7/27/2019 18. circle-3.ppt
1/62
Mathematics
7/27/2019 18. circle-3.ppt
2/62
Circle Sessions - 3
Session
7/27/2019 18. circle-3.ppt
3/62
Session Objectives
7/27/2019 18. circle-3.ppt
4/62
Session Objective
1. Equation Of Family Of Circles
2. Equation Of Chord Whosemid Point is Given
3. Equation Of Chord Of Contact
4. Angle Between Two Circles5. Orthogonal Intersection
6. Equation Of Common Chord
7/27/2019 18. circle-3.ppt
5/62
Equation Of Family Of Circles
Family of circles passing throughthe intersection of a circle S = 0and a line L = 0 is
S = 0 L = 0
A
B
S+L = 0
7/27/2019 18. circle-3.ppt
6/62
Questions
7/27/2019 18. circle-3.ppt
7/62
Illustrative Problem
Write the family of circles passing
through the intersection of x2+ y29 =0 and x + y 1 = 0. Find that memberof this family which passes through theorigin.
Solution:
Family of required circle is
S + L = 0
7/27/2019 18. circle-3.ppt
8/62
Solution Cond.
2 2(x y 9) (x y 1) 0 ...(1)
Since the required circle passesthrough the origin, we have
(0+0-9) + (0+0-1) = 0
= -9
Substituting value of in (1) we get
2 2
x y 9x 9y 0
7/27/2019 18. circle-3.ppt
9/62
Equation Of Family Of CirclesPassing through points (x1,y1)and (x2,y2)
A(x1,y1)
B(x2,y2)
S (x-x1)(x-x2) + (y-y1)(y-y2) = 0
2 11
2 1
y yL y y 0
x x
Family of circles is S + L = 0
7/27/2019 18. circle-3.ppt
10/62
Questions
7/27/2019 18. circle-3.ppt
11/62
Illustrative Problem
Find the equation of the circle passing
through the (4,1), (6,5) and havingits centre on 4x+3y-16=0
Solution:
Circle whose end points of diameter are
(4,1), (6,5) is
(x 4)(x 6) (y 1)(y 5) 0
Equation of line passing through(4,1) and (6,5) is 2x-y-7=0.
Therefore family of circles is
2 2x y 10x 6y 29 (2x y 7) 0
7/27/2019 18. circle-3.ppt
12/62
Solution Cond.
6centre is 5, lies on 4x 3y 16 02
6 264( 5) 3 16 0
2 5
2 2Circle is 5(x y 10x 6y 29) 26(2x y 7) 0
2 25x 5y 2x 56y 37 0
7/27/2019 18. circle-3.ppt
13/62
Equation of family of circlewhich touches a given circle S
at a given point (x1,y1)
S=0
L=0
A(x1,y1)
S + L = 0 where L = 0 is tangent at thegiven point on it.
7/27/2019 18. circle-3.ppt
14/62
Questions
7/27/2019 18. circle-3.ppt
15/62
Illustrative problem
Find the equation of the circle
which touches the circle x2+y2=25at (3,4) and passes through (1,1).
Solution:
Tangent at (3,4) is 3x+4y-25=0
Therefore family of circletouching x2+y2-25 at (3,4) is
2 2
x y 25 (3x 4y 25) 0
7/27/2019 18. circle-3.ppt
16/62
Solution Cond.
