Q.B on Straight line and circle [44] - Weebly · Q.B on Straight line and circle ... Q.152 416/cir Locus of the middle points of the portion of the tangent to the circle terminated

Q.B on Straight line and circle [44]

[MATCH THE COLUMN]

[MATCH THE COLUMN]

Q.174181

Column-I Column-II

(A) If the straight line y = kx ∀ K ∈ I touches or passes outside (P) 1

the circle x2 + y2 – 20y + 90 = 0 then | k | can have the value

(B) Two circles x2 + y2 + px + py – 7 = 0 (Q) 2

and x2 + y2 – 10x + 2py + 1 = 0 intersect each other orthogonally

then the value of p is

(C) If the equation x2 + y2 + 2λx + 4 = 0 and x2 + y2 – 4λy + 8 = 0 (R) 3

represent real circles then the value of λ can be

(D) Each side of a square is of length 4. The centre of the square is (3, 7). (S) 5

One diagonal of the square is parallel to y = x. The possible abscissae

of 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 [11th, 04-01-2009, P-2]

x2(1 + k2) – 20kx + 90 = 0

D ≤ 0

400k2 – 4 × 90(1 + k2) ≤ 0

10k2 – 9 – 9k2 ≤ 0

k2 – 9 ≤ 0 ⇒ k ∈ [–3, 3]

(B) 2

×+× p

2

p5

2

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

(C) r1

2 = λ2 – 4 ≥ 0

λ ∈ (– ∞, – 2] ∪ [2, ∞) ....(1)

r2

2 = 4λ2 – 8 ≥ 0

λ2 – 2 ≥ 0

λ ∈ (– ∞, – 2 ] ∪ [ 2 , ∞) ....(2)

(1) ∩ (2) is λ ∈ (– ∞, – 2] ∪ [2, ∞) Ans.

(D) Ans. {1, 5}]

Q.17576

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] [11th, 23-12-2007]


Q.160505/cir

x x− 1

cosθ =

y y− 1

sinθ = r , represents :

(A*) equation of a straight line , if θ is constant and r is variable

(B*) equation of a circle , if r is constant and θ is a variable

(C*) a straight line passing through a fixed point and having a known slope

(D*) a circle with a known centre and a given radius.

Q.162509/cir

A circle passes through the points (− 1, 1) , (0, 6) and (5, 5) . The point(s) on this circle, the

tangent(s) at which is/are parallel to the straight line joining the origin to its centre is/are :

(A) (1, − 5) (B*) (5, 1) (C) (− 5, − 1) (D*) (− 1, 5)

[ Hint : Note that ∆ is right angled at (0, 6) . Centre of the circle is (2, 3) . Slope of the line joining origin to the

centre is 3/2. Take parametric equation of a line through (2, 3) with

tan θ = − 2

3 as

x − 2

cosθ =

y − 3

sinθ = ± r where r = 13 .

Get the co−ordinates on the circle. Also see that the circle passes through the origin. ]

Q.163510/cir

Consider the circles S1 : x2 + y2 = 4 and S

2 : x2 + y2 – 2x – 4y + 4 = 0 which of the following

statements are 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 S

2 are orthogonal.

[Sol. S1 : x2 + y2 = 4 and S

2 = x2 + y2 – 2x – 4y + 4 = 0 [11th pqrs J, 21-1-2007]

centre: (0, 0); radius = 2 centre : (1, 2); radius = 1

(A) d = distance between centres = 5

r1 + r

2 = 3 ⇒ | r

1 – r

2 | = 1

∴ | r1 – r

2 | < d < r

1 + r

2

∴ 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 = 0

x + 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]


Paragraph for question nos. 150 to 152

Consider a circle x2 + y2 = 4 and a point P(4, 2). θ denotes the angle enclosed by the tangents from P on

the circle and A, B are the points of contact of the tangents from P on the circle.

Q.150414/cir

The value of θ lies in the interval

(A) (0, 15°) (B) (15°, 30°) (C) 30°, 45°) (D*) (45°, 60°)

Q.151415/cir

The intercept made by a tangent on the x-axis is

(A) 9/4 (B*) 10/4 (C) 11/4 (D) 12/4

Q.152416/cir

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. Tangent [11th, 12-10-2008, P-1]

y – 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 = 3

4 → tan θ =

3

4 ⇒ θ ∈ (45°, 60°)

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

or y – 2 = 3

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

x-intercept = 4

10 Ans.(ii) ⇒ (B)

Variable line with mid point (h, k)

