MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

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MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

1. Find the equation of the circle with centre (-7, -3) and radius ‘4’.
Sol:Given Centre, C (h, k) = ( -7, -3),
radius, r = 4
 The equation of the circle is
(x - h)2 + (y - k)2 = r2
 (x + 7)2 + (y + 3)2 = 42
 x2 + 49 + 14x + y2 + 9 + 6y = 16
2 2
 x + y + 14x + 6y + 42 = 0

(H/W)Find the equation of the circle with centre C = (-1, -8) and radius r = 5.
Ans: x2 + y2 + 2x + 16y + 40 = 0.

2. Find the equation of the circle passing through (3, 4) and having the centre at (-3, 4).
Sol:Given centre C (h, k) = (-3, 4)
Let the given point A = (3, 4)
Since ‘A’ is the point on the circle.
 Radius, r = CA

 3  3    4  4
2 2


 6   0
2 2


 36  0  36  6
 The equation of the required circle is
 x  h   y  k   r2
2 2

 (x + 3)2 + (y - 4)2 = 62
 x2 + 9 + 6x + y2 + 16 - 8y = 36
 x2 + y2 + 6x - 8y - 11 = 0.

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MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

3. Find the equation of the circle passing through the origin and has centre at (-4, -3).
Sol:Given centre C = (-4, -3), origin O = (0, 0)
Radius = distance between (0, 0) and (-4, -3)
= 16+9 = 2 5 =5.
Equation of the circle which has centre at (-4, -3) and radius 5 is
(x - x1)2 + (y - y1)2 = r2
2 2 2
 (x + 4) + (y + 3) = 5

 x2 + 8x + 16 + y2 + 6y + 9 = 25
 x2 + y2 + 8x + 6y = 0.

4. Find the equation of the circle passing through (2, - 1) having the centre at (2, 3).
Sol:Given C = (2, 3), P = (2, -1)
radius = r = CP =  2  2  3  1  4
2 2

Equation of the required circle with centre

C =(2, 3) = (x1 , y1) and the radius (r) = 4
Equation of circle is (x - x1)2 + (y - y1)2 = r2
2 2 2
 (x - 2) + ( y - 3) = 4
2 2
 x + 4 - 4x + y + 9 - 6y = 16
2 2
x + y - 4x - 6y - 3 = 0

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
5. Find the equation of the circle passing through (-2,3) having the centre at (0,0).
Sol:Given centre C(h,k) = (0,0)
Let the given point A (- 2,3)
Since A(- 2,3) is a point on the circle.
Radius r = CA = (  2  0)2  (3  0)2

 (  2)2  (3 )2  4  9  13
 The equation of the required circle is
(x - h)2 + (y - k)2 = r2.
 (x - 0)2 + (y - 0)2 = (13)2  x2 + y2 = 13.

(H/W)Find the equation of the circle whose centre is (-1,2) and which passes through (5,6).
Ans: x2 + y2 + 2x - 4y - 47 = 0.
6. Find the values of a,b if ax2 + bxy + 3y2- 5x + 2y- 3 = 0 represents a circle. Also find the radius and centre
of the circle.
Sol:ax2 + bxy + 3y2 - 5x + 2y - 3 = 0
represents a circle
 Coeff. x2 = Coeff.y2 and Coeff.of xy = 0
a = 3 and b = 0
Equation of the circle is
 3x2 + 3y2 - 5x + 2y - 3 = 0 3
5 2
 x2 + y2 - 3 x + 3 y -1 = 0
5 1
centre C =  6 , 3  , Radius= g2  f 2  c
 
2 2
 5   2  25 4 25  4  36 65
=  6    6    1  
36 36
1   .
    36 6

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
7. Find the value of ‘a’of 2x2 + ay2 - 3x + 2y - 1 = 0 represents a circle and also find its radius.
Sol:2x2 + ay2 - 3x + 2y - 1 = 0 represents a circle
 Coe. of x2 = coe. of y2  a = 2
Equation of the circle is
2x + 2y2 - 3x + 2y - 1 = 0
2
 2

3 1
 x2 + y2 - x+y- 2= 0
2
3 1 1
2g = 2
 g = -3/4, 2f = 1  f = 2
, c= 2
2 2
 3   1  1 9 1 1 948 21
Radius = g  f c
2 2
=  4    2   2  =     .
      16 4 2 16 4

8. If x2 + y2 + 2gx + 2fy - 12 = 0 represents a circle with centre (2, 3) find g, f and its radius.
Sol:Centre of the circle
x2 + y2 + 2gx + 2fy - 12 = 0 is (-g, -f) = (2, 3)
 g = -2, f = -3 and c = -12
Radius = g2  f 2  c =   2 2    3 2    12   4  9  12  5 .

9. If the centre of the circle x2 + y2 + ax + by - 12 = 0 is (2, 3), find the values of a, b and the radius of the circle.
Sol: Comparing x2 + y2 + ax + by - 12 = 0 with
x2 + y2 + 2gx + 2fy + c =0,
a b
2g = a  g = 2
, 2f = b  f = 2
, c = -12.
 -a -b 
Now centre (-g, -f) =  2, 2 = (2, 3)  a = - 4, b = - 6.
 
Radius = g2 + f 2 - c = 4 + 9 + 12 = 5.

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MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

10. If x2 + y2 - 4x + 6y + c = 0 represent a circle with radius 6 then find the value of c.

Sol:Given equation of circle x2 + y2 - 4x + 6y + c = 0
center c =(-g, -f)= (2, -3) , Radius = 6
Radius = g2  f 2  c =  2 2    3 2  c = 6
 4 + 9 -c = 36  c = 13 - 36  c = - 23.

