MATHEMATICSCLASS TEST

Target IIT JEE 2014

[1]

[SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.9 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. [9 × 3 = 27, �1]

Q.1 One of the diagonals of a square is the portion of the line 23

y

2

x intercepted between the axes, then

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)

[Sol: Extremities of the given diagonal are (4, 0) and (0, 6)

Slope of this diagonal = 2

3

Slope of other diagonal = 3

2

Equation of the other diagonal is 13

23y

13

32x

= r

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 linepx + 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 = 2hy1 = 2ksatisfy x1y1 in px + qy = rp(2h) + q(2k) = r2px + 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) arebisected by the x axis, then :(A) p2 = q2 (B) p2 = 8q2 (C) p2 < 8q2 (D*) p2 > 8q2

[Sol:

CLASS : XI-PTP CLASS-TEST ON CIRCLES MM: 73 TIME: 80 MIN

[2]

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 = 02h2 � 3ph + p2 + q2 = 0

D > 0

9p2 � 8 (p2 + q2) > 0p2 � 8q2 > 0p2 > 8q2 D ]

Q.5 The locus of the foot of the perpendicular from the origin upon chords of the circlex2 + 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

[Sol: Equation to the chord AB is

(y � y1) = 1

1

y

x (x � x1)

xx1 + yy1 = x12 + y1

2 ...........(i)Where M (x1, y1) is the foot of perpendicular from the origin.Now homogenising the equation of the given circle, we get

0yyxx4yxyyxxy4x2yxyx 211

21

2111

221

21

22

This represents a pair of perpendicular lines passing through the originHence coefficient of x2 + coefficient of y2 = 0

0yx4]yxy4yxx2[yx2 21

21

21

211

21

211

22

or 02y2xyx 1121

21

Hence locus of M (x1, y1) is x2 + y2 � x � 2y � 2 = 0 ]

Q.6 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

[Sol: Given circles areS1 x2 + y2 + 4x + 22y + c = 0 ....(1)S2 x2 + y2 � 2x + 8y � d = 0 ....(2)The equation of their common chord isS1 � 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 ]

[3]

Q.7174/cir If a circle of radius 3 units is touching the lines 0x3xy4y3 22 in the first quadrant

then the length of chord of contact to this circle, is

(A) 2

13 (B)

2

13 (C*)

2

133 (D)

2

133

[Sol. Given equation of lines 0x3xy4y3 22

0x3xyxy3y3 22

0x3yxy3 15°

15°

O (0, 0)

75°

y=x

y= x3

y= 3 x

x

y

A

B

M

P3

3

x3y,3

xy

APO = 75°

In AMP, sin 75° = 3

AM AM = 3 sin 75°

Now length of chord of contact AB = 2AM

2

133

22

13675sin675sin32

Ans.]

Q.8 If a variable line L is passing through a point P(2, 3) and intersecting the circlex2 + y2 � 7x � 7y + 22 = 0 at A1, A2, A3, .......... An then all lines perpendicular to L throughA1, 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

[Sol. 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) whoseco-ordinates are (5, 4).

P(2,3)

72

, 72

A1

A2A3

(x , y )0 0

So,(x0 + y0) = 5 + 4 = 9. Ans.]Q.9 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. ]

[4]

[COMPREHENSION TYPE]

Q.10 to Q.15 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. [6 × 3 = 18, �1]

Paragraph for question nos. 10 to 12Two fixed points A and B are 4 units apart, and are on the same side of a moving line L. If perpendiculardistances of A and B say p1 and p2 from the line L are such that p1 + 3p2 = k, k being a constant, thenthe line L always touches a fixed circle C.

Q.10 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

Q.11 If k = 4 then the radius of the circle is(A*) 1 (B) 2 (C) 4 (D) 8

Q.12 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[Sol.(i) Let A = (0, 0) and B = (4, 0)

and the line be ax + by = 1

p1 = 22 ba

1

; p2 = 22 ba

1a4

p1 + 3p2 = k

22 ba

1

+ 3 22 ba

1a4

= k

now (0, 0) and (4, 0) must give the same sign i.e. � ve with the line L (4a � 1 < 0)

2222 ba

)a41(3

ba

1

= k

22 ba

)a31(4

= k ; 4

k

ba

1a322

hence centre of the fixed circle is (3, 0) which lies on the line segment AB (A)(ii) If k = 4 r = 1 Ans.