1k2

y

h2

x=+ , it touches the circle x2 + y2 = 4

∴

22k4

1

h4

1

1

+

− = 2 ⇒

4

1

k4

1

h4

122

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

Paragraph for question nos. 153 to 155

Let A, B, C be three sets of real numbers (x, y) defined as

A : {(x, y): y ≥ 1}

B : {(x, y): x2 + y2 – 4x – 2y – 4 = 0}

C : {(x, y): x + y = 2 }

Q.153417/cir

Number of elements in the A ∩ B ∩ C is

(A) 0 (B*) 1 (C) 2 (D) infinite

Q.154418/cir

(x + 1)2 + (y – 1)2 + (x – 5)2 + (y – 1)2 has the value equal to

(A) 16 (B) 25 (C*) 36 (D) 49


family of circles touches the line x – 1 = 0 at (1, 0) is

(x – 1)2 + (y – 0)2 + λ(x – 1) = 0

passing through (3, 2) ⇒ 4 + 4 + 2λ = 0 ⇒ λ = – 4

∴ x2 + y2 – 6x + 5 = 0

∴ radius 59 − = 2 Ans. ]

Paragraph for questions nos. 141 to 143

Consider the two quadratic polynomials

Ca : y = 2aaax

4

x 22

−++− and C : y = 2 – 4

x2

Q.141408/cir

If the origin lies between the zeroes of the polynomial Ca then the number of integral value(s) of 'a'

is

(A) 1 (B*) 2 (C) 3 (D) more than 3

Q.142cir

If 'a' varies then the equation of the locus of the vertex of Ca , is

(A*) x – 2y – 4 = 0 (B) 2x – y – 4 = 0 (C) x – 2y + 4 = 0 (D) 2x + y – 4 = 0

Q.143cir

For a = 3, if the lines y = m1x + c

1 and y = m

2x + c

2 are common tangents to the graph of C

a and

C then the value of (m1 + m

2) is equal to

(A) – 6 (B*) – 3 (C) 1/2 (D) none

[Sol. y = f (x) = 2aaax4

x 22

−++− [13th, 08-03-2009, P-2]

(9) for zeroes to be on either side of origin

f (0) < 0

a2 + a – 2 < 0 ⇒ (a + 2)(a – 1) < 0 ⇒ – 2 < a < 1 ⇒ 2 integers i.e. {–1, 0} ⇒ (B)

(10) Vertex of Ca is (2a, a – 2)

hence h = 2a and k = a – 2

h = 2(k + 2)

locus x = 2y + 4 ⇒ x – 2y – 4 = 0 Ans.

(11) Let y = mx + c is a common tangent to y = 10x34

x2

+− ....(1) (for a = 3)

and y = 2 – 4

x2

....(2) where m = m1 or m

2 and c = c

1 or c

2

solving y = mx + c with (1)

mx + c = 10x34

x2

+−

or4

x2

– (m + 3)x + 10 – c = 0

D = 0 gives

(m + 3)2 = 10 – c ⇒ c = 10 – (m + 3)2 ....(3)

|||ly mx + c = 2 – 4

x2

⇒4

x2

+ mx + c – 2 = 0

D = 0 gives

m2 = c – 2 ⇒ c = 2 + m2 ....(4)


[COMPREHENSION TYPE] [3 × 3 = 9]

Paragraph for Question Nos. 129 to 131

Let 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 line

through P intersects C at Q and R, S is the midpoint of QR.

Q.129401/cir

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.130 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.131 Let XYZ be an equilateral triangle inscribed in C. If α, β, γ denote the distances of D from vertices X,

Y, Z respectively, the value of product (β + γ – α) (γ + α – β) (α + β – γ), is

(A*) 0 (B) 8

αβγ(C)

6

3333 αβγ−γ+β+α(D) None of these

[Sol.

(i) Locus of S is a part of circle with OP as diameter passing inside the circle 'C' [12th, 22-06-2008]

MP

R

Q

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) {Q (XY = YZ = ZX) }

⇒ β = γ + α ⇒ (A)

X

D

ZY

60°

60°

30° 30°

α

γβ

Alternatively:

2

1 =

)YD(2

)XY()YD( 222

α

−+αfrom ∆XYD

2

1 =

)YD(2

)YZ()YD( 222

γ

−+γfrom ∆YZD

α(YD) = α2 + (YD)2 – (XY)2 ....(1)

γ(YD) = γ2 + (YD)2 – (YZ)2 ....(2)

(1)-(2) ⇒ (α – γ)β = α2 – γ2

⇒ β = α + γ ]


Q.119309/cir

Let C be a circle with centre 'O' and HK is the chord of contact of pair of the tangents from point

A. OA intersects the circle C at P and Q and B is the midpoint of HK, then

Statement-1 : AB is the harmonic mean of AP and AQ.

because

Statement-2 : AK is the Geometric mean of AB and AO and OA is the arithmetic mean of AP and AQ.