It passes thorough (1,1)
1 + 1 25 + (3x+4y-25) = 0
23
18
Therefore the required equation ofthe circle is
2 2 23x y 25 (3x 4y 25) 018
7/27/2019 18. circle-3.ppt
17/62
Equation of circles whichtouches a given line at a givenpoint on it
A(x1,y1)
L = 0
2 2
1 1S (x x ) (y y ) 0
family of circle is S + L=0
7/27/2019 18. circle-3.ppt
18/62
Questions
7/27/2019 18. circle-3.ppt
19/62
Illustrative Problem
Find the equation of the circle
passing through origin(0,0) andtouching the line 2x+y-1=0 at (1,-1)
Solution:
Family of the circles touching
2x+y-1 = 0 at (1,-1) is
2 2(x 1) (y 1) (2x y 1) 0
It passes through (0,0)
= 2
2 2the equation of the circle is x y 2x 4y 0
7/27/2019 18. circle-3.ppt
20/62
Equation of Circle Passing throughthe points of intersection of twocircles
S1+ S2= 0 (-1)
S1= 0S2= 0
7/27/2019 18. circle-3.ppt
21/62
Question
7/27/2019 18. circle-3.ppt
22/62
Illustrative Problem
Find the equation of circle passing
through origin and the points ofintersection of the two circles x2 +y2 - 4x - 6y 3 = 0 and x2 + y2 +4x - 2y 4 = 0
Solution:
Equation of family of circle is
x2+ y2 - 4x - 6y 3 + (x2+ y2+ 4x - 2y - 4) = 0
It passes through (0,0) 34
=>x2 + y2 - 28x - 18y = 0
7/27/2019 18. circle-3.ppt
23/62
Equation of chord whosemid-point is (x1,y1)
S=0
C=(0,0)
A B(x1,y1)D
xx1+ yy1= x12+ y1
2
T = S1
7/27/2019 18. circle-3.ppt
24/62
Questions
7/27/2019 18. circle-3.ppt
25/62
Illustrative Problem
Find the equation of the chord of
the circle (x-1)2+ (y-2)2= 4 whosemid point is (2,1).
Equation of circle is2 2
x y 2x 4y 1 0
x 2 y 1T 2(x) 1(y) 2 4 1 0
2 2
T x y 3 0
Solution: Method 1
2 2
1S (2) (1) 2(2) 4(1) 1
1T S x y 3 2 x y 1
7/27/2019 18. circle-3.ppt
26/62
Illustrative Problem
Find the equation of the chord of
the circle (x-1)2
+ (y-2)2
= 4 whosemid point is (2,1).
Solution Method 2:
C=(1,2)
A B(2,1)D
2 1Slope of CD 1
1 2
Slope of AB = 1
Equation of AB is
x y 1 = 0
7/27/2019 18. circle-3.ppt
27/62
Equation of chord of contact oftangents drawn from a point(x1,y1)
P(x1,y1)
A(x2,y2)
B(x3,y3)
Equation of chord ofcontact is
xx1+yy1=a2
or
T = 0
7/27/2019 18. circle-3.ppt
28/62
Question
7/27/2019 18. circle-3.ppt
29/62
Illustrative problem
Find the equation of chord of contact of
a point (-2,-3) with respect to circle x2+ y2- 2x - 6y + 1=0
Required circle is T = 0
x(-2) + y(-3) 2(-2) 6(-3) + 1 = 0
2x + 3y - 23 = 0
Solution :
7/27/2019 18. circle-3.ppt
30/62
Angle at which two circlesintersect
2 2 2
1 2
1 2
d r r
cos 2r r
S1=0 S2= 0
C1 C2
P
r1 r2
d = distance(c1,c2)
7/27/2019 18. circle-3.ppt
31/62
Question
7/27/2019 18. circle-3.ppt
32/62
Illustrative Problem
Find the angle at which the
circles x2+ y28x 2y - 9 =0and x2 + y2 + 2x + 8y - 7 = 0intersect.
C1= (4,1); r1= 16 1 9 26
C2= (-1,-4); r2= 1 16 7 24
2d 25 25 50
50 26 24cos 0
2 26 24
90
Solution :
7/27/2019 18. circle-3.ppt
33/62
Orthogonal Intersection
C1C2
r1 r2
S1=0 S2=0
90
d = distance (c1,c2)
Method 1:
d2= r12+r2
2
7/27/2019 18. circle-3.ppt
34/62
Orthogonal Intersection
C1C2
r1 r2
S1=0 S2=0
90
Method 2:
2g1g2+ 2f1f2= c1+ c2
7/27/2019 18. circle-3.ppt
35/62
Question
7/27/2019 18. circle-3.ppt
36/62
Illustrative Problem
If two circles of equal radii a withcentre (2,3) and (5,6) respectively cuteach other orthogonally then find thevalue of a.