(H/W)If the cirlcle x2 + y2 - 4x + 6y + a = 0 has radius 4 then find ‘a’. ( Ans: a = - 3)

11. Find the centre and radius of circle x2 + y2 + 6x + 8y - 96 = 0
sol: Given circle is x2 + y2 + 6x + 8y - 96 = 0
centre = (-3, -4)
Radius = g2 + f 2 - c = 3 2 + 4 2 + 96 = 9 +16 + 96 = 121  11 .
12. Find the centre and radius of the circle 1+ m
2
(x2 + y2) - 2cx - 2mcy = 0 (c > 0).
Sol: Given equation of the circle is
1+ m
2 (x2 + y2) - 2cx - 2mcy = 0  1+ m
2

2c 2mc
x2 + y2 - x- y=0
1 + m2 1 + m2
 c mc 
Coordinates of centre = (-g, -f) =  , 
 1 + m2 1 + m2 

c2 m2c 2 c 2 (1 + m2 )
Radius = 2
g +f -c
2 = + = = c.
1 + m2 1 + m2 1 + m2

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

13. If one end of the diameter of the circle x2 + y2 - 2x + 4y = 0 is (3, -1), then find the other end of the
diameter.
Sol:Centre of the circle x2 + y2 - 2x + 4y = 0 is C(1, -2)
Let B(a, b) be the other end of the diameter through A(3, -1).
 x  x 2 y1  y 2   3 + a -1+ b 
Here C is the midpoint of  1 ,  2 , 2  = (1, -2)
AB  2 2   
3+a -1+ b
2
=1 2
= -2
3+a=2 -1 + b = -4
a = -1 b = -3.
Required point B is (-1, -3).
14. Show that A (- 3, 0) lies on x2 + y2 + 8x + 12y + 15 = 0 and find the other end of diameter
through A.
Sol :Given equation of circle
x2 + y2 + 8x + 12y + 15 = 0 ............... (1)
given A(-3,0)
verify A on (1)
( - 3)2 + 02 + 8(- 3) + 12 (0) + 15 = 0
 9 + 0 - 24 + 0 + 15 = 0  24 - 24 = 0  A lies on (1)
Centre of circle (- 4, - 6) = C
Let B (a,b) be other end of diameter through A
 x 1  x 2 y1  y 2  3 a 0 b
 C = mid point of AB=(- 4, - 6)   ,  = , 
 2 2   2 2 

a3
4  a-3=-8  a=-8+3=-5
2
b
 = - 6  b = - 12  B( - 5, - 12).
2

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15. Obtain the parametric equations of the circle. x2 + y2 = 4.

Sol:Given circle is x2 + y2 = 4(Standard form)
Centre = (-g1 - f) = (0, 0) radius = r = 2
Parametric equations of the circle
x = - g + r cos  , y = - f + r sin 
x = 0 + 2 cos  , y = 0 + 2 sin 
x = 2 cos  , y = 2 sin  (0 <  < 2  )

(H/W)Obtain the parametric equations of the circle 4(x2 + y2) = 9.

3 3
Ans: x = cos  , y = sin  ,0 <  < 2 .
2 2

16. Obtain the parametric equations of the circle (x - 3)2 + (y - 4)2 = 82.
Sol:Given circle is (x - 3)2 + (y - 4)2 = 82 ( Central form)
Centre (h,k) = (3, 4), r = 8
Parametric equations is
x = h + r cos  , y = k + r sin  (0 <  < 2  )
 x = 3 + 8 cos  , y = 4 + 8 sin  (0 <  < 2  )

17. Obtain the parametric equations of the circle x 2 + y2 - 6x + 4y - 12 = 0.

Sol:For the circle x2 + y2 - 6x + 4y - 12 = 0(genral equation)
Centre (-g, -f) = (3, -2)
Radius = g2 + f 2 - c = (-3)2 + 22 + 12 = 5 = r
Parametric equations of the circle are
x = -g + r cos  , y = -f + r sin  . 0 <  < 2 
x = 3 + 5 cos  , y = -2 + 5 sin  0 <  < 2  .

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

18. Find equation of the circle which is concentric with x2 + y2 - 6x - 4y - 12 = 0

and passing through (-2, 14).
Sol:Equation of a circle concentric with
x2 + y2 - 6x - 4y -12 = 0. is in the form of

x2 + y2 - 6x - 4y +k =0

Since, it is passing through (-2, 14).

 4 + 196 + 12 - 56 + k = 0.
 156 + k = 0  k = -156
Required equation of a circle
x2 + y2 -6x - 4y - 156 = 0.
(H/W)Find the equation of the circle passing through (2, 3) and concentric with the circle x2 + y2 + 8x + 12y
+ 15 = 0. Ans :x2 + y2 + 8x + 12y - 65 = 0.

19. Find the equation of the circle having line joining (-4, 3) (3, - 4) as a diameter.
Sol:Equation of the circle with A (- 4, 3), B (3, - 4) as ends of diameter is

(x - x1) (x - x2) + (y - y1) (y- y2) = 0

 (x+4) (x - 3) + (y - 3) (y + 4) = 0

 x2 - 3x + 4x - 12+y2 + 4y - 3y - 12 = 0

 x2 + y2 + x + y - 24 = 0.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

20. Find the power of the point P(2, 3) with respect to the circle x2 + y2 - 2x + 8y -23 = 0.
Sol:Given P(2,3) and S  x2 + y2 - 2x + 8y - 23 = 0.(power=PA.PB)
S11 = (2)2 + (3)2 - 2(2) + 8(3) - 23 = 4 + 9 - 4 + 24 - 23 = 10
Power of P w.r.t.S = 0 is 10.
(H/W)Find the power of the point (2, 4) with respect to x2 + y2 - 4x - 6y - 12 = 0Ans: - 24
7. Find the position of the point (3, 2) with respect to the circle x2 + y2 - 4x - 6y - 12 = 0.
Sol: S = x2 + y2 - 4x - 6y - 12 = 0, P(3, 2) = (x1, y1) S  0, point lies on the circle.
11
S11= 32 + 22 - 4(3) - 6(2) - 12 S11  0, point lies inside the circle.
= 9 + 4 - 12 - 12 - 12 = -23. S11  0, point lies outside the circle
p lies inside the circle
21. Find the length of the tangent from (1,3) to the circle x2 + y2 - 2x + 4y - 11 = 0.
Sol:Here (x1, y1) = (1, 3) and
S = x2 + y2 - 2x + 4y - 11 = 0
The length of the tangent is S11

1   3 21  4 3 11  1  9 21211  9  3 .