(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

[Sol.)OA(

)AK( = cos θ =

AK

AB[12th, 07-12-2008, P-1]

⇒ (AK)2 = (AB) (OA) = (AP) (AQ) [AK2 = AP · AQ using power of point A]

Also OA = 2

AQAP +

[AQ – AO = r = AO – AP ⇒ 2AO = AQ + QP]

⇒ (AP) (AQ) = AB

+

2

AQAP θ

K

B

O

H

AP Q

⇒ AB = )AQAP(

)AQ()AP(2

+ ]

Q.121305/cir

Statement-1: Angle between the tangents drawn from the point P(13, 6) to the circle

S : x2 + y2 – 6x + 8y – 75 = 0 is 90°.

because

Statement-2: Point P lies on the director circle of S.





[Hint: Equation of director's circle is (x – 3)2 + (y + 4)2 = 200 and point (13, 6) satisfies the given circle

(x – 3)2 + (y + 4)2 = 100 ] [11th, 25-11-2007]

Q.123314/cir

Let C1 denotes a family of circles with centre on x-axis and touching the y-axis at the origin.

and C2 denotes a family of circles with centre on y-axis and touching the x-axis at the origin.

Statement-1: Every member of C1 intersects any member of C

2 at right angles at the point other than

origin.

because

Statement-2: If two circles intersect at 90° at one point of their intersection, then they must intersect at

90° on the other point of intersection also.




(D) Statement-1 is false, statement-2 is true. [11th, 16-11-2008, P-1]


Q.1035cir

The number of tangents that can be drawn from the point

1,

2

5 to the circle passing through the

points ( )3,1 , ( )3,1 − and ( )3,3 − is

(A) 1 (B*) 0 (C) 2 (D) None

[Sol. The triangle is right angled at ( )3,1 − . Its circumcircle is x2 + y2 – 4x = 0

2

2

5

+ 1– 4 ·

2

5 < 0 The

point is inside the circle.] (IIT-JEE 1985)

Q.104161/cir

A straight line l1 with equation x – 2y + 10 = 0 meets the circle with equation x2 + y2 = 100 at B in

the first 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 =

2

1

slope of l2 = – 2

equation of l2

y = – 2(x – 10) ⇒ y + 2x = 20

Hence t = 20 Ans.] [18-12-2005, 12th 13th]

Q.107107/cir

B and C are fixed points having co−ordinates (3, 0) and (− 3, 0) respectively . If the vertical angle

BAC is 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 ]

[REASONING TYPE]

Q.111302/cir

Passing through a point A(6, 8) a variable secant line L is drawn to the circle

S : x2 + y2 – 6x – 8y + 5 = 0. From the point of intersection of L with S, a pair of tangent lines are

drawn which intersect at P.

Statement-1: Locus of the point P has the equation 3x + 4y – 40 = 0.

because

Statement-2: Point A lies outside the circle.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.



(D*) Statement-1 is false, statement-2 is true.

[Sol. Locus of P is the polar of A(6, 8) w.r.t. the circle S = 0 i.e. x2 + y2 – 6x – 8y + 5 = 0

xx1 + yy

1 + g(x + x

1) + f (y + y

1) + c = 0 [12th, 20-07-2008]

6x + 8y – 3(x + 6) – 4(y + 8) + 5 = 0

3x + 4y – 18 – 32 + 5 = 0

3x + 4y – 45 = 0 ⇒ S-1 is correct

Also the point A (6, 8) lies outside S.

Hence S-1 is false and S-2 is true. ]


Q.8712cir

The locus of the centre of circle which touches externally the circle x2 + y2 –6x –6y + 14 = 0 and also

touches the y -axis is (IIT-JEE 1993)

(A) x2 – 6x – 10y + 14 = 0 (B) x2 – 10x – 6y + 14 = 0

(C) y2 – 6x – 10y + 14 = 0 (D*) y2 – 10x – 6y + 14 = 0

[Sol. |2h|)3k()3h(22 +=−+−

⇒ (x – 3)2 + (y – 3)2 = (x + 2)2 ⇒ y2 – 10x – 6y + 14 = 0 ]

Q.88131/cir

A circle is drawn touching the x−axis 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. 11ah

bk−=−×

−

−⇒ k – b = h – a or a – b = h – k [ME with Q.123]

also 2

ha + and

2

kb + lies on y-axis ⇒

2

ha + =

2

kb +

⇒ a – b = k – h

and touches the x-axis ⇒ cg22 − = 0

∴ centre (b, a)

g2 = c = b2

∴ equation x2 + y2 – 2hx – 2ay + b2 = 0 ]

Q.9035/cir

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 θ, sin θ); H(h, k) is the orthocentre of the ∆ ABC

h = 1 + cos θk = 1 + sin θ O

A(1, 0)

B(0, 1)

C(cos

, s

in

)

θ

θ

(x – 1)2 + (y – 1)2 = 1

x2 + y2 – 2x – 2y + 1 = 0 ] [13th, 14-09-2008, P-2]

Q.94139/cir

A line meets the co-ordinate axes in A and B. A circle is circumscribed about the triangle OAB. If d1

and d2 are the distances of the tangent to the circle at the origin O from the points A and B respectively,

the diameter of the circle is :

