Jee 2014 Booklet2 Hwt Solutions Circles

Vidyamandir Classes

Solutions to Home Practice Test/Mathematics Circles 1.(A)

Chord will be farthest from centre when (2, 3) is the centre of the chord. 

2.(B)

HWT - 1

equation is 2 x  3 y   2    3  13 2

2

Since (1, 4) lies inside the circle, we have,

12   4 2  6 1  10  4   P  0



P < 29

And circle do not touch or intersect coordinate axes 

6.(B)

g2  c

 32  P



and

f2 c



 5 2  P

Let u be y = mx Since intercepts made by both lines on the circle are equal 



7.(C)

8.(D)

distance of centre of the circle from both lines are equal

m  3   2  2  m 1 2

1 3  1 2 2   2 2



m 2  9  6 m  8m 2  8



m

 

m3 m2  1

2 2

7 m 2  6m  1  0

6  64 6  8 1   1, 14 14 7

From geometry we get A0A1 = 1 and

A0 A2  A0 A4  3



product is 1

 3  3   3

On every Ck we have k = k () 

( is angular displacement)

 = 1 radian

Now to cross x axis  must be 2. 

If we add all displacement,  will just exceed 2 in the 7th circle.

Circles 1.(B)

HWT - 2

There will be two such circles, one above the line and other below the line. Coordinates of their centres are given be x1  1  5 cos  , y1  2  5 sin 

and

x2  1  5 cos  , y2  2  5 sin 

where tan is the slope of the line perpendicular to the given line 

x1  5, y1  5



equations for circles are

C1 :

 x  52   y  52  25

 tan 



x2  3, y2  1

C2 :

 x  32   y  12  25

3 4

But since circle should lie in all quadrants 

VMC/Circles

g 2  c and f 2  c





reject C1

6

C2 is the correct answer.

HWT-Solutions/Mathematics

Vidyamandir Classes

3.(A)

PR  S1

tan

5.(C)

2  1 r S1 S1  r 2 2  cos     r 2 S1  r 2 1  tan 2 1 2 S1

1  tan 2

 r  2 S1

and

r2  g 2  f 2  c

and

According to the question n

 x  x 

2

i

  y  yi   k 2 2

i 1

centre of the circle 



7.(A)

In module.

9.(C)

In circle’s equation

  coefficient of x 



2 coefficient of x 2





2  xi  xi  2 n n

8.(D)

coefficient of xy = 0

2 2  a  b a 2  b2  Use :  x     y      2  2 4  

2.(D)

2

coefficient of x2 = coefficient of y2

and

Circles 1.(C)

   

HWT - 3

Normal passes through centre of circle i.e.

normal passes through (2, 3) and (1, 1)



equation is



2x  2  y 1



2x  y  1

Use

0  0  tan   cos 



1  tan 2  cos

2

cos 



2

2

sec 

y 1  3 1  2   x  1  2  1  1

2



cos   2

and

f 2  c , it touches x-axis.

Therefore no possible solution. g 2  c , it touches y-axis

4.(C)

Since

5.(B)

Equation of circle having radius r and touching both coordinate axes is

 x  r 2   y  r 2  r 2 Since it touches x + y = 1, we have 

8.(A)

r  r 1 2

r

2r  1  2 r



2r  1   2 r

1 2 2



two values of r.



r



two such circles exist.

Since r1  r2  C1 C2 Circles touch externally.

VMC/Circles

7

HWT-Solutions/Mathematics

Vidyamandir Classes

Circles 1.(A)

HWT - 4

These circles touch each other externally at origin.  003 3  3 0 , Centroid     3 3  

Centroid  (1, 0)

5.(A)

Common chord must pass through the centre of the circle having smaller radius. i.e. C1 C2 :

 x  a 2   y  b 2  25 C1  C2 represents common chord.



a 2  b 2  2ax  2ay  9



a  b  2a  0   2a  0   9 2

which passes through (0, 0) 

2

a 2  b2  9

Circles 1.(A)

HWT - 5

Let other end of the diameter be  x1 , y1  

equation of the circle is  x  p   x  x1    y  q   y  y1   0

Since it touches x-axis, g 2  c 2

 P  x1   2   qy1  px1  

5.(B)



P 2  x12  2 x1 p  4qy1



 P  x1 2  4qy1

Since f 2  c and g 2  c , circle cuts both the axes. and

(0, 0) lies outside the circle.

Now since centre lies in fourth quadrant, circle can’t lie in second quadrant. 10.(D)

Let centres of the circle be  x1 , y1 

3 x1  4 y1  4

5



3 x1  4 y1  1  25



 

5

3 x1  4 y1  1  25 3 x1  4 y1  26  0 and 3 x1  4 y1  24  0 are parallel lines.

Circles

HWT - 6

2.(D)

Length of the side of an equilateral triangle is 3R where R is radius of the circle.

3.(A)

Equation of the circle is  x  2    y  3   3 2

2

2

If touches x-axis at (2, 0).

