MATHEMATICS
CLASS TEST
Target IIT JEE 2014
CLASS : XI-PTP
CLASS-TEST ON CIRCLES MM: 73 TIME: 80 MIN
[SINGLE CORRECT CHOICE TYPE]
Q.1 to Q.9 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.1
[Sol:
One of the diagonals of a square is the portion of the line
[9 × 3 = 27, –1]
x y
  2 intercepted between the axes, then
2 3
the extremities of the other diagonal are
(A*) (5, 5), (–1, 1)
(B) (0, 0), (4, 6)
(C) (0, 0), (–1, 1)
(D) (5, 5) , (4, 6)
Extremities of the given diagonal are (4, 0) and (0, 6)
3
 Slope of this diagonal = 
2
2
 Slope of other diagonal =
3
x 2 y3

2 =r
 Equation of the other diagonal is 3
13
13
For the extremities of the diagonal r =  13

x – 2 = +3, y – 3 = +2 ;

x = 5, –1 and y = 5, 1 ]
Q.2
Tangents TP and TQ are drawn from a point T to the circle x2 + y2 = a2. If the point T lies on the line
px + qy = r, the locus of centre of the circumcircle of triangle TPQ is
(A*) 2px + 2qy = r
(B) px + qy = r
(C) 2px + qy = r
(D) px + 2qy = r
[Sol:
x1 = 2h
y1 = 2k
satisfy x1y1 in px + qy = r
p(2h) + q(2k) = r
2px + 2qy = r ]
Q.3
The triangle PQR is inscribed in the circle, x2 + y2 = 25. If Q and R have co-ordinates (3, 4) &
(–4, 3) respectively, then  QPR is equal to :
(A)

2
(B)

3
(C*)

4
(D)

6
[Hint: Angle subtended at the centre is 90º]
Q.4
If two distinct chords, drawn from the point (p, q) on the circle x2 + y2 = px + qy (where pq  0) are
bisected by the x  axis, then :
(A) p2 = q2
(B) p2 = 8q2
(C) p2 < 8q2
(D*) p2 > 8q2
[Sol:
[1]
x2 – px + y2 – qy = 0
x(x – p) + y(y – q) = 0
This is the circle with (p, q) & (0, 0) as diameter.
Let (h, 0) lies on x–axis
now (2h – p, –q) lies on circle ( x axis bisect the two distinct chord  h will have two values)
(2h – p) (2h – 2p) – q(–2q) = 0
(2h – p) (h – p) + q2 = 0
2h2 – 3ph + p2 + q2 = 0
D>0 
9p2 – 8 (p2 + q2) > 0
p2 – 8q2 > 0
p2 > 8q2

Q.5
[Sol:
D
]
The locus of the foot of the perpendicular from the origin upon chords of the circle
x2 + y2 – 2x – 4y – 4 = 0, which subtend a right angle at the origin is
(A*) x2 + y2 – x – 2y – 2 = 0
(B) 2(x2 + y2) – 2x – 4y – 3 = 0
(C) x2 + y2 – 2x – 4y + 4 = 0
(D) x2 + y2 + x + 2y – 2 = 0
Equation to the chord AB is
 x1
(y – y1) = y (x – x1)
1
 xx1 + yy1 = x12 + y12
...........(i)
Where M (x1, y1) is the foot of perpendicular from the origin.
Now homogenising the equation of the given circle, we get
2
2
2
 y 2 x12  y12  2 x  4 y xx1  yy1  x12  y12  4xx1  yy1   0
This represents a pair of perpendicular lines passing through the origin
Hence coefficient of x2 + coefficient of y2 = 0
x




2x 2  y 2   [2 x1 x12  y12   4 y1 x12  y12 ]  4x12  y12   0
or x12  y12   x1  2 y1   2  0
Hence locus of M (x1, y1) is x2 + y2 – x – 2y – 2 = 0
Q.6
[Sol:
]
If the circle x2 + y2 + 4x + 22y + c = 0 bisects the circumference of the circle x2 + y2 – 2x + 8y – d = 0,
then c + d is equal to
(A) 30
(B*) 50
(C) 40
(D) 56
Given circles are
S1  x2 + y2 + 4x + 22y + c = 0
....(1)
S2  x2 + y2 – 2x + 8y – d = 0
....(2)
The equation of their common chord is
S1 – S2 = 6x + 14y + c + d = 0
S2 bisects the circumference of S2, the centre of S2 i.e.