(A) 2

dd221

+(B)

2

d2d21

+(C*) d

1 + d

2(D)

21

21

dd

dd

+


Q.6943/cir

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 – 1/2 [11th, 23-12-2007]

and passing through the centre (2, 1) of the circle

y – 1 = – 2

1(x – 2)

2y – 2 = – x + 2 ⇒ x + 2y – 4 = 0 Ans. ]

Q.70108/cir

If aa

,1

, b

b,

1

, c

c,

1

and d

d,

1

are four distinct points on a circle of radius 4 units then,

abcd is equal to

(A) 4 (B) 1/4 (C*) 1 (D) 16

[Sol. Let us assume that circle : x2 + y2 = 16

points are of form

t

1,t ⇒ t2 + 2

t

1 = 16 should satisfy

⇒ t4 – 16t2 + 1 = 0

∴ product of roots = 1 ]

Q.72119/cir

The radical centre of three circles taken in pairs described on the sides of a triangle ABC as diametres

is the :

(A) centroid of the ∆ ABC (B) incentre of the ∆ ABC

(C) circumcentre o the ∆ ABC (D*) orthocentre of the ∆ ABC

[Hint : observe by making a figure or proceed analytically by taking (xr, y

r).

r = 1, 2, 3 as the co−ordinates of the vertices ]

Q.75120/cir

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 3

4(B) cos −1

4

5(C)

π

2(D) tan−1 1

[Hint : m1 = ∞ ; m

2 = ∞ . The two circles are

x2 + y2 − 2x + 2y + 1 = 0 and x2 + y2 − 10x − 6y + 9 = 0 (use family of circles) ]

Q.77124/cir

A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the

equation 2x – 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) [12th, 06-01-2008]

slope of PN:24

13

−

− = 1

equation of normal through PN is

y – 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 + 1

x2 + y2 – 2x – 1 = 0 Ans. ]


Q.5477/cir

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 of the given line = − 5/2

⇒ −+

+

5

2

3

2.

f

g = − 1 ⇒ 15 + 5f = 4 + 2g

⇒ locus is 2x − 5y + 11 = 0 ]

Q.5574/cir

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 all

non-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.5776/cir

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: 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.5813cir

If the angle between the tangents drawn from P to the circle x2 + y2 + 4x – 6y + 9 sin2 α + 13 cos2α=0 is 2α, 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 sin2α – 13 cos2α = 4 sin2αR = 2 sin α

2 sin α

αα

(–2,3)C

P(x,y)

y

Now use sin α = CP

sin2 α ⇒ CP = 2

Hence locus is (x + 2)2 + (y – 3)2 = 4 ]

Q.6078/cir

The locus of the mid points of the chords of the circle x² + y² + 4x − 6y − 12 = 0 which subtend an

angle of π

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 ]


i.e. |x1

2 + y12 – ax

1| = 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= – y

1

1

+ve sign r2 = 2

2

k4

a− (not possible as r2 becomes – ve)

–ve sign r2 = 4

a 2

+ k2 Ans. ]

Q.4036/cir

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 = 5 [12th 11-05-2008]

perpendicular distance from (3, – 4) on

3x + 4y – 25 = 0 is

p = 5

25169 −− =

5

32

d = 5

32 – 5 =

5

7 Ans. ]

Q.4263/cir

If the circle C1 : x2 + y2 = 16 intersects another circle C

2 of radius 5 in such a manner that the

common chord is of maximum length and has a slope equal to 3/4, then the co-ordinates of the centre of

C2 are

(A) ± ±

9

5

12

5, (B*) ±

9

5

12

5, m (C) ± ±

12

5

9

5, (D) ±

12

5

9

5, m

[Hint : Equation of common chord is y = 3

4x ⇒ 3x – 4y = 0

family of circle x2 + y2 – 4 + λ (3x – 4y) = 0

equate radius of this circle as 5. Alternately ...............]

Q.4487/cir

The points (x1, y

1) , (x

2, y

2) , (x

1, y

2) and (x

2, y

1) 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 − x

2) + (y − y

1) (y − y

2) = 0

Note that points form a rectangle]

Q.4767/cir

Let C be the circle of radius unity centred at the origin. If two positive numbers x1 and x

2 are such that

the line passing through (x1, – 1) and (x

2, 1) is tangent to C then

(A*) x1x

2 = 1 (B) x

1x

2 = – 1 (C) x

1 + x

2 = 1 (D) 4x

1x

2 = 1

[Sol. Let the equation tangent in parametric form is [12th 13th 26-3-2006]

x cos θ + y sin θ = 1

put (x1, – 1)

x1 cos θ – sin θ = 1

x1 cos θ = 1 + sin θ ....(1)

now put (x2, 1)

x2 cos θ = 1 – sin θ ....(2)

(1) × (2) x1x

2 cos2θ = 1 – sin2θ = cos2θ

x1 x

2 = 1 Ans.