2 2
=

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22. If the length of the tangent from (2, 5) to the circle x2 + y2 - 5x + 4y + k = 0 is 37 , then find k.
Sol:x2 + y2 - 5x + 4y + k = 0, P(2, 5)
Given S11 = 37
 S11 = 37
2 2
 2 + 5 - 5(2) + 4(5) + k = 37
 k = 37 - 39
2 2
 k = -2.(H/W)If the L.T from (5, 4) to the circle x + y + 2ky = 0 is 1 then find k. ( Ans: - 5)
23. Find the length of the chord formed on thecircle x2 + y2 = a2 on the line x cosa + ysina = p.
Sol:Given equation of circle x2 + y2 = a2 ........... (1)
Centre = C (0,0), radius (r) = a
given equation of chord x cos a + y sin a - p = 0 .... (2)
d = perpendicular distance from C (0,0) to (2)

ax1  by1  c

a2  b2

0  0 p p
  p
2
cos   sin  2 1

Length of the chord = 2 r2  d2  2 a  p units

2 2
.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

24. Find the equation of tangent and normal of x2 + y2 - 6x + 4y - 12 = 0 at (-1, 1).

Sol:Eqation of tangent to x 2 + y2 - 6x + 4y - 12 = 0 at (-1, 1) is
xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0.
 x(-1) + y(1) - 3(x - 1) + 2(y + 1) - 12 = 0
 -4x + 3y - 7 = 0
 4x - 3y + 7 = 0.
Equation of normal at (-1, 1) is in the form
b(x- x1)-a(y-y1)=0
 -3(x+1)-4(y-1) = 0.
 -3x-3-4y+4=0
Equation of normal at (-1, 1) is 3x + 4y - 1 = 0.

25. Find the equation of the normal at P(3,5) of the circle S = x2 + y2 - 10x - 2y + 6 = 0.
Sol:Given circle is x2 + y2 - 10x - 2y + 6 = 0
centre C = (- g, - f) = (5,1)
Given point P = (3,5)
equation of the normal passing through P is equation of CP.
1 5
 y-5= 53
(x - 3)
4
 y-5= (x - 3)
2
 y - 5 = - 2 (x - 3)
 y - 5 = - 2x + 6  2x + y - 11 = 0.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
26. Find the area of the triangle formed with the coordinate axes and the tangent drawn at the point (x1, y1)
on the circle x2 + y2 = a2.
So l: Equation of tangent to x2 + y2 = a2 at (x1, y1) is
xx1 + yy1 = a2
x y
 2
+ 2 = 1.
a / x1 a / y1
Area of the triangle formed by this tangent with coordinate axes.

1
= 2 |(x - intercept) (y - intercept)|

1  a2  a2  a4
=     =
2  x1  y1  2 | x1y1 | square units.
s.

27.Find the area of the triangle formed by the normal at (3, - 4) to the circle x2 + y2 - 22x - 4y + 25 = 0
with the coordinate axis.
Sol: Given circle is x2 + y2 - 22x - 4y + 25 = 0
centre ( - g, - f) = C(11,2)
Equation of normal passing through (3, - 4) (11,2) is
24
y+4= 11  3
(x - 3)

6
y+4= (x - 3)
8
 4y + 16 = 3x - 9
 3x - 4y - 25 = 0
Area of the triangle formed by the normal with coordinate axes is
c2 (  25)2 625
  sq.units .
2 | ab | 2 | 3(  4) | 24

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
28. State the necessary and sufficient condition for lx+ my + n =0 to be a normal to the circle x2 + y2 + 2gx
+ 2fy + c = 0.
Sol:The straight line lx + my + n = 0 is normal to the circle S º x2 + y2 + 2gx + 2fy + c = 0
the centre (-g, -f) of the circle lies
on lx + my + n = 0

 l(-g) + m (-f) + n = 0
 lg + mf = n.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

29. Find the polar of (1,2) w.r.t to x2 + y2 = 7.

Sol: Given circle S  x2 + y2 - 7 = 0.
Given point is P(1,2)
Equation of polar of P with respect to S = 0 is
S1 = 0  x(1) + y(2) - 7 = 0
 x + 2y - 7 = 0.

(H/W)Find the polar of (3, - 1) with respect to 2x2 + 2y2 = 11. (Ans:6x - 2y - 11 = 0.)

30. Find the polar of (1, -2) with respect to x2 + y2 - 10x - 10y + 25 = 0.
Sol.Given x2 + y2 - 10x - 10y + 25 = 0......... (1)
point = (1, -2)
Equation of polar of P(1,-2) w.r.t. (1) is S1 = 0
xx1 + yy1 - 5(x + x1) - 5(y + y1) + 25
x(1) + y (-2) -5 (x +1) - 5 (y-2) + 25 = 0
x - 2y -5x - 5 - 5y + 10 + 25 = 0
- 4x - 7y + 30 = 0
4x + 7y - 30 = 0.