(1, –4) will lie on (3)

6 × 1 + 14 × (–4) + c + d = 0

c + d = 50
]
[2]
Q.7174/cir If a circle of radius 3 units is touching the lines
then the length of chord of contact to this circle, is
3 1
2
(A)
[Sol.
(B)
 3 1

3
(C*) 

2


3 1
2


A

3y  x y  3 x  0
In AMP, sin 75° =
3P
M
15°
15°
x
,y  3 x
 y 
3
 APO = 75°

y= 3 x
3 y 2  3xy  xy  3 x 2  0

(D) 3
y
3 y 2  4 xy  3 x 2  0
Given equation of lines

3 y 2  4 xy  3 x 2  0 in the first quadrant

3 1
2
y=x
75°
3
y=
x
3
B
x
O (0, 0)
AM
 AM = 3 sin 75°
3

Q.8
[Sol.

 3 1
3 1
3
Now length of chord of contact AB = 2AM  23 sin 75  6 sin 75  6
Ans.]

2
 2 2 
If a variable line L is passing through a point P(2, 3) and intersecting the circle
x2 + y2 – 7x – 7y + 22 = 0 at A1, A2, A3, .......... An then all lines perpendicular to L through
A1, A2, ........ An passes through a fixed point (x0, y0). The value of (x0 + y0), is equal to
(A) 5
(B) 7
(C*) 9
(D) 11
The point (2, 3) lies on the circle x2 + y2 – 7x – 7y + 22 = 0. From the figure it is clear that every line
perpendicular to variable line L passing through other end of the diameter through P(2, 3) whose
co-ordinates are (5, 4).
A3
A2
A1
(x0, y0)
7,7
2 2
P(2,3)
So,(x0 + y0) = 5 + 4 = 9. Ans.]
Consider the lines L : (k + 7)x – (k – 1)y – 4(k – 5) = 0 where k is a parameter
and the circle C : x2 + y2 + 4x + 12y – 60 = 0
Statement-1: Every member of L intersects the circle 'C' at an angle of 90°
Statement-2: Every member of L is tangent to the circle C.
(A) Statement-1 is true, statement-2 is true; statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true; 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.
[Exp. Centre (– 2, –6). Substituting in L
– 2(k + 7) + 6(k – 1) – 4(k – 5) = (– 2k + 6k – 4k) – 14 – 6 + 20 = 0
Hence every member of L passing through the centre of the circle

cuts it at 90°.
Hence S-1 is true and S-2 is false. ]
Q.9
[3]
[COMPREHENSION TYPE]
Q.10 to Q.15 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.
Q.10
Q.11
Q.12
[Sol.
(i)
[6 × 3 = 18, –1]
Paragraph for question nos. 10 to 12
Two fixed points A and B are 4 units apart, and are on the same side of a moving line L. If perpendicular
distances of A and B say p1 and p2 from the line L are such that p1 + 3p2 = k, k being a constant, then
the line L always touches a fixed circle C.
The centre of the circle C lies on
(A*) line segment joining AB
(B) perpendicular bisector of AB
(C) one of A or B
(D) nothing definite can be said
If k = 4 then the radius of the circle is
(A*) 1
(B) 2
(C) 4
(D) 8
If A and B are (–2, 0) and (2, 0) respectively, then the centre of the circle C is
(A) (0, 1)
(B*) (1, 0)
(C) (3/2, 0)
(D) can not be found
Let
A = (0, 0) and B = (4, 0)
and the line be ax + by = 1
1
p1 =
2
a b
;
2
4a  1
p2 =
a 2  b2
p1 + 3p2 = k
1
4a  1
+3
a 2  b2
a 2  b2
=k
now (0, 0) and (4, 0) must give the same sign i.e. – ve with the line L (4a – 1 < 0)
1