Q.2339/cir

Triangle ABC is right angled at A. The circle with centre A and radius AB cuts BC and AC internally

at D 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) [13th, 09-09-2007]

Also CD · CB = CE · CX (using power of the point C)

16 · 36 = (b – r)(b + r)

∴ b2 – r2 = 16 · 36

from (1) and (2)

2b2 = 36(36 + 16) = 36 · 52

b2 = 36 · 26 ⇒ b = 266 Ans. ]

Q.27143/cir

The equation of a line inclined at an angle π

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 + c

y – x = c ....(1)

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

c− =

2

c

In ∆AON,

22

2

c2

− = AN

and in ∆CPM,2

c232 −− = CM

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

c2 −

Given AN = CM = 2

c4

2

− = 9 – 2

)c2( 2− ⇒ c = –

2

3

∴ equation of line y = x – 2

3 of 2x – 2y – 3 = 0 ]

Q.2847/cir

If x = 3 is the chord of contact of the circle x2 + y2 = 81, then the equation of the corresponding pair

of tangents, 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 = 0

x2 – 8y2 – 54x + 729 = 0 Ans.]


Q.1426/cir

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

+

2

3,

2

33 ?

(A) 92

3y

2

33x

22

=

−+

−− (B) 9

2

3y

2

3x

22

=

−+

−

(C) x2 + (y − 3)2 = 9 (D*) (x − 3)2 + y2 = 9

[Hint: Line must pass through the centre of the circle ]

Q.157/cir

Let C be a circle x2 + y2 = 1. The line l intersects C at the point (–1, 0) and the point P. Suppose that

the slope of the line l is a rational number m. Number of choices for m for which both the coordinates of

P are rational, is

(A) 3 (B) 4 (C) 5 (D*) infinitely many

[Sol. Equation of the line l is [13th, 16-12-2007]

y – 0 = m(x + 1) ....(1)

solving it with x2 + y2 = 1

x2 + m2(x + 1)2 = 1

(m2 + 1)x2 + 2m2x + (m2 – 1) = 0, m ∈ Q

x1x

2 =

1m

1m2

2

+

−

but x1 = – 1 ⇒ x

2 = 2

2

m1

m1

+

− = x coordinate of P

since m ∈ Q hence x will be rational.

If x is rational then y is also rational from (1) ]

Q.17100/cir

Three concentric circles of which the biggest is x2 + y2 = 1, have their radii in A.P. If the line y = x + 1

cuts all the circles in real and distinct points. The interval in which the common difference of the A.P. will

lie is

(A)

4

1,0 (B)

22

1,0 (C*)

−

4

22,0 (D) none

[Sol. Radius of circle are r1, r

2 and 1

line y = x + 1

perpendicular from (0, 0) on line y = x + 1

= 2

1

now r1 >

2

1 but r

1 = 1 – 2d

hence 1 – 2d > 2

1;

2

12 − > 2d; d <

22

12 −

∴ d = 22

12 −

Aliter :Equation of circle are

x2 + y2 = 1; x2 + y2 = (1 – d)2 ; x2 + y2 = (1 – 2d)2


7m = – 1 ⇒ m = – 7

1 (rejected)

hence the line is

y – 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 – m

2

∴ 2A = (2 – m)

−

m

21 (m < 0)

2A = 2 – m – m

4 + 2 = 4 +

−−

m

4m

let – m = M (M > 0)

2A = 4 + M + M

4 = 4

M

2M4

2

+

−+ =

2

M

2M8

−+

area is minimum if M = 2 ⇒ m = – 2

2A|min

= 8 ⇒ A|min

= 4 Ans. ]

[MULTIPLE OBJECTIVE TYPE]

Q.161511/st.line

If x

c

y

d+ = 1 is a line through the intersection of

x

a

y

b+ = 1 and

x

b

y

a+ = 1 and the lengths of

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

(A*) 1 12 2

a b+ =

1 12 2

c d+ (B)

1 12 2

a b− =

1 12 2

c d−

(C*) 1 1

a b+ =

1 1

c d+ (D) none

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

Q.165529/st.line

A and B are two fixed points whose co-ordinates are (3, 2) and (5, 4) respectively . The co-

ordinates of a 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.168531/st.line

Straight lines 2x + y = 5 and x − 2y = 3 intersect at the point A . Points B and C are chosen on

these two 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 the

bisectors of the given lines, the slopes of the bisectors being − 1/3 and 3 ]