(iii) p1 = 22 ba

1a2

; p2 = 22 ba

1a2

hence, with the same argument, 2a � 1 < 0

and � 2a � 1 < 0

p1 + 3p2 = k

2222 ba

)a21(3

ba

a21

= k

22 ba

)a1(4

= k

4

k

ba

|1a|22

hence centre is (1, 0) Ans.]

[5]

Paragraph for question nos. 13 to 15Consider 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 theorigin.

Q.13 cos AOB equals

(A*) 5

2(B)

5

2(C)

26

5(D)

26

5

Q.14 Area of OAB equals(A) 1 (B) 2 (C) 3 (D*) 5

Q.15 The length of external common tangent of S1 and S2, is(A) 6 (B) 8 (C*) 10 (D) 12

[Sol. Slope of line perpendicular to 4x + 3y � 10 = 0 is tan = 43

Equation of AB in parametric form is 53

2y

54

1x

= 4

5

5

B(�3,�1)

A(5,5)

M(1,2)

4x+3y=10

so, any point on it is

r

5

32,r

5

41

A (5, 5) and B(� 3, � 1)

(i)

A(5, 5) B(�3, � 1)

O(0, 0)

10

1052

cos AOB = 10252

1001050 = 5·225

20 =

5

2 Ans.

(ii) Area of OAB = AOBsin)OB()OA(21

=

5

411025

2

1 =

5

12·525

21

area (OAB) = 5 Ans.(iii) As, S1 and S2 have same radius 5 (each)

so length of external common tangent = distance between their centres = 10 Ans.]

[6]

[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 Consider the circles S1 : x2 + y2 = 4 and S2 : x

2 + 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.

[Sol. S1 : x2 + y2 = 4 and S2 = x2 + y2 � 2x � 4y + 4 = 0

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

(A) d = distance between centres = 5r1 + 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 = 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.17 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.

[Sol. (A) if two circles are concentric then infinite common normals False(B) perpendicular bisector only if two circles have equal radius False

(C)

Now 221

21 rp l and 22

222 rp l obviously p1 < p2 True

(D) Radical axis does not exist in case of two concentric circles False]

[7]

[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

infinite G.P. with 1st term p and common ratio p

1 is

2

9. The circle touches the straight line segment

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.

[Sol: given

p1

1

p

= 2

9

1p

p2

=

2

9 2p2 � 2p 9 = 0

= 2

3 = 3 Thus circle passes through

3,

2

3, mid point of (1, 0) (3, �2) is (2, �1), equation of line

joining (1, 0) (3, �2) x + y � 1 = 0

Thus equation of circle is (x � 2)2 + (y + 1)2 + (x + y �1) = 0

it passes through

3,

2

3

= 14

65 & equation of circle becomes

14x2 + 14y2 �121x � 37y + 135 = 0

a = b = 14, g = �121, f = �37, c = 135

a + b � g � f = 186 ]

Q.2 Let S : 2x2 + 2y2 2x + 6y 3 = 0 and a point P (1, 1) lying outside to the circle S. PA and PB aretangents 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 = q

p, (p, q N) where p and q

are expressed in their lowest form, then find the value of (p + q). [Ans. 63][Sol. 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 = 2

5, 1= 22

3

LR

RL

, 2 = RL

12 = 22

42

LR

LR

=

25

4

425

4

= 13

50.

p = 50, q = 13 (p + q) = 63. Ans.]

[8]

Q.3 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).[Ans. 12]

[Sol. Using 2nd degree curve, we get(3x � y + 3) (x � 3y � 6) + xy = 0

if it is a circle then coefficient of xy = 0 � 10 = 0 = 10.

A(�1,0)

(0, � 2)

(0, 6)x

y

B(0,3)

equation of circle is(3x � y + 3) (x � 3y � 6) + 10xy = 0

3(x2 + y2) � 15x � 3y � 18 = 0

x2 + y2 � 5x � y � 6 = 0

Hence, (a + b + c) = 12. Ans.]

Q.4 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]

[Sol: (x � 4) (x � 6) = �(y � 5) (y � 7)

4x

5y

6x

7y = �1

locus is obviously circle

again AB = 22 DP = 2

so, Ar (ABP) = ½ × 22 × 2 = 2as Ar (ABP) > Ar (ABC) possible position of c will be 4 ]