(H/W)Find the equation of the polar of (2,3) with respect to the circle x2 + y2 + 6x + 8y -96 = 0.
Ans: 5x + 7y - 78 = 0

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
31. Find the angle between the tangents drawn from (3, 2) to the circle x2 + y2 - 6x + 4y - 2 = 0.
Sol:Given circle x2 + y2 - 6x + 4y - 2 = 0
center C = (3, -2)
r =  32   22  2  942  15

3 22 6 3 4 2 2  941882 = 1

2
S11 

 r 15
ta n  
2 S11 1

1  tan2 
cos   2

1  tan2 
2

1  15 14 7 7
 cos  =     = cos-1  .
1  15 16 8 8
32. Find the pole of ax + by + c = 0 with respect to x2 + y2 = r2.
Sol:Given equation of circle
S  x2 + y2 - r2 = 0 .......................... (1)
Given equation of line ax + by + c = 0 ........... (2)
Let P(x1,y1) be the pole.
Equation of polar P with respect to S = 0 is S1 = 0
xx1 + yy1 - r2 = 0 .......................... (3)
(2), (3) represents same line
a1 b c x1 y1  r 2
 1  1 =  
a2 b2 c2 a b c
x1  r 2 ar 2 y1  r 2 br 2
   x1   and   y1  
a c c b c c
 ar 2 br 2 
Coordinates of pole =   ,  

 c c .

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
34. Find the value of k if the points (1,3) and (2, k) are conjugate with respect to the circle x2 +
y2 = 35.
Sol:Here (x1 , y1) = (1, 3), (x2, y2) = (2, k) and
x2 + y2 - 35 = 0 since the given points are conjugate we have S12 = 0.
x1 x2 + y1 y2 - 35 = 0
 (1) (2) + 3 (k) - 35 = 0
 2 + 3k - 35 = 0
 3k = 33  k = 11.
35. Show that (4, -2) and (3, -6) are conjugate with respect to the circle x2 + y2 - 24 = 0.
Sol: Given (x1, y1) = (4, - 2), (x2 , y2) = (3, -6)
s= x2 + y2 - 24 = 0
For conjugate points, S12 = 0  x1 x2 + y1 y2 - 24 = 0
S12 = 4(3) + (-2) (-6) - 24 = 0
The given points are conjugate with respect to the given circle.
(H/W)Show that the points (- 6,1) and (2,3) are conjugate points with respect to the circle
x2 + y2 - 2x + 2y + 1 = 0. Ans: Verify S12 = 0.
36. Find the value of k, if the points (4,2) and (k, - 3) are conjugate points with respect to the circle x2 +
y2 - 5x + 8y + 6 = 0.
Sol:Given equation of circle
S  x2 + y2 - 5x + 8y +6 = 0 .............. (1)
given P(4,2) Q (k,- 3) are conjugate points with respect to the circle S = 0
5
 S12 = 0  x1x2 + y1y2 - (x1 + x2) + 4(y1 + y2) + 6 = 0
2
5
 4(k) + 2( - 3) - (4 + k) + 4 (2 - 3) + 6 = 0
2
5 5 5
 4k - 6 - (4) - k - 4 + 6 = 0  4k - k - 14 = 0  8k - 5k - 28 = 0
2 2 2
28
 3k = 28  k = .
3

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

37. Find the chord of contact of (1,2) w.r.t to x2 + y2 = 7.

Sol: Given circle S  x2 + y2 - 7 = 0.
Given point is P(1,2)
Equation of chord of contact of P with respect to S = 0 is
S1 = 0  x(1) + y(2) - 7 = 0
 x + 2y - 7 = 0.
38. Find the number of common tangents that exist for the pair of circles x2 + y2 = 4,
x2 + y2 - 6x - 8y + 16 = 0.
Sol:Given equations of the circles are
x2 + y2 = 4 .................. (1), x2 + y2 - 6x - 8y - 16 = 0 ............ (2)
For the circle (1), centre, C1 = (0,0), Radius r1 = 2
For the circle (2), centre, C2 = (3,4)
Radius r2 = ( 3)2  (  4)2  16 = 9  16  16 = 9 = 3.
| C1C2 |  = r1 + r2
3 2  4 2  5 | C1C2 |
The given circles touch each other externally, No. of common tangents = 3.
39. Find the number of common tangents that exist for the pair of circles x2 + y2 +6x+6y+14=0,
x2 + y2 - 2x- 4y - 4 = 0.
Sol:Given equations of the circles are
x2 + y2 + 6x+6y+14= 0 .................. (1), x2 + y2 - 2x -4y - 4 = 0 ............ (2)
For the circle (1), centre, C1 = (-3, -3), Radius r1 = ( 3)2  (  3)2  14 = 9  9  14 = 4 2
For the circle (2), centre, C2 = (1, 2)
Radius r2 = (1) 2  ( 2 ) 2  4 = = 9 = 3.
| C1C2 |  ( 3  1)2  ( 3  2)2  16  25  41 r1 + r2 | C1C2 |
Thus one circle lies outside the other, Hence No. of common tangents =4.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

40.Find the internal centre and external centres of similitude for the circles
x2 + y2 +6x+6y+14=0, x2 + y2 - 2x- 4y - 4 = 0.
ANS:
41.Find the equation of the circle with centre (2, 3) and touching the line 3x - 4y + 1 = 0.
Sol: Radius = perpendicular distance from (2, 3) to the line 3x - 4y + 1 = 0
3(2)  4 3 1 6121
r=  =1
5 5
Equation of the required circle
(x - 2)2 + (y - 3)2 = 12  x2 + 4 - 4x + y2 + 9 - 6y = 1  x2+ y2 - 4x - 6y + 12 = 0.
42.Find the equation of the circle with centre (-3,4) and touching y - axis.
Sol: Centre = (h, k) = (-3, 4), Radius = |h| = | - 3| = 3
Equation of the required circle is
(x +3)2 + (y - 4)2 = 32  x2 + 9 + 6x + y2 + 16 - 8y = 9  x2 + y2 + 6x - 8y + 16 = 0.