Two circles cut orthogonally
2 2 2
1 2d r r
2 2d (2 5) (3 6) 3 2
Therefore a2 + a2 = 18 => a = 3
Solution :
7/27/2019 18. circle-3.ppt
37/62
Common Chord Of two circlesS1 = 0 and S2= 0
S1=0 S2=0
A
B
Equation of common chord is S1 - S2= 0
7/27/2019 18. circle-3.ppt
38/62
Question
7/27/2019 18. circle-3.ppt
39/62
Illustrative Problem
Find the equation of common chord of
two circles x2+ y2=25 and4x2 + 4y2 - 40x + 91=0
2 2
1here S x y 25 0 2 2
2
91S x y 10x 0
4
1 2S S 0 40x 191 0
Solution :
7/27/2019 18. circle-3.ppt
40/62
Class Test
7/27/2019 18. circle-3.ppt
41/62
Class Exercise - 1
A variable chord is drawn through the
origin to the circle x2
+ y2
2ax = 0.The locus of the centre of the circledrawn on this chord as diameter is
2 2(a) x y ax 0 2 2(b) x y ay 0
2 2(c) x y ax 0 2 2(d) x y ay 0
7/27/2019 18. circle-3.ppt
42/62
Solution
Let (h, k) be the centre of the circle.
Then (h, k) being the mid-point ofthe chord of the given circle
2 2hx ky a x h h k 2ah.
Since it passes through (0, 0) 2 2ah h k 2ah 2 2h k ah 0
Locus is 2 2x y ax 0
Hence, answer is (c).
7/27/2019 18. circle-3.ppt
43/62
Class Exercise - 2
If the circle passes through the point
(a, b) and cuts the circle orthogonally,equation of the locus of its centre is
(a) 2ax + 2by = a2+ b2+ k2
2 2 2(b) ax by a b k
2 2 2(c) x y 2ax 2by k 0
2 2 2 2 2(d) x y 2ax 2by a b k 0
7/27/2019 18. circle-3.ppt
44/62
Solution
Let (h, k) be the centre
2 22 2 2h k h a k b k
2 2 2 2 2 2 2h k h 2ah a k 2bk b k
2 2 22ah 2bk a b k 0
Locus is 2ax + 2by 2 2 2a b y 0
Hence, answer is (a).
7/27/2019 18. circle-3.ppt
45/62
Class Exercise - 3
Equation of the circle which passes
through the origin, has its centre onthe line x + y = 4 and cuts the circlex2+ y24x + 2y + 4 = 0 orthogonally is
2 2(a) x y 2x 6y 0
2 2(b) x y 6x 3y 0
2 2(c) x y 4x 4y 0
(d) None of these
7/27/2019 18. circle-3.ppt
46/62
Solution
Let centre of the circle is (g, 4 g)
[its centre is on x + y = 4]
Equation of circle is 2 2x y 2gx 2 4 g y 0
[it passes through origin]
Since it cuts the given circle orthogonally,
2 g 2 2 (4 g) = 4 6g = 12 g = 2
Equation of the required circle is
2 2x y 4x 4y 0 Hence, answer is (c).
7/27/2019 18. circle-3.ppt
47/62
Class Exercise - 4
If O is the origin and OP, OQ are
distinct tangents to the circlex2+ y2+ 2gx +2fy + c = 0, thecircumcentre of the triangle OPQ is
(a) (g, f) (b) (g, f)
(c) (f, g) (d) None of these
7/27/2019 18. circle-3.ppt
48/62
Solution
O
(0,0)
P
Q
(g, f)
PQ is chord of contact of (0, 0)Equation of PQ is gx + fy + c = 0
Family of circles passing through PQand given circle is
2 2x y 2gx 2fy c gx fy c 0
7/27/2019 18. circle-3.ppt
49/62
Solution Cont.
It passes through (0, 0) 1
2 2x y gx fy 0
g fCentre is ,
2 2
Hence, answer is (d).
7/27/2019 18. circle-3.ppt
50/62
Class Exercise - 5
Prove that the circle
x2
+ y2
6x 4y + 9 = 0 bisects thecircumference of the circlex2+ y28x 6y + 23 = 0.