2
a b
2

4(1  3a )
3(1  4a )
a 2  b2
=k ;
a 2  b2
=k
3a  1
a 2  b2

k
4
(ii)
hence centre of the fixed circle is (3, 0) which lies on the line segment AB
If k = 4

r = 1 Ans.
(iii)
p1 =
 2a  1
2
a b
2
;
p2 =

(A)
2a  1
a 2  b2
hence, with the same argument, 2a – 1 < 0
and
– 2a – 1 < 0

p1 + 3p2 = k
1  2a
3(1  2a )

=k
a 2  b2
a 2  b2
4(1  a )
=k
a 2  b2
| a 1|
2
2

k
4
a b
hence centre is (1, 0) Ans.]
[4]
Q.13
Paragraph for question nos. 13 to 15
Consider two circles S1 and S2 such that the line 4x + 3y = 10 is a common tangent at M(1, 2)
and radius of each of the circle is 5. Given A and B are centres of circle S1 and S2 and O is the
origin.
cos AOB equals
(A*)
Q.14
Q.15
[Sol.
2
5
(B)
2
5
(C)
Area of OAB equals
(A) 1
(B) 2
5
26
(D)
(C) 3
(D*) 5
The length of external common tangent of S1 and S2, is
(A) 6
(B) 8
(C*) 10
(D) 12
Slope of line perpendicular to 4x + 3y – 10 = 0 is tan  =

5
26
3
4
A(5,5)
x 1 y  2
Equation of AB in parametric form is 4 5  3 5 = 4
=10
4x+3y
5
M(1,2)
5
3 
 4
so, any point on it is 1  r , 2  r 
5 
 5

A (5, 5) and B(– 3, – 1)
B(–3,–1)
O(0, 0)
52

10
(i)
A(5, 5)

(ii)
cos AOB =
Area of OAB =
=
(iii)
B(–3, – 1)
10
50  10  100
 20
2
=
=
Ans.
2 5 2 10
5 2 2· 5
5
    

1
(OA ) (OB) sin AOB
2
1
5 2
2
  10 
1
 4
1   = 5 2
2
 5
 
 1 
5· 2 

 5


area (OAB) = 5 Ans.
As, S1 and S2 have same radius 5 (each)
so length of external common tangent = distance between their centres = 10 Ans.]
[5]
[MULTIPLE CORRECT CHOICE TYPE]
Q.16 to Q.17 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [2 × 4 = 8, 0]
Q.16
[Sol.
Consider the circles S1 : x2 + y2 = 4 and S2 : 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 S2 are orthogonal.
S1 : x2 + y2 = 4 and
S2 = x2 + y2 – 2x – 4y + 4 = 0
centre: (0, 0); radius = 2 centre : (1, 2); radius = 1
(A)
(B)
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
P(h, k) power of point P is same w.r.t. these two circles.

(C)

h 2  k 2  4 = h 2  k 2  2h  4k  4
–4
= – 2h – 4k + 4
2h + 4k – 8 = 0
x + 2y – 4 = 0 
(B) is correct
y intercept of S1 is 2
4 =4
y intercept of S2 is 2 4  4 = 0

sum of y-intercept = 4 
(C) is incorrect
(D)
Q.17
[Sol.
2(0 + 0) = – 4 + 4

circle is orthogonal.

