 = 1 + m2

3m2 – 4m = 0

m = 0 or m = 34

tan = 34

(45°, 60°)Hence equation of tangent is y = 2 and (with infinite intercept on x-axis)

or y – 2 = 34

(x – 4) 3y – 6 = 4x – 16 4x – 3y – 10 = 0

x-intercept = 410

Ans.(ii) (B)

Variable line with mid point (h, k)

1k2y

h2x

, it touches the circle x2 + y2 = 4

22 k41

h41

1

= 2 41

k41

h41

22 locus is x–2 + y–2 = 1 Ans.(iii) (A)]

[MULTIPLE OBJECTIVE TYPE]

Q.54 Let L1 be a line passing through the origin and L2 be the line x + y = 1. If the intercepts made by the circlex2 + y2 – x + 3y = 0 on L1 and L2 are equal then the equation of L1 can be(A) x + y = 0 (B*) x – y = 0 (C*) x + 7y = 0 (D) x – 7y = 0

[Sol. The chords are of equal length, then the distances of the centre from the lines are equal.

Let L1 be y – mx = 0. Centre is

23,

21

2

123

21

1m2m

23

2

7m2 – 6m – 1 = 0 m = 1, 7

1 ]


Q.55 Consider the circles S1 : x2 + y2 = 4 and S2 : x2 + y2 – 2x – 4y + 4 = 0 which of the following statementsare correct?(A*) Number of common tangents to these circles is 2.(B*) If the power of a variable point P w.r.t. these two circles is same then P moves on the

line x + 2y – 4 = 0.(C) Sum of the y-intercepts of both the circles is 6.(D*) The circles S1 and S2 are orthogonal.

[Sol. S1 : x2 + y2 = 4 and S2 = x2 + y2 – 2x – 4y + 4 = 0centre: (0, 0); radius = 2 centre : (1, 2); radius = 1(A) d = distance between centres = 5

r1 + r2 = 3 | r1 – r2 | = 1 | r1 – r2 | < d < r1 + r2 these 2 circles are intersecting. number of common tangents is 2. (A) is correct

(B) P(h, k) power of point P is same w.r.t. these two circles.

4kh 22 = 4k4h2kh 22 – 4 = – 2h – 4k + 4

2h + 4k – 8 = 0x + 2y – 4 = 0 (B) is correct

(C) y intercept of S1 is 2 4 = 4

y intercept of S2 is 2 44 = 0 sum of y-intercept = 4 (C) is incorrect

(D) 2(0 + 0) = – 4 + 4 circle is orthogonal. (D) is correct]

Q.56 If al2 bm2 + 2 dl + 1 = 0, where a, b, d are fixed real numbers such that a + b = d2 then the linelx + my + 1 = 0 touches a fixed circle :(A*) which cuts the xaxis orthogonally(B) with radius equal to b(C*) on which the length of the tangent from the origin is d b2 (D) none of these .

[Hint : (d2 b) l2 + 2 dl + 1 = bm2 d2l2 + 2 dl + 1 = b (l2 + m2)

d

m

12 2

= b2 centre (d, 0) and radius b (x d)2 + y2 = b

2 ]

Q.57 Three distinct lines are drawn in a plane. Suppose there exist exactly n circles in the plane tangent to allthe three lines, then the possible values of n is/are(A*) 0 (B) 1 (C*) 2 (D*) 4

[Sol. Case-1: If lines form a triangle then n = 4 i.e. 3 excircles and 1 incircle

Case-2: If lines are concurrent or all 3 parallel then n = 0

Case-3: If two are parallel and third cuts then n = 2

hence (A), (C), (D) ]


Q.58 Consider the circlesS1 : x2 + y2 + 2x + 4y + 1 = 0S2 : x2 + y2 – 4x + 3 = 0S3 : x2 + y2 + 6y + 5 = 0

Which of this following statements are correct?(A) Radical centre of S1, S2 and S3 lies in 1st quadrant.(B*) Radical centre of S1, S2 and S3 lies in 4th quadrants.(C*) Radius of the circle intersecting S1, S2 and S3 orthogonally is 1.(D*) Circle orthogonal to S1, S2 and S3 has its x and y intercept equal to zero.