Solution:
Equation of common chord of the given circle isS1S2= 0
2x 2y 14 0 x y 7 0
7/27/2019 18. circle-3.ppt
51/62
Solution Contd..
Centre of circle
is (4, 3) 2 2x y 8x 6y 23 0
which lies on x + y 7 = 0, first circlebisects the circumference of the second
circle because common chord passesthrough the end points of a diameterof the second circle.
7/27/2019 18. circle-3.ppt
52/62
Class Exercise - 6
If OA and OB are two equal chordsof the circle x2+ y22x + 4y = 0
perpendicular to each other andpassing through the origin then findthe equation of OA and OB.
Solution:
Let chords be y = mx and . Since chords are
of equal lengths, perpendicular distance fromthe centre to the chords will be same.
1
y xm
2 2
1 2m 2 m
1 m 1 m
7/27/2019 18. circle-3.ppt
53/62
Solution Contd..
m 2 2m 1
1
m 2 2m 1 m 3,3
Equations of OA and OB are x + 3y = 0 and 3x y = 0
7/27/2019 18. circle-3.ppt
54/62
Class Exercise - 7
The coordinate of two points P and Qare (2, 3) and (3, 2) respectively.
If circles are described on OP and OQas diameters, O being the origin thenfind the length of their common chord.
Solution:
(0,0)
O
Q
(3,2)
P(2,3)
R
ORP , ORQ2 2
Therefore OR is perpendicularto PQ. We have to find the length
of OR. From figure it is clear thatOR is length from origin to theline PQ.
7/27/2019 18. circle-3.ppt
55/62
Solution Contd..
Equation of PQ is y 2 = 1 (x 3)
x y 5 0
5
OR2
l
7/27/2019 18. circle-3.ppt
56/62
Class Exercise - 8
Determine the equation of the circlewhose diameter is the chord x + y = 1
of the circle x2+ y2= 4.
Solution:
Equation of family of circles passingthrough the intersection of circle andline is
2 2x y 4 x y 1 0 ...(i)
2 2x y x y 4 0
S l i C d
7/27/2019 18. circle-3.ppt
57/62
Solution Contd..
Its centre lies on x + y 1 = 0
,
2 2
1 0 1
2 2
Substituting value of in the equation (i), we get
2 2x y x y 3 0
Cl E i 9
7/27/2019 18. circle-3.ppt
58/62
Class Exercise - 9
Consider a family of circles passingthrough two fixed points A (3, 7)and
B (6, 5). Show that the chords in whichthe circle x2 + y24x 6y 3 = 0cuts the members of the family areconcurrent at a point. Find the coordinateof this point.
Solution:
Family of circles is(x 3) (x 6) + (y 7) (y 5)
5 7y 7 x 3 0
6 3
x 3 x 6 y 7 y 5 2x 3y 27 0
S l ti C td
7/27/2019 18. circle-3.ppt
59/62
Solution Contd..
2 2x y 2 9 x 3 12 y 53 27 0
Common chord isS1S2= 0
2 5 x 3 6 y 56 27 0
( 5x 6y 56) (2x 3y 27) 0
This chord is intersection of 5x 6y + 56 = 0and 2x + 3y 27 = 0
Solving these equations we get .
232,3
Cl E i 10
7/27/2019 18. circle-3.ppt
60/62
Class Exercise - 10
If two chords, drawn from the point(p, q) on the circle x2 + y2 = px + qy
(where pr # 0) are bisected by the x-axis,then
(a) p2= q2 (b) p2= 8q2(c) p
2< 8q2 (d) p2> 8q2
Solution:
Let the chord is bisected at A (h, 0)
P is (p, q)
P(p,q) A(h,o)
S l ti C td
7/27/2019 18. circle-3.ppt
61/62
Solution Contd..
Q (2h p, q) lies on circle
2 22h p q p 2h p q q
2 2 2 2 24h p 4ph q 2ph p q
2 2 24h 6ph 2 p q 0
This is quadratic in h and h is real and distinct
D 0
2 2 236p 32 p q 0 2 24p 32q 0
2 2p 8q
7/27/2019 18. circle-3.ppt
62/62
Thank you