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.B on Straight line and circle [42]
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]
Q.B on Straight line and circle [40]
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
Q.B on Straight line and circle [38]
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)
Q.B on Straight line and circle [36]
[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.B on Straight line and circle [34]
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.
(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.
[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.
(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. [11th, 16-11-2008, P-1]
Q.B on Straight line and circle [32]
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.
(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. 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.B on Straight line and circle [30]
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.B on Straight line and circle [28]
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.B on Straight line and circle [26]
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 ]
Q.B on Straight line and circle [24]
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.B on Straight line and circle [22]
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.B on Straight line and circle [20]
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
Q.B on Straight line and circle [18]
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
Q.B on Straight line and circle [16]
[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.
(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.
[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.
Q.B on Straight line and circle [14]
[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.
(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.
[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.
(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.
[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
(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.
Q.B on Straight line and circle [12]
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.B on Straight line and circle [10]
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 ]
Q.B on Straight line and circle [8]
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 mid- point 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)]
Q.B on Straight line and circle [6]
[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.B on Straight line and circle [4]
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. ]
Q.B on Straight line and circle [2]
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]