(H/W)Find the equation of the circle with centre (-3,4) and touching X - axis.

43. Find the equation of the tangent at the point 300 (Parametric value of  ) of the circlee
x2 + y2 + 4x + 6y - 39 = 0.
Sol: Given circle x2 + y2 + 4x + 6y - 39 = 0
Here g = 2, f = 3, r  4  9  39  52  2 13
The required equation of tangent at P(  ) is (x + g) cos  + (y + f) sin  = r
(x + 2) cos 300 + (y + 3) sin 300 = 2 13
 3  1
 (x + 2)  2  + (y + 3)  2  = 2 13
   

 3 x  2 3  y  3  4 13  3 x  y  3  2( 3  2 13)  0 .

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
1. If a point P is moving such that the lengths of tangents drawn from P to the circles x2 + y2 - 4x - 6y - 12 = 0
and x2 + y2 + 6x + 18y + 26 = 0 are in the ratio 2 : 3, then find the equation of the locus of P.
Sol: Given equations of circles are
S = x2 + y2 - 4x - 6y - 12 = 0, S = x2 + y2 + 6x + 18y + 26 = 0
Let P(x1, y1) be any point on the locus.
S11 2
Given 
S11 : S11 =2:3  
S11
= 3

 3 S11 = 2 S11
 squaring on both sides
es
 )
 9(S11) = 4( S11

 9(x12+y12-4x1-6y1-12) = 4(x12 + y12 + 6x1 + 18y1+26)

 9x12 + 9y12 - 36x1-54y1-108 = 4x12 + 4y12 + 24x1 + 72y1 + 104
 5x12 + 5y12 - 60x1 - 126y1 - 212 = 0
Hence the required equation of locus of P(x1, y1) is 5x2 + 5y2 - 60x - 126y - 212 = 0.
2. If a point P is moving such that the lengths of tangents drawn from ‘P’ to the circles
x2 + y2 + 8x + 12y + 15 = 0 and x2 + y2 - 4x - 6y - 12 = 0 are equal, then find the equation of the
locus of P.
Sol:Given equations of the circles are S  x2 + y2 + 8x + 12y + 15 = 0, S  x2 + y2 - 4x - 6y - 12 = 0
Let P(x1, y1) be any point on the locus.
PA, PB be the lengths of the tangent from P to the circles (1) & (2) respectively.
Given condition is S11  S11  S11  S11
 x12 + y12 + 8x1 + 12y1 + 15= x12 + y12 - 4x1 - 6y1 - 12
 8x1 + 12y1 + 15 + 4x1 + 6y1 + 12 = 0.
 12x1 + 18y1 + 27 = 0.
The equation of the locus of ‘P’ is 12x + 18y + 27 = 0  4x + 6y + 9 = 0.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
3. Find the length of the chord intercepted by the circle x2 + y2 - 8x - 2y - 8 = 0 on the line
x + y + 1 = 0.
Sol: Given circle x2 + y2 - 8x - 2y - 8 = 0 ------ (1)
Centre (C) = (4, 1)
Radius (r) = 42 + 12 + 8 = 16 + 1 + 8 = 25 = 5
Given chord, x + y + 1 = 0 --------- (2)
d = perpendicular distance from C to (2)
| ax1  by1  c | 4 + 1+ 1 6
2
a b 2 = 2 2 = 2
1 +1
2
 6  36 50  36 14
Length of chord = | AB | = 2 2
r -d 2 = 2 25 -   = 2 25 -
2
= 2
2
= 2
2
= 2 7 units.
s.
 2
4. Find the length of the chord intercepted by the circle x2 + y2 - x + 3y - 22 = 0 on the line y = x - 3.
Sol:Given equation of the circle is
x2 + y2 - x + 3y - 22 = 0
 1 3 
Centre C = (- g, - f) =  2, 2  .
 
2 2
  1 3 49
Radius r = =  2    2   22 
2 2
g f c
    2

 1 3 
Now, d = The perpendicular distance from the centre C  2 ,  to the line x - y - 3 = 0
 2 

 1   3 
1   1 3 1
| ax1  by1  c | 2  2 
d=  =
a2  b2 (1)2  ( 1)2 2

2 2
 49   1  49 1 48
Length of the chord = 2
2 r d 2
= 2     2  = 2  2 24  4 6 units.
s.
 2   2 2 2 2

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MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

5. The line y = mx + c and the circle x2 + y2 = a2 intersect at A and B. If AB = 2l,

then show that c2 = (1 + m2) (a2 -  2)
Sol:Given equation of circle x2 + y2 = a2 .............. (1)
Centre C (0,0) radius, r = a
given equation of line y = mx + c, mx - y + c = 0 ............(2) and (1), (2) intersect at A and B.
| ax1  by1  c | 00c c
d = perpendicular distance from centre C to (2). = 2 2
  .
a b m 1
2
m2  1

Given |AB| = 2   2 r 2  d  2  r 2  d2  
squaring on both sides  r2 - d2 = 
2

c2 c2

2
a -  2 
2
a - 2
= (a2 -  2) (1 + m2) = c2.
1  m2  1  m2 
6. Find the equation of the circle with centre (-2, 3) cutting a chord of length 2 units on 3x + 4y + 4 = 0.
Sol: Given centre C(-2, 3)
Given equation of chord, 3x + 4y + 4 = 0 ------------- (1)
d = perpendicular distance from C (-2, 3) to (1)
| ax1  by1  c | 3(-2) + 4(3) + 4 -6 +12 + 4 10 10
= a2  b 2
= 2 2 = 9 + 16
= 25
= 5
=2
3 +4

