07-circle-and-conics.pdf

Chapter 07: Circle and Conics

07 1. 

Circle and Conics

Since the triangle is equilateral. centroid of the triangle is same as the circumcentre

4.

According to the given condition, x1  x2  2a , x1 x2  b 2

2 and radius of the circumcircle = (median) 3

y1  y2  2p , y1 y2  q 2

2 (3a) = 2a 3 Hence, the equation of the circumcircle whose centre is at (0, 0) and radius 2a is x 2  y 2  (2a)2

The equation of the circle with A (x1, y1) and B (x2, y2) as the end points of diameter is (x  x1) (x  x2) + (y  y1) (y  y2) = 0  x2 + y2  x(x1 + x2)  y(y1 + y2) + x1x2 + y1y2 = 0  x2 + y2 + 2ax + 2py  b2  q2 = 0

=

 x2 + y2 = 4a2 2.

Let the other end be (t,3  t) .



the equation of the circle in diameter form is ( x  1)( x  t)  ( y  1)( y  3  t)  0

Let A  ( x1 , y1 ) and B  ( x2 , y2 ) .

5.

Let p be the altitude, then a p  a sin 60o  3. 2 A

 x  y  (1  t) x  (4  t) y  3  0 2



3.

2

We have, x 2  8 x  12  0  (x  6) (x  2) = 0  x = 2, 6 and y2  14y + 45 = 0  (y  5) (y  9) = 0  y = 5, 9 Y A

y=9

B(2, 5)

O



a B



y=5

x S

r x

O Gr R

D

C

Since, the triangle is equilateral, therefore centroid, orthocentre, circumcentre and incentre all coincide. radius of the inscribed circle p a   r 3 2 3 Let x be the side of the square inscribed, then angle in a semicircle being a right angle,

x 2  x 2  (2r) 2  4r 2 D(6, 9)

 2x2 

x=6

x=2



P

the centre (h, k) is given by 1 t 4t h ,k  2 2  2h + 2k = 5 Hence, the locus is 2 x  2 y  5 .

C

X

Since, centre of circle is inscribed in square. BD is the diameter of circle  26 59 centre = (h, k) =  ,  = (4, 7) 2   2

4a 2 a 2  12 3 a2 6



required area  x 2 

6.

Given, parabola y = x2 ....(i) Straight line y = 2x  4 ....(ii) From (i) and (ii), x2  2x + 4 = 0 Let f(x) = x2  2x + 4 f (x) = 2x  2



1

Std. XI : Triumph Maths

For least distance, f (x) = 0  2x  2 = 0 x=1 y = (1)2 = 1 ….[From (i)] So the point least distance from the line is (1, 1).



From (i), we get 16  y 2  25



2



For parabola, y = 8x  4a = 8  a = 2 vertex of y2 = 8x is O  (0, 0) Now, end points of latus rectum are L(a, 2a);L '(a, 2a)  L(2, 4);L '(2, 4) the circle passes through the points (0,0), (2,4) and (2,–4). All the three points are satisfied by the option (C). Option (C) is the correct answer.

8.

Eccentricity of

7.



Eccentricity

x2 y 2 a 2  b2  2  1 is e  2 a b a2 of conjugate hyperbola



9.



is

a b b2 The given equation of hyperbola can be written as x2 y 2  1 1 1 3 1 Here, a 2  1, b 2  3 1 1 3  4 2 e'  1 3 e' 

2

10.

2



11.

 

7,0 ,  7,0







2

7  0   0  3 = 2

79 = 4

The equation of the ellipse is

The foci of the ellipse are S





Eccentricity of the hyperbola = 2 4  a b2 = a2  1 = 3  3

x2  y 2  8x  7  0 ….(ii) Solving (i) and (ii), we get coordinates of A and B. Subtracting (i) from (ii), we get 8 x  32  0  x  4

2



3,0 and



1 2 = e 3

….(i)

The hyperbola passes through S ….(i)



S  3 ,0 .

B

x 2  y 2  25

x2 y 2   1. 4 1

Let e be its eccentricity. 3 1 Then, e = 1  = 4 2



C



Centre of the circle is (0, 3) radius of the circle is =

A

O

Here a2 = 16, b2 = 9 But b2 = a2(1  e2) 9 = 16(1  e2) 9 7   e2 = 1  16 16 7 e= 4 Now, foci are (ae, 0), (ae, 0) i.e.,

Let AB be the line of intersection of the two circles

L M

 y 2  9  y  3 Thus coordinates of A and B are (4, 3) and (4, –3). y 3 x 4   x40 equation of L is 33 4 4 Also coordinates of centre C of second circle is (4, 0). Hence, CM  Length of perpendicular from 44 0 C to the line L  1





3 ,0 .

