Show that circle A: x
2 
2 x
 y
2 
6 y

10

0 touches circle B: ( x

2 )
2 
( y

3 )
2 
5

Starter Activity
Show that circle A: touches circle B: x
2 
2
( x

2 )
2 x


( y
2 
6 y

3 )
2 y

10

5

0
First has centre at (-1, 3) with radius
Second centre at (2, -3) with radius 5
20
5
But distance between centres =
3
2 
6
2 
Since distance to centres = sum of radii then circles must touch
▪
45

3 5

The student should be able to :
• Find the equation of a tangent to a circle;
• Find the equation of a normal to a circle;
• Find the equation of a circle through 3 points; and to gain the high grades, to :
• Find the length of a tangent from a point;
• Find the equation given a chord and tangent;
• Prove that a line is a tangent to a circle

The best way is to intersect 2 perp bisectors.
eg. Find the equation of the circle passing through A(3, 1), B(8 , 2), and C(2, 6).
Grdt AB = y
2 x
2

 y
1 x
1

2
8

1

3

1
5 so perp grdt = -5
Grdt AC = y
3 x
3

 y x
1
1
Midpoint AB =

(
6
2

1
3


3
8

2
,

5
1
1


2
2

5
)
 so perp grdt = 1/5
(
11
2
,
3
2
)
Midpoint AC = (
3

2
2
,
1

2
6
)

(
5
2
,
7
2
)

So perp bisr of AB is
& Perp bisr of AC is y y

3

2
7
2



5 (
1
5
( x x


11
5
2
)
2
) or y = - 5x + 29 or y

1
5 x

3
Hence 2 values of y must be equal
So 
5 x

29

1 x

3 from which x = 5
5
Subst in the second equation, y = 4
So centre at (5, 4) and dist from A to centre is
D = ( 5

3 )
2 
( 4

1 )
2 
4

9

13
So circle is ( x

5 )
2 
( y

4 )
2 
13

We have just seen that Grdt AB x Grdt AC = -1 so Angle BAC = 90 degrees !
Hence BC is a diameter

eg. B(8, 2) and C(2 , 6) mark the end points of the diameter of a circle. Find the equation of the circle.
Diameter BC = ( 8

2 )
2 
( 2

6 )
2 
36

16

52

2 13
Hence radius = 13
Centre is midpoint of AB =
Equation is then
8

2
2
,
2

6
2

( 5 , 4 )
( x

5 )
2 
( y

4 )
2 
52

• Find the equation of the tangent and normal
First we need the circle in CTS: ( x

2 )
2 
( y

5 )
2 
100
Well, the grdt of the line joining (10, 11) to the centre (2, 5) is so grdt of nml = 3/4
10


5
2

6
8

4
Hence, normal is y

11

3
4
( x

10 )
And tangent is y

11


3
4
( x

10 )

eg. Find the equation of the tangent to the circle
( x

3 )
2 
( y

2 )
2 
13 at the point (1, 5).
Well, the grdt of the line joining (1, 5) to the centre is
2
3


5
1
 
3
2 so grdt of tgt = 2/3
Hence, tangent is y

5

2
3
( x

1 ) eg. For the same circle find the length of the tangent
from (10, 11).
We will need Pythagoras !
Distance from (3, 2) to (10, 11) = 7
2 
9
2 
130

eg.
Given that AB is chord of a circle where A is at (1, 3) and B is at (4,4). The tgt to the circle at A is the line y = 2x + 1.
Find the equation of the circle.

3
 
1
Normal at A has grdt -½ so eqtn of normal is
2

 
1
2 x
2
But grdt AB = 3
1

1
3 y

7
2
 
3 (
So perp bisector of AB is (2)
 x
4
4


  
3 x y

3

5
2
7 and midpoint AB = So
2
 x 11
1
Hence, So 7 – x = - 6x + 22
2
7
2

So 5x = 15 and then x = 3 and so y = 2 so centre is at (3, 2) x

5
2
)
Radius is distance from (3,2) to A i.e
Circle is ( x

3 )
2 
( y

2 )
2 
5
( 3

1 )
2 
( 2

3 )
2 
4

1

5

eg. Show that the line y = 7x + 10 is a tangent to the circle x
2  y
2 
2 and find the point of contact.
Well at intersection, x
2 
( 7 x

10 )
2 
2
Expanding the brackets,
So
So
Discrim = x
2 
49 x
2 
100

140 x

2

0
50 x
2
25 x
2

140 x

70 x

98

0

49

0 b
2 
4 ac

70
2 
4 .
25 .
49

0
Hence equal roots because discrim = 0 so line is tangent and point of contact at x = -b/2a = - 70/50 = -7/5, y = 7(-7/5) + 10 = 1/5 contact at (-7/5, 1/5)