Given length of chord = 2  2 r 2 - d2 = 2  r 2 - d2 = 1

Squaring on both sides  r2 - d2 = 1
substituting d = 2  r2 - 22 = 1
r2 = 1 + 4 = 5
Required equation of the circle is (x - h)2 + (y - k)2 = r2
(x + 2)2 + (y - 3)2 = 5  x2 + 4x + 4 + y2 - 6y + 9 - 5 = 0
x2 + y2 + 4x - 6y + 8 = 0.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

7. Find the pole of 3x + 4y - 45 = 0 with respect to x 2 + y2 - 6x - 8y + 5 = 0.

Sol: Given circle x2 + y2 - 6x - 8y + 5 = 0
2g = -6  g = -3, 2f = -8  f = -4
r = (-3)2 + (-4)2 - 5 = 9 + 16 - 5 = 20
Given line 3x + 4y - 45 = 0
l = 3, m = 4, n = -45 and lg+mf-n= 3(-3) + 4(-4) + 45 = -9 - 16 + 45=20
 r 2 mr 2   3 (2 0 ) 4 (2 0 ) 
pole =  -g + ,-f +    3 + , 4 + 
  g + mf - n g + mf - n   2 0 2 0 
= (3 + 3, 4 + 4) = (6, 8).
(H/W)Find the pole of x + y + 2 = 0 with respect to the circle x2 + y2 - 4x + 6y - 12 = 0.Ans: (- 23, - 28).
8. F i n d t he va lue o f k , i f k x + 3 y - 1 = 0 , 2x + y + 5 = 0 are conjugate lines with respect to
the circle x 2 + y2 - 2x - 4y - 4= 0.
Sol: Given equation of circle
x2 + y2 - 2x - 4y - 4 = 0 ------------- (1) 2g = -2  g = -1, 2f = -4  f = -2 and c = -4
r = (-1) + (-2) - (-4) = 1 + 4 + 4 = 9 = 3
2 2

given lines kx + 3y - 1 = 0, 2x + y + 5 = 0
l1 = k, m1 = 3, n1 =-1, l2 = 2, m2 = 1, n2 = 5
Conjugate lines condition : r2 (l1 l2 + m1m2) = (l1g + m1f - n1) (l2g + m2f - n2)
9(k.2 + 3.1) = [k(-1) + 3(-2) + 1] [2(-1) + 1(-2) - 5) 9  (2k + 3) = (-k - 5) ( - 9 )
2k + 3 = k + 5
2k - k = 5 - 3
k = 2.
(H/W)F i n d t he va lue o f k , i f x + y - 5 = 0 , 2x + ky - 8 = 0 are conjugate lines with respect to
the circle x 2 + y2 - 2x - 2y - 1 = 0.Ans: k = 2
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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
9. Find the equations of the tangents to the circle x 2 + y2 - 4x + 6y - 12 = 0
which are parallel to x + y - 8 = 0.
Sol: Given equation of circle x2 + y2 - 4x + 6y - 12 = 0 ---------- (1)
Centre (2, -3), radius = (2)2 + (-3)2 - (-12) = 4 + 9 + 12 = 25 = 5.
Given line x + y - 8 = 0 ---------- (2)
Equation of a line parallel to (2) is,
x + y + k = 0 ---------- (3)
Given (3) is tangent to (1)
Condition : r = d [perpendicular distance from C to (3)]
| ax1  by1  c | 2 + (-3) + k -1 + k
 5= a b2 2 = 2 2  5= 2
1 +1
 5 2= |k - 1|  k - 1 = + 5 2
k=1+ 5 2
Hence required equations of tangents are x+y+1+ 5 2 = 0.
10. Find the equations of the tangents to the circle x2 + y2 + 2x - 2y - 3 = 0
which are perpendicular to 3x - y + 4 = 0.
Sol: Let the equation of the tangent perpendicular to 3x - y + 4 = 0 is x + 3y + k = 0, ________ (1)
centre of x2 + y2 + 2x - 2y - 3 = 0 is (-1, 1)
Radius = (1)2  (1)2  3  5
Given (1) is tangent of a given cirlce
Now r = d
Radius = length of the perpendicular from (-1, 1) to x + 3y + k = 0
| ax1  by1  c | | ( 1)  3(1)  k |
 r=  5  5 2 | k  2 |  k = -2 + 5 2.
a 2  b2 10

Required tangents are x + 3y - 2 + 5 2 = 0.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
11. x2 + y2 - 2x + 4y = 0 at (3, - 1). Also find the equation of tangent parallel to it.
Sol:Given equation of the circle is x2 + y2 - 2x + 4y = 0 ......... (1)
The given point P(x1, y1) = (3, - 1)
 The equation of tangent at P is S1 = 0.
x(3) + y (- 1) - 1 (x + 3) + 2 (y - 1) = 0
 3x - y - x - 3 + 2y - 2 = 0  2x + y - 5 = 0
The equation of the straight line parallel to the tangent 2x + y - 5 = 0 is 2x + y + k = 0 ......... (2)
If (2) is a tangent to the given circle
where r = d (perpendicular distance from centre (1, - 2) to the line (2)).
| ax1  by1  c | | 2(1)  1(  2)  k | |k |
(  1)2  22  5
 r= 2 2  
a b 2 1
2 2
5
 |k| = 5  k = + 5
The equations of the tangents to the circle are 2x + y + 5 = 0
One of these tangnets namely 2x + y - 5 = 0 is the tangent at (3, - 1).
The other tangent parallel to it is 2x + y + 5 = 0.
12. Find the area of the triangle formed by the normal at (3, - 4) to the circle x2 + y2 - 22x - 4y + 25 = 0
with the coordinate axis.
Sol: Given circle is x2 + y2 - 22x - 4y + 25 = 0
centre ( - g, - f) = C(11,2)
Equation of normal passing through (3, - 4) (11,2) is
24 6
y+4= 11  3
(x - 3)  y + 4 = (x - 3)  4y + 16 = 3x - 9
8
 3x - 4y - 25 = 0
c2 (  25)2 625
Area of the triangle formed by the normal with coordinate axes is   sq.units .
2 | ab | 2 | 3(  4) | 24