(D) is correct]
Which of the following statements is/are incorrect?
(A*) Two circles always have a unique common normal.
(B*) Radical axis is always perpendicular bisector to the line joining the centres of two circles.
(C) Radical axis is nearer to the centre of circle of smaller radius.
(D*) Two circles always have a radical axis.
(A)
if two circles are concentric then infinite common normals

False
(B)
perpendicular bisector only if two circles have equal radius

False
(C)
Now p12  r12  l 2 and p 22  r22  l 2 obviously p1 < p2

(D)
Radical axis does not exist in case of two concentric circles
True

False]
[6]
[INTEGER TYPE]
Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [3 × 5 = 15, 0]
Q.1
Let a circle passes through the point (, ) where  <  and ,  are values of p such that sum of
1
9
is . The circle touches the straight line segment
p
2
joining
(1, 0), (3, –2) at mid point of the segment. If the equations of the circle is
ax2 + by2 + gx + fy + 135 = 0 then find the value of a + b – g – f.
infinite G.P. with 1st term p and common ratio
[Sol:
given
p
1
1
p
=
p2
9
9

=
2p2 – 2p 9 = 0
p 1 2
2
3 
3
 = 3 Thus circle passes through  , 3  , mid point of (1, 0) (3, –2) is (2, –1), equation of line
2
2 
joining (1, 0) (3, –2) x + y – 1 = 0
Thus equation of circle is (x – 2)2 + (y + 1)2 + (x + y –1) = 0
=
3 
it passes through  , 3 
2 


Q.2
65
& equation of circle becomes
14
14x2 + 14y2 –121x – 37y + 135 = 0
a = b = 14, g = –121, f = –37, c = 135
a + b – g – f = 186
]
=
Let S : 2x2 + 2y2  2x + 6y  3 = 0 and a point P (1, 1) lying outside to the circle S. PA and PB are
tangents drawn to S. If  denotes area of triangle PAB and 2 denotes the area of quadrilateral
PACB, where C is the centre of the circle S and the value of 1 · 2 =
[Sol.
are expressed in their lowest form, then find the value of (p + q).
S : 2x2 + 2y2  2x + 6y  3 = 0,
R = radius of circle = 2
L = length of the tangent from P(1, 1) to the circle S = 0
L=
5
,
2
1=
RL3
R 2  L2
,
2 = RL

25
R L
4 = 50 .
12 = 2
2 =
5
13
R L
4
2


p = 50, q = 13
(p + q) = 63. Ans.]
2 4
p
, (p, q  N) where p and q
q
[Ans. 63]
4
[7]
Q.3
[Sol.
Q.4
[Sol:
The lines 3x – y + 3 = 0 and x – 3y – 6 = 0 cut the co-ordinate axes at A, B and C, D respectively.
If the equation of the circle through these four points of intersection is x2 + y2 – ax – by – c = 0,
then find the value of (a + b + c).
y
[Ans. 12]
Using 2nd degree curve, we get
(3x – y + 3) (x – 3y – 6) + xy = 0
if it is a circle then coefficient of xy = 0
B(0,3)
 – 10 = 0  = 10.

equation of circle is
(3x – y + 3) (x – 3y – 6) + 10xy = 0 A(–1,0)
x
3(x2 + y2) – 15x – 3y – 18 = 0
(0, 6)
x2 + y2 – 5x – y – 6 = 0
Hence, (a + b + c) = 12. Ans.]
(0, – 2)
If the vertices of  ABC are A (4, 5), B (6, 7) and C (x, y). If the coordinate of C satisfy the equation
(x – 4) (x – 6) + (y – 5) (y – 7) = 0 and Area ( ABC) = 1 then find the number of possible
position of C.
[Ans. 0004]
 y  7  y  5 

 = –1
(x – 4) (x – 6) = –(y – 5) (y – 7)  
 x  6  x  4 
locus is obviously circle
again AB = 2 2
so,
as

 DP =
2
Ar (ABP) = ½ × 2 2 × 2 = 2
Ar (ABP) > Ar (ABC)
possible position of c will be 4 ]
[8]