Q.171534/st.line

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


[Sol. (2cos θ1 cos θ

2 + 2cos θ

2 cos θ

3 + 2cos θ

3 cos θ

1 + cos2θ

1 + cos2θ

2 + cos2θ

3) +

(sin2θ1 + sin2θ

2 + sin2θ

3 + 2sin θ

1sin θ

2 + 2sin θ

2 sin θ

3 + 2sin θ

3sin θ

1) = 0

⇒ (cos θ1 + cos θ

2 + cos θ

3)2 + (sin θ

1 + sin θ

2 + sin θ

3)2 = 0

⇒ cos θ1 + cos θ

2 + cos θ

3 = 0 & sin θ

1 + sin θ

2 + sin θ

3 = 0

⇒ centroid and circumcentre of ∆ ABC is at origin [11th, 12-10-2008, P-2]

⇒ ABC is equilateral

∴ Orthocentre of ABC is also origin. ]

Q.126318/st.line

Consider the line L: = 3x + y + 4 = 0 and the points A(–5, 6) and B(3, 2)

Statement-1: There is exactly one point on the line L which is equidistant from the point A and B.

because

Statement-2: The point A and B are on the different sides of the line.


(B*) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for

statement-1.



[Hint: The points A and B may be on same side also [11th, 16-11-2008, P-2]

]

[COMPREHENSION TYPE]

Paragraph for Question Nos. 132 to 134

Consider a general equation of degree 2, as

λx2 – 10xy + 12y2 + 5x – 16y – 3 = 0

Q.13251st.line

The value of 'λ' for which the line pair represents a pair of straight lines is

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

Q.13352

For the value of λ obtained in above question, if L1 = 0 and L

2 = 0 are the lines denoted by the given

line pair then the product of the abscissa and ordinate of their point of intersection is

(A) 18 (B) 28 (C*) 35 (D) 25

Q.13453

If θ is the acute angle between L1 = 0 and L

2 = 0 then θ lies in the interval

(A) (45°, 60°) (B) (30°, 45°) (C) (15°, 30°) (D*) (0, 15°)

[Sol. [11th, 30-11-2008, P-2]

(i) a = λ; h = – 5; b = 12; g = 2

5; f = – 8, c = – 3

λ(12)(–3) + 2(–8)

2

5(– 5) – λ(64) –

4

25· 12 + 3 · 25 = 0

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


[REASONING TYPE]

Q.112303/st.line

Consider the lines, L1: 1

4

y

3

x=+ ; L

2 = 1

3

y

4

x=+ ; L

3 : 2

4

y

3

x=+ and L

4 : 2

3

y

4

x=+

Statement-1: The quadrilateral formed by these four lines is a rhombus.

because

Statement-2: If diagonals of a quadrilateral formed by any four lines are unequal and intersect at right

angle then it is a rhombus.



(C*) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

[Hint: It can be kite also ⇒ S-2 is false] [12th, 06-01-2008]

Q.114308/st.line

Given the lines y + 2x = 3 and y + 2x = 5 cut the axes at A, B and C, D respectively.

Statement-1 : ABDC forms quadrilateral and point (2, 3) lies inside the quadrilateral

because

Statement-2 : Point lies on same side of the lines.

(A) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D*) Statement-1 is False, Statement-2 is True

[Hint: ; P lies outside the quadrilateral ]

Q.116309/st.line

Statement-1: Centroid of the triangle whose vertices are A(–1, 11); B(– 9, – 8) and C(15, – 2)

lies on the internal angle bisector of the vertex A.

because

Statement-2: Triangle ABC is isosceles with B and C as base angles.



(C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

[Hint: AB = 22)19()8( + = 425 ;AC = 22

)13()16( + [11th, 25-11-2007]

∴ ∆ is isosceles ]

Q.118310/st.line

Consider the following statements

Statement-1: The equation x2 + 2y2 – 32 x – 4y + 5 = 0 represents two real lines on the cartesian

plane.

because

Statement-2: A general equation of degree two

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

denotes a line pair if

abc + 2fgh – af2 – bg2 – ch2 = 0




(D*) Statement-1 is false, statement-2 is true.


Q.9115st.line

The distance of the point P(x1, y

1) from each of the two straight lines through the origin is d. The

equation of the two straight lines is

(A*) (xy1 – yx

1)2 = d2(x2 + y2) (B) d2(xy

1 – yx

1)2 = x2 + y2

(C) d2(xy1 + yx

1)2 = x2 + y2 (D) (xy

1 + yx

1)2 = d2(x2 + y2)

[Sol. Let R (h, k) be any point on OM [11th, 27-01-2008]

Area of ∆ OPR = 1001kh

1yx

2

1 11

= 2

1)hykx(

11−

also area of ∆ OPR = 2

d·kh22 +

∴2

1)hykx(

11− =

2

d·kh22 +

locus of (h, k) is

(xy1 – yx

1)2 = d2(x2 + y2) Ans.