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
13. Show that x + y + 1 = 0 touches the circle x 2 + y 2 - 3x + 7y + 14 = 0 and find its point of
contact.
Sol: Given equation of circle
x2 + y2 - 3x + 7y + 14 = 0 ---------- (1)
 3 -7 
Centre (C) =  2, 2 
 
2 2
 -3   7  9 49 9 + 49 - 56 2 1
Radius (r) =  2  +  2  - 14 = + - 14 = = = 2
    4 4 4 4
Given line x + y + 1 = 0 ---------- (2)
d = perpendicular distance from C to (2)
| ax1  by1  c | 3 - 7 +1
2 2  3 - 72 + 2  -1 1
d= 2
a b 2 = 12 + 12
= 2 = 2
= 2
= r..
The line (2) touches the circle (1)
The point of contact is the foot of the perpendicular from C to (2)
h - x1
=
k - y1
=
-(ax1 + by1 + c)

h- 3 k+ 7 - 3
2 = 2 = 2
- 7 +1
2  
2 2
a b a +b 1 1 1 +12
2

h-
3
=k +
7
= 
- 3-7+2
2 
2 2 1+1
3 7 1
h-
2
=k +
2
= 2
3 1 3 1
 h-
2
=
2
 h= +
2 2
 h=2
7 1 -7 1
 k+
2
= 2
 k= +
2 2
 k=-3

Coordinates of point of contact (h, k) = (2, - 3).

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
14. Show that the tangent at (-1, 2) of the circle x 2 + y 2 - 4x - 8y + 7 = 0 touches the circle
x 2 + y 2 + 4x + 6y = 0 and also find its point of contact.
Sol: Equation of tangent at (-1, 2) to the circle x2 + y2 - 4x - 8y + 7 = 0 is
S1 = xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0.
 x(-1) + y(2) - 2(x - 1) - 4(y + 2) + 7 = 0
 -3x - 2y + 1 = 0  3x + 2y - 1 = 0 ---------- (1)

For the circle x2 + y2 + 4x + 6y = 0

Centre = (-2, -3), radius = 2 2
2 +3 -0 = 13

| ax1  by1  c | |3(-2) + 2(-3) - 1| |-13|

Perpendicular distance from (-2, -3) to the line (1) d= 2
a b 2 = 2
3 +2 2 = 13
= 13 =r

So line (1) also touches the 2nd circle.

Let (h, k) be the required point of contact.
So it is the foot of the perpendicular from the centre (-2, -3)
h - x1 k - y1 -(ax1 + by1 + c)
= =
2 2
a b a +b

h+2 k+3 -[3(-2) + 2(-3) - 1]

Þ = =
2 2
3 2 3 +2
h+2 k +3
=1 =1
3 2
h+2=3 k+3=2
h=1 k = -1
Coordinates of point of contact = (1, -1).

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
15. Find the mid point of the chord intercepted by x2 + y2 - 2x - 10y + 1 = 0 on the line x - 2y + 7 = 0.
Sol: Given equation of circle x2 + y2 - 2x - 10y + 1 = 0 -------- (1)
Centre = C(1, 5)
Given equation of chord, x - 2y + 7 = 0 ------ (2)
Midpoint of chord is foot of the perpendicular from centre to (2)
h - x1 k - y1 -(ax1 + by1 + c) h - 1 k - 5 -(1- 2(5) + 7)
= =  = = 2
a b a +b
2 2 1 -2 1 + (-2)2
h - 1 k - 5 -(-2) h -1 k - 5 2
= =  = =
1 -2 1+ 4 1 -2 5
h -1 2 k-5 2
= =
1 5 -2 5
5h - 5 = 2 5k - 25 = -4
5h = 2 + 5 = 7 5k = -4 + 25 = 21
7 21
h= 5
k= 5
 7 21 
Coordinates of mid point (h, k) =  , .
5 5 
16. Find the inverse point of (-2, 3) with respect to the circle x2 + y2 - 4x - 6y + 9 = 0.
Sol: Given equation of circle x2 + y2 - 4x - 6y + 9 = 0--------- (1)
Centre C(2, 3) P(-2, 3)
Equation of CP, y = 3 ------ (2)
Equation of polar of P is S1 = 0.
x(-2) + y(3) -2(x - 2) - 3(y + 3) + 9 = 0
-2x + 3y - 2x + 4 - 3y - 9 + 9 =0
-4x + 4 = 0  -4x = -4
x = 1 ------------- (3)
P.I. of (2), (3) is (1, 3) . The inverse point of P is (1, 3).

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
17. Find the condition that the tangents drawn from (0,0) to S  x2 + y2 + 2gx + 2fy + c = 0,
be perpendicular to each other.
Sol:Given equation of the circle is
S  x2 + y2 + 2gx + 2fy + c = 0
Centre C = (- g, - f), Radius, r = g2  f 2  c
Let, the given point P(x1, y1) = (0,0)
Angle between the tangents  = 900
The length of the tangent = S11 = 02  02  2(0)  2f(0)  c  c.
If ‘  ’ is the angle between the tangents then
 r
tan = S11
2
g2  f 2  c
tan 450 =
c
2 2
g f c
1=
c
c  g2  f 2  c
Squaring on both sides, we get
c = g2 + f2 - c
2 2
 g + f - 2c = 0.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

18. Find the locus of ‘P’ where the tangent drawn from ‘P’ to x2 + y2 = a2 are perpendicular to each other.
Sol:Given equation of the circle x2 + y2 = a2
Radius r = a
Let P(x1y1) be any point on the locus.
Length of the tangent = S11
= x 21  y 21  a2
Given that angle between the tangents q = 900.
If ‘q’ is the angle between the tangents through ‘P’ to the given circle then
 r
tan  2   S11

a
 tan 450 = x1  y12  a2
2

a
 1= x12  y12  a2

 x12  y12  a2  a
Squaring on both sides
 x21 + y21 - a2 = a2
 x12 + y12 = a2 + a2 = 2a2
2 2 2
 The locus of P(x1, y1) is x + y = 2a .