6.(B)

2   40,  20 9 9

AD  DB 

AB 1  1 2 2

OB = r

sin 20  7.(C)

1 r

Radius  g 2  r 2  c

VMC/Circles

8

HWT-Solutions/Mathematics

Vidyamandir Classes 8.(C)

Touches x-axis if g 2  c  25  a 2  9 If circle touches x = y 3 4

Line, then



9

2

a

1 2

Circles 2.(C)

a 2  ab  16  0 and a=b 

as coefficient of xy = 0 for a circle. as coefficient of x2 = coefficient of y2

2a 2  16





2 2 x y

3.(D)

2

4

a2 2

2 

radius = 2

Side = 2r

a 2

8.(B)

2

x2  y 2  2



HWT - 7



 

4  25  21  2 2



2 4

r 1  3 r 2

By geometry

2r  3  r



r=3



 x  1  r 2   y  0 2  r 2



x2  y 2  8x  7  0

Circles 1.(A)

HWT - 8

The equation of line AB is x y  1 a b  2ab 2 2a 2 b The reflection of origin in the above line is  2 ,  a  b2 a 2  b2 

2.(C)

   

Let the equation of circle be x 2  y 2  2 gx  2 fy  c  0 According to the question,

2g 0  2 f 0  c  a2 Also,



VMC/Circles

 a2 p  2  p  q2  a2 p  q2 2

2

  a2q   2   p  q2  

1





c  a2

2

   a2   p2  q2  a2





9

p

a2 p2 2

 q2



 p 2

a2q2 2

 q2



2

1

(p, q) lies on the circle.

HWT-Solutions/Mathematics

Vidyamandir Classes 3.(B)

The centre of circle is (1, 0) The image of (1, 0) in the line x  y  2 is (2, 1). The radius of circle which is 1 unit remains the same. 

5.(A)

2

2

Maximum area will be possible if circle touches internally as  3,  2  lies inside the circle x 2  y 2  4 x  y  0 

9.(A)

The equation of the circle is  x  2    y  1  12  1

r1  r2  C1C2

17 9 13  r  1  4 4 4



x 2  y 2  2 x  4 y  3   x  1   y  2   0 2

2



17  13 2

r

 x  y  12  x  y  3  0



Solving these two lines we find the coordinates of the centre of the circle as (1, 2). As circle passes through the point (1, 1) its radius is 1 unit.  10.(A)

equation of circle is  x  1   y  2   12  1 2

2

For an equilateral triangle the circumcentre coincides with the centroid.  (1, 1) is the circumcentre of the triangle. As  1, 2  is a vertex of the triangle the circumradius is 

equation of circumcircle is  x  1   y  1  2

2

 1  12   2  12

 5

2

5

Circles 1.(C)

 4 1  5

HWT - 9

Let r the inradius. Now,



r

 , where  is area and s is semi-perimeter of the triangle. s

1 3  2 1 r 2 2  3 5 2 2 2 2 2

1 1 incentre   ,  2 2 So, the equation of incircle is 

2

2

1  1  1 x 2  y 2  2       6.(A)

R cos 30 

R 3 a  2 2

Area of equilateral triangle 

 8.(C)

2



aR 3

3 2 a 4

3  3R 2 4

Centre must lie at perpendicular bisector of the line-segment joining (0, 0) and (4, 0) and the line-segment joining (0, 0) and (0, –6). The equation of the first perpendicular bisector is x = 2 and that of the second is y = –3. 

10.(D)

centre of required circle is (2, –3).

The corners of the square (6, 3), (6, 6), (9, 3) and (9, 6). The centre of the circle is same as the point of intersection of the diagonals of the square. 

VMC/Circles

 15 9  The required point is  ,   2 2

10

HWT-Solutions/Mathematics

Vidyamandir Classes

Circles 1.(A)

HWT - 10

 4  20  is a general point lying on the line 4 x  5 y  20  , 5    4  20  The chord of contact of this point is  x   y9 5   Now, let (h, k) be the mid-point of the chord of contact of the circle. 



hx  ky  h 2  k 2

The equation of chord of contact is :

. . . .(i)



. . . .(ii)

 4  20 9   2 h 5k h  k2

From (i) and (ii), we get :

Eliminating  from above equations and by replacing h  x and k  y , we get equation of locus. 2.(B)

OP 2  OA2  AP 2  16  4  12 

OP  12  2 3



Locus of P is :  h  0    k  0   2 3 2

or 4.(A)

2



x2  y 2  2 3

 , 4  2 



2



2

 12

is a general point lying on the line 2 x  y  4 .

Chord of contact of this point is :  x   4  2  y  1

5.(A)

 4 y  1    x  2 y   0

or

1 1 The above equation represents a family of lines passing through point of intersection of 4 y  1  0 and x  2 y  0 or  ,  . 2 4 The family of conics passing through points of intersection of given conics is represented by the equation.

 sin    cos   x   cos    sin   y   2h  2h   xy  32  16  x  16  32  y  19  19   0 . 2

2

2

2

2

2

2

As the points of intersection are con-cyclic it should be the equation of a circle. 

sin 2    cos 2   cos 2    sin 2 

. . . .(i)

and

2h  2h  0

. . . .(ii)

From (i) we have  = 1 Putting the value of  in (ii) we get :

8.(C)

PA 

 4 2   0 2  4  2

h  h  0

3

Area of PAOB = 2   area of PAO 

1   2    PA  AO  2  1   2  2 3  2  4 3 2  9.(BC)

The required points are the points of contact of the tangents drawn from the point (–2, 0) to the circle. So, find the equation of chord of contact of (–2, 0) with respect to the circle and solve it with the equation of the circle to get the coordinates of the required points.

10.(D)

Equation of common chord of the two given circles is 6 x  14 y  c  d  0 . According to the question the common chord passes through the centre of the circle x 2  y 2  2 x  8 y  d  0 which is (1, –4). 

VMC/Circles

6  1  14  4   c  d  0 or c + d = 50.

11

HWT-Solutions/Mathematics