Alternatively: Let the line through (0, 0) be y = mx

∴ d = 2

11

m1

ymx

+

− = m2( 2

1x – d2) – 2mx

1y

1 + 2

1y – d2 = 0

replacing m by y/x

x2( 21

y – d2) – 2 xy x1y

1 + y2( 2

1x – d2) = 0

(xy1 – yx

1)2 = d2(x2 + y2) Ans. ]

Q.93137/st.line

Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR

is 2x + 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.979st. line

The greatest slope along the graph represented by the equation 4x2 – y2 + 2y – 1 = 0, is

(A) – 3 (B) – 2 (C*) 2 (D) 3

[Hint: y2 – 2y + 1 = 4x2 [13th, 09-09-2007]

(y – 1) = 2x or – 2x

y = 2x + 1 or y = 1 – 2x

greatest slope = 2 Ans. ]

Q.99139/st.line

If the straight lines joining the origin and the points of intersection of the curve

5x2 + 12xy − 6y2 + 4x − 2y + 3 = 0 and x + ky − 1 = 0

are equally inclined to the co-ordinate axes then the value of k :

(A) is equal to 1 (B*) is equal to − 1

(C) is equal to 2 (D) does not exist in the set of real numbers .

[Hint : Homogenise and put co-efficient of xy = 0 ]


Q.76120/st.line

Let (x1, y

1) ; (x

2, y

2) and (x

3, y

3) are the vertices of a triangle ABC respectively. D is a point on BC

such that BC = 3BD. The equation of the line through A and D, is

(A)

1yx

1yx1yx

22

11 + 2

1yx

1yx1yx

33

11 = 0 (B) 3

1yx

1yx1yx

22

11 +

1yx

1yx1yx

33

11 = 0

(C)

1yx

1yx1yx

22

11 + 3

1yx

1yx1yx

33

11 = 0 (D*) 2

1yx

1yx1yx

22

11 +

1yx

1yx1yx

33

11 = 0

[12th 13th (27-11-2005)]

Q.7831st.line

Area of the triangle formed by the line x + y = 3 and the angle bisectors of the line pair

x2 – 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 = 2

1·1 =

2

1 Ans. ] [13th, 27-07-2008]

Q.80122/st.line

If the straight lines , ax + amy + 1 = 0 , b

x + (m + 1) b

y + 1 = 0 and cx + (m + 2)cy + 1 = 0,

m ≠

0 are concurrent then a, b, c are in :

(A) A.P. only for m = 1 (B) A.P. for all m

(C) G.P. for all m (D*) H.P. for all m.

[Sol. D = 1)2m(cc

1)1m(bb1ama

++ = 0 ;

1c2c1bb10a

= 0

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

b = ca

ac2

+⇒ a, b, c are in H.P. ]

Q.82124/st.line

If in triangle ABC , A ≡ (1, 10) , circumcentre ≡

( )− 13

23

, and orthocentre ≡ ( )11

343

, then the

co-ordinates of mid-point of side opposite to A is :

(A*) (1, − 11/3) (B) (1, 5) (C) (1, − 3) (D) (1, 6)

[Hint: O, G, C are collinear. Get G ]


put x = 0, get y = – 5

(0, – 5) Ans. ]

Q.6191/st.line

Given A ≡ (1, 1) and AB is any line through it cutting the x-axis in B. If AC is perpendicular to AB

and meets 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(m

1−−

2h = 1 m

1−

2k = 1 + m

1

____________

locus is x + y = 1

____________ ]

Q.6495/st.line

In a triangle ABC, if A (2, – 1) and 7x – 10y + 1 = 0 and 3x – 2y + 5 = 0 are equations of an

altitude and an angle bisector respectively drawn from B, then equation of BC 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

tanθ =

10

31

5

1

2

3

+

−

=

2

m31

m2

3

+

−

⇒ 1 = m32

m23

+

− or + 1 =

m32

m23

+

−

⇒ m = 1/5 (rejected) or – 5

equation of BC = y + 2 = – 5 (x + 3) ⇒ 5x + y + 17 = 0

Alternatively : Take image of (2, –1) in the line BD to get a point on BC]

Q.66102/st.line

AB is the diameter of a semicircle k, C is an arbitrary point on the

semicircle (other than A or B) and S is the centre of the circle inscribed

into triangle ABC, then measure of

(A) angle ASB changes as C moves on k.

(B) angle ASB is the same for all positions of C but it cannot be determined without knowing the radius.

(C*) angle ASB = 135° for all C.