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
19. Find the condition that the tangents drawn from the exterior point (g, f) to the circle
x + y2 + 2gx + 2fy + c = 0 are perpendicular to each other.
2

Sol:Given equation of the circle is

x2 + y2 + 2gx + 2fy + c = 0, Radius r = g2  f 2  c

Let the given point P(x1, y1) = (g, f)

given   900
Length of the tangent = S11

 x12  y12  2gx1  2fy1  c

 g2  f 2  2g2  2f 2  c

 3g2  3f 2  c

If ‘  ’ is the angle between the tangents then

 r g2  f 2  c
tan     tan 450 
2 S11 3g2  3f 2  c

g2  f 2  c
 1
3g2  3f 2  c

 3g2  3f 2  c  g2  f 2  c
squaring on both sides
 3g2 + 3f2 + c = g2 + f2 - c
 2g2 + 2f2 + 2c = 0
 g2 + f2 + c = 0.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
20. Show that the area of the triangle formed by the two tangents through P(x1, y1) to the circle S º x2 + y2 +
r(S11 )3/2
2gx + 2fy + c = 0 and the chord of contact of P with respect to S = 0 is S11 + r 2
, where r is radius of the circle.
Sol: Given equation of circle
S = x 2 + y 2 + 2gx + 2fy + c = 0 ----- (1)

P(x1, y1) be an external point

PA, PB be the tangents |PA| = |PB| = S11

AB is chord of contact of P w.r.to S = 0

If ‘  ’ be the angle between the tangentss

r
tan (  /2) = S11 (where r is radius of circle)
e)
1
Area of DPAB = 2
|PA| . |PB| sin 

 
 r   r   r 
 S11   S11   S11 
1 2 tan (θ / 2)    
= 2
S11 S11 .
1 + tan2 (θ / 2)
=S  
11  1 +  r
2
  = S11 1 + r2  = S11  S11+ r2 
 
  S
 11
 
   S11   S11 

r . S11 r(S11) S11 r(S11 )3/2

= S11 2
S11 (S11 + r ) = S11 + r 2 = .
S11 + r 2

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial
21. Find the pair of tangents from the origin to the circle x2 + y2 + 2gx + 2fy + c = 0
and hence deduce a condition for these tangents to be perpendicular.
Sol:Given equation of the circle is
x2 + y2 + 2gx + 2fy + c = 0
Let the given point P(x1, y1) = (0,0)
The equation of the pair of tangents drawn from (0,0) to the given circle is S12 = SS11
 [x(0) + y(0) + g(x + 0) + f(y + 0) + c]2= (x2 + y2 + 2gx + 2fy + c) (0 + 0 + 0 + 0 + c)
 (gx + fy + c)2 = (x2 + y2 + 2gx + 2fy + c)c
 g2x2 + f2y2 + c2 + 2fgxy + 2fcy + 2gcx = cx2 + cy2 + 2gcx + 2fcy + c2
 (g2 - c)x2 + 2gf xy + (f2 - c)y2 = 0
Given that these tangents to be perpendicular then coefficient of x2 + coefficient of y2 = 0
 g2 - c + f2 - c = 0  g2 + f2 = 2c.

22. Find the equation of pair of tangents drawn from (0, 0) to x2 + y2 + 10x + 10y + 40 = 0.
Sol: Given equation of circle S= x2 + y2 + 10x + 10y + 40 = 0
P(0, 0) = (x1, y1)
S1 = x(0) + y(0) + 5(x + 0) + 5(y + 0) + 40
S1 = 5x + 5y + 40
S11 = 02 + 02 + 10(0) + 10(0) + 40= 40
Equation of pair of tangents, S12 = SS11
(5x + 5y + 40)2 = (x2 + y2 + 10x + 10y + 40) 40
52(x + y + 8)2 = (x2 + y2 + 10x + 10y + 40) 40
2 2 2 2 2
5
25 [x + y + 8 + 2xy + 2.y.8 + 2.8.x] = (x + y + 10x + 10y + 40) 40 8

5x2 + 5y2 + 320 + 10xy + 80y + 80x = 8x2 + 8y2 + 80x + 80y + 320
 3x2 - 10xy + 3y2 = 0.

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 MATHEMATICS-2B CIRCLE(SAQ’S & VSAQ’S) Aimstutorial

23. If the abscissae of points A, B are the roots of the equation x2 + 2ax - b2 = 0 and ordinates of A, B are the
roots of y2 + 2py - q2 = 0, then find the equation of a circle for which AB as a diameter..
Sol: Let A be (x1, y1) and B be (x2, y2).

Given x1, x2 be the roots of x2 + 2ax - b2 = 0

 (x - x1) (x - x2) =x2 + 2ax - b2

Given y1, y2 be the roots of y2 + 2py - q2 = 0

 (y - y1) (y - y2) = y2 + 2py - q2

Equation of circle with AB as a diameter is (x - x1) (x - x2) + (y - y1) (y - y2) = 0

 (x2 + 2ax - b2) + (y2 + 2py - q2) = 0
 x2 + y2 + 2ax + 2py - b2 - q2 = 0
is the equation of required circle.

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