3  0 = 1  a2 = 3 2 a Putting a2 = 3 in (i), we get b2 = 1 Hence, the equation of the hyperbola is x2 y 2   1 i.e., x2  3y2 = 3. 3 1

Chapter 07: Circle and Conics

12.

13. 



14.



Semi minor axis = b = 2 Semi major axis = a = 4 x2 y 2 Equation of ellipse is 2  2 = 1 a b 2 2 x y  =1  16 4  x2 + 4y2 = 16 Given equation of circle is x2 + y2  4x  8y  5 = 0 Centre = (2, 4) and radius = 4  16  5 = 5 the circle is intersecting the line 3x  4y = m at two distinct points. length of perpendicular from centre on the line < radius 6  16  m  5 5  |10 + m| < 25   25 < m + 10 < 25   35 < m < 15 Let P = (1, 0) , Q(1, 0) and A = (x, y) AP BP CP 1 Now, = = = 3 AQ BQ CQ AP 1   AQ 3  3AP = AQ  9AP2 = AQ2  9(x  1)2 + 9y2 = (x + 1)2 + y2  9x2  18x + 9 + 9y2 = x2 + 2x + 1 + y2  8x2  20x + 8y2 + 8 = 0 5  x2 + y2  x + 1 = 0 ....(i) 2 Since, points A, B and C lies on the circle Circumcentre of ABC = Centre of Circle (i) 5  =  ,0  4 

 



16.

xy+2=0



Y

15.

P(2,1) X

1

V

2

A

V A 2 (4,0) X

Y

Given equation of ellipse is x2 y 2 x2 + 4y2 = 4  = 1  a = 2, b = 1  4 1  P(2, 1)

Let the required equation of ellipse is x2 y 2  1 a 2 b2 Since, the ellipse passes through (4, 0). a=4 Also, it is passes through P(2, 1). 4 1  1 16 b 2 1 1  2 1 b 4 4  b2 = 3 x2 3 y 2 equation of ellipse becomes  1 16 4  x2 + 12y2 = 16

 

 

xy2=0

The lines x  y  2 = 0 and x  y + 2 = 0 are parallel, and tangent to the circle. Distance between them = diameter of the circle 2  (2) 4 = = =2 2 2 2 2 1 1 Let (h, k) be the centre of the circle. Since, x + y = 0 is the diameter. h+k=0  h = k ....(i) Now, perpendicular drawn from (h, k) to the x  y  2 = 0 is equal to radius. hk2  2 2 k  k  2  2 2

....[From (i)]

 2k + 2 = 2 k=0 h=0 ....[From (i)] required equation of circle is

(x  0)2 + (y  0)2 =

 2

2

 x2 + y2 = 2 3

Std. XI : Triumph Maths

17.

Centre of the given circle = C(2, 5) Radius of the circle CN = CT = =

2 2  52  7 =

Now, PC =

N

36 = 6

62  82 = 100 = 10

(2, 5) C

g2  f 2  c

T

P(4,3)



Given parabola is y2 = 2ax Focus (a/2, 0) and directrix is given by x = a/2 Since, circle touches the directrix. Radius of circle = distance from the point a Y (a/2, 0) to the line x  2 a a  X X 2 2 (a/2, 0) Radius   a (–a/2, 0) 1 Equation of circle is Y



a  2 2 ....(i) x   y a 2   Also, y2 = 2ax ....(ii) Solving (i) and (ii), we get a 3a x , 2 2 Putting these values in y2 = 2ax we get y   a and x = 3a/2 gives imaginary values of y. Required points are (a / 2,  a) .

20. 





2

We join the external point, (4, 3) to the centre of the circle (2, 5). Then PT is the minimum distance, from external point P to the circle and PN is the maximum distance. Minimum distance = PT = PC  CT = 10  6 =4 Maximum distance = PN = PC + CN = 10 + 6 = 16 So, sum of minimum and maximum distance = 16 + 4 = 20 18.