[Hint: radical centre is (1, –1)radius of the circle orthogonally to S1 , S2 , S3 is 1 and has the equation

x2 + y2 – 2x + 2y + 1 = 0 ]

Q.59 Locus of the intersection of the two straight lines passing through (1, 0) and (–1, 0) respectively andincluding an angle of 45° can be a circle with(A) centre (1, 0) and radius 2 . (B) centre (1, 0) and radius 2.

(C*) centre (0, 1) and radius 2 . (D*) centre (0, – 1) and radius 2 .

[Sol. m1 = 1hk

; m2 = 1hk

; tan 45° =

1hk1

1hk

1hk

2

2

= 22 k1h)1h1h(k

h2 + k2 – 1 = ± 2k locus is x2 + y2 ± 2y – 1 = 0

x2 + (y ± 1)2 = 22 (0, 1) or (0, – 1) and radius = 2 (C) and (D) Ans. ]

[MATCH THE COLUMN]Q.60 Column-I Column-II

(A) If the straight line y = kx K I touches or passes outside (P) 1the circle x2 + y2 – 20y + 90 = 0 then | k | can have the value

(B) Two circles x2 + y2 + px + py – 7 = 0 (Q) 2and x2 + y2 – 10x + 2py + 1 = 0 intersect each other orthogonallythen the value of p is

(C) If the equation x2 + y2 + 2x + 4 = 0 and x2 + y2 – 4y + 8 = 0 (R) 3represent real circles then the value of can be

(D) Each side of a square is of length 4. The centre of the square is (3, 7). (S) 5One diagonal of the square is parallel to y = x. The possible abscissaeof the vertices of the square can be

[Ans. (A) P, Q, R; (B) Q, R; (C) Q, R, S; (D) P, S][Sol. (A) x2 + k2x2 – 20kx + 90 = 0

x2(1 + k2) – 20kx + 90 = 0


D 0400k2 – 4 × 90(1 + k2) 010k2 – 9 – 9k2 0k2 – 9 0 k [–3, 3]

(B) 2

p

2p5

2p

= – 6 – 5p + p2 + 6 = 0 p2 – 5p + 6 = 0 p = 2 or 3 Ans.

(C) r12 = 2 – 4 0

(– , – 2] [2, ) ....(1)r2

2 = 42 – 8 02 – 2 0 (– , – 2 ] [ 2 , ) ....(2)(1) (2) is (– , – 2] [2, ) Ans.

(D) Ans. {1, 5}]

Q.61 Column-I Column-II(A) Two intersecting circles (P) have a common tangent(B) Two circles touching each other (Q) have a common normal(C) Two non concentric circles, one strictly inside (R) do not have a common normal

the other(D) Two concentric circles of different radii (S) do not have a radical axis.

[Ans. (A) P, Q; (B) P, Q; (C) Q; (D) Q, S]

Q.62 Column-I Column-II(A) Let 'P' be a point inside the triangle ABC and is equidistant (P) centroid

from its sides. DEF is a triangle obtained by the intersectionof the external angle bisectors of the angles of the ABC.With respect to the triangle DEF point P is its

(B) Let 'Q' be a point inside the triangle ABC (Q) orthocentre

If (AQ)sin 2A

= (BQ)sin 2B

= (CQ)sin 2C

then with respect to

the triangle ABC, Q is its(C) Let 'S' be a point in the plane of the triangle ABC. If the point is (R) incentre

such that infinite normals can be drawn from it on the circle passingthrough A, B and C then with respect to the triangle ABC, S is its

(D) Let ABC be a triangle. D is some point on the side BC such that (S) circumcentrethe line segments parallel to BC with their extremities on ABand AC get bisected by AD. Point E and F are similarly obtainedon CA and AB. If segments AD, BE and CF are concurrent ata point R then with respect to the triangle ABC, R is its

[Ans. (A) Q; (B) R; (C) S; (D) P]

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