(D) angle ASB = 150° for all C. [12th 13th (27-11-2005)]


[Sol. D = 0 [12 13th test (29-10-2005)]

x2 = 4(x – y)2

x = 2(x – y) or x = –2(x – y)

x = 2y or 3x = 2y ⇒ line pair with slope 3/2 and 1/2 ⇒ D ]

Q.4364/st.line

If L is the line whose equation is ax + by = c. Let M be the reflection of L through the y-axis, and

let N be the reflection of L through the x-axis. Which of the following must be true about M and N for all

choices 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 whose

equation is ax – by = c, while a reflection through the y-axis

results in the line N whose equation is –ax + by = c. Both clearly

have slope equal to a/b (from, say, the slope-intercept form of

the equation.) [13th, 09-03-2008] ]

Q.4667/st.line

Vertices of a parallelogram ABCD are A(3, 1), B(13, 6), C(13, 21) and D(3, 16). If a line passing

through the origin divides the parallelogram into two congruent parts then the slope of the line is

(A) 12

11(B*)

8

11(C)

8

25(D)

8

13

[Hint: as shown [12 13th test (29-10-2005)]

m = 13

x21− =

3

1x +

63 – 3x = 13x + 13

16x = 50

x = 8

25; Hence m =

3

1·1

8

25

+ =

24

33 =

8

11 Ans. ]

Q.5179/st.line

The distance between the two parallel lines is 1 unit . A point 'A' is chosen to lie between the lines

at a distance 'd' from one of them . Triangle ABC is equilateral with B on one line and C on the other

parallel line . The length of the side of the equilateral triangle is

(A) 1dd3

2 2 ++ (B*) 3

1dd2

2 +−(C) 1dd2 2 +− (D) 1dd2 +−

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

dividing 3 cot θ = 1

1

+

−

d

d , squaring equation (2) and putting

the value of cot θ, x2 = 1

3(4d2 − 4d + 4) ⇒ x = 2

d d2 1

3

− +]


Q.1620/st.line

If A (1, p2) ; B (0, 1) and C (p, 0) are the coordinates of three points then the value of p for which

the area of the triangle ABC is minimum, is

(A) 3

1(B) –

3

1(C)

3

1 or –

3

1(D*) none

[Sol. A = 10p1101p1

2

12

= 2

1[1(1 – 0) + p(p2 – 1)] =

2

1(p3 – p + 1) [12th Test (16-01-2005)]

Hence A = 2

1| p3 – p + 1 |

Now, minimum value of modulus is zero. Since A (p) is a cubic it must vanish for some p other than

given as A, B, C ⇒ (D) ]

Q.2125/st.line

Each member of the family of parabolas y = ax2 + 2x + 3 has a maximum or a minimum point

depending upon the value of a. The equation to the locus of the maxima or minima for all possible values

of 'a' is

(A*) a straight line with slope 1 and y intercept 3.

(B) a straight line with slope 2 and y intercept 2.

(C) a straight line with slope 1 and x intercept 3.

(D) a circle

[Hint: maxima / minima occurs at =

−−

a

13,

a

1(f (x)= ax2 + bx + c has a maxima or minima if x= –

a2

b)

h = – a

1and k =

a

13 − ; elliminating 'a' we get [11th, 26-02-2009, P-2]

k = 3 + x

locus is y = x + 3 ⇒ (A) ]

Q.2632/st.line

A ray of light 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 of AB 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.2938/st.line

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 the line 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

reflection of 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 [12 13th test (29-10-2005)]

Area of ∆ ABC = 2

)nm(m2 −

= m2 – mn

∴ area of pentagon = 4mn + m2 – mn

= m2 + 3mn Ans. ]


STRAIGHT LINE

[STRAIGHT OBJECTIVE TYPE]

Q.11/st.line

If the lines x + y + 1 = 0 ; 4x + 3y + 4 = 0 and x + αy + β = 0, where α2 + β2 = 2, are concurrent

then

(A) α = 1, β = – 1 (B) α = 1, β = ± 1 (C) α = – 1, β = ± 1 (D*) α = ± 1, β = 1

[Sol. Lines are x + y + 1 = 0 ; 4x + 3y + 4 = 0 and x + αy + β = 0 , where α2 + β2 = 2

βα1434111

= 0

1 (3β – 4α) – 1 (4β – 4) + 1 (4α – 3) = 3β – 4α – 4β + 4 + 4α – 3

= – β + 1 = 0 ⇒ β = 1

∴ α = ± 1 ]

Q.32/st.line

Given the family of lines, a (3x + 4y + 6) + b (x + y + 2) = 0 . The line of the family situated at the

greatest distance 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 is

perpendicular to the line joining (− 2, 0) and (2, 3) or taking (2, 3) as centre and radius equal to PA draw

a circle, the required line will be a tangent to the circle at (− 2, 0) ]

Q.58/st.line

A variable rectangle PQRS has its sides parallel to fixed directions. Q and S lie respectively on the

lines x = 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 PS

0 = h + x1

⇒ x1 = –h hence p (–h, 0)

equation of PQ ≡ y = mx + c

passes through (–h, 0)

c = mh

∴ y = mx + mh

since Q lies on it

⇒ y 1 = 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]

Documents

Q.B on Straight line and circle [44] - Weebly · Q.B on Straight line and circle ... Q.152 416/cir Locus of the middle points of the portion of the tangent to the circle terminated