General equation of circle f(x, y) = (x  h)2 + (y  k)2  a2 = 0 f(0, ) = h2 + (  k)2  a2 = 0  2  2k + k2 + h2  a2 = 0 Equation has equal roots (1, 1) b  Sum of roots =   2 = 2k  k = 1 a Also, f(, 0) = (  h)2 + (0  k)2  a2 = 0 2  2h + h2 + k2  a2 = 0 1 and 2 Equation has roots  = 2 b 5 5 Sum of roots =   = 2h  h = 2 4 a

21.

P(t12 , 2t1 ) O X (0,0)

19.

Here, y = x2  4x + 3  y + 1 = x2  4x + 4  y + 1 = (x  2)2 So, the vertex is (2, 1). and for the circle, x2 + (y  3)2 = 9 Centre = (0, 3) Distance between vertex and centre =

4

22  (4) 2  2 5

90°

R(h,k) X

Q(t 22 , 2t 2 )

5  Centre (h, k) =  ,1 4 



Let the coordinates of P and Q are (t12 , 2t1 ),(t 22 ,2t 2 ) on parabola y2 = 4x.



Slope of OP =

2t1  0 2  t12  0 t1

Slope of OQ =

2t 2  0 2  t 22  0 t 2

Slope of OP  slope of OQ = 1 2 2 ….(i)   = 1  t1t2 = 4 t1 t 2 Let the coordinates of mid point of PQ are (h, k) t12  t 22  2h ….(ii) t1 + t2 = k ….(iii) 2 2 2 Now, (t1 + t2) = t1  t 2  2 t1t 2  k2 = 2h + 2(4) ….[From (i), (ii) and (iii)]  y2 = 2x  8, which is required locus.

Chapter 07: Circle and Conics

22.

Also, b2 = a2 (1  e2)

According to the given condition, ( x  2) 2  y 2  ( x  2) 2  y 2  8

2

.…(i)

 ae  2 2   = a (1  e )  3

Squaring on both sides, we get (x  2)2 + y2 = 64 + (x + 2)2 + y2

 e2 = 3  3e2

 2  8 ( x  2) 2  y 2  4x = 64 + 4x  16 ( x  2)  y 2

25.

Put y = 3 in (i), we get x 8 = 2 ( x  2) 2  9 Squaring on both sides, we get x2 + 64 + 16x = 4(x + 2)2 + 36  3x2 = 12  x2 = 4  x = 2 23.



Equation of auxiliary circle is x2 + y2 = 9 ….(i) x y Equation of AB i.e., AM is   1 ….(ii) 3 1  12 9  Solving (i) and (ii), we get M   ,   5 5 Y

 12 9  M  ,   5 5

X

N

24. 

In the given figure, S is focus whose coordinates are (ae, 0). ABS is an equilateral triangle.



Area of ABS =

1 3  AB  OS  (side)2 2 4 3 1 (2b) 2   2b  ae = 4 2 Y A(0, b) X

O

27.

Locus of the point P, if A and B are fixed and

 28.



 , is a circle with diameter AB. 2

But, we have PA2 + PB2 = constant. Locus of the point P is a circle.

y = 7x – 25 ....(i) 2 2 and x + y = 25 x2 + (7x  25)2 = 25  x2 + 49x2 + 625 – 350x = 25 50x2  350x + 600 = 0  x2  7x +12 = 0  x = 3, 4 Substituting x = 3,4 in (i), we get y = 21 – 25  y = 4, y = 28 – 25  y = 3 Let A  (3, 4), B  (4, 3) Using distance formula, we get AB =

X

S

b2 3  2 4b 2



APB 

1  OA  MN 2 1 9 27 =  3 = sq. unit. 2 5 10

b2 a2

Locus of the point P, if A and B are fixed and PA + PB = constant, is an ellipse. We have, PA + PB = 4, which is a constant. Locus of the point P is an ellipse.

Y

Now, area of AOM =

x2 y 2  1 a 2 b2

is x2 + y2 = a2 Area of this circle = a2 a2 = 2  ab  a = 2b

 1

B(0,1)

X A(3,0)

The auxiliary circle of the ellipse

Eccentricity of ellipse = 1 

26.

O

3 2

e=

2

 x 8 = 2 ( x  2) 2  y 2

....[From (i)]

(3  4) 2  (4  3) 2

 1  49  50  5 2

B(0,b)

 ae =

3b

Y

….(i) 5