1. SECTIONS
Lecture 3 of 9
1.2 CIRCLES
CONIC

2. LEARNING OBJECTIVES (a) Find the equation of a tangent and
normal to a circle.
(b) Find the length of a tangent from a fixed
point to a circle.

3. A tangent to a circle is a line that touches the circle at a point.
A normal to a circle is a line that is perpendicular to the tangent and passing through the centre. C
normal
tangent

4. Example 1
Find the tangent and normal line to a circle
at the point K(-1,-5).
x2 + y2 - 6x – 10y -82 = 0
Solution
x2 + y2 - 6x – 10y -82 = 0
g = -3 and f = -5
Centre = (-g,-f) = (3,5)

5. The gradient of the normal is
• C(3,5)
(-1,-5)•
Thus, the equation of the normal is

6. The gradient of the tangent is
Thus, the equation of the tangent is

7. xx1 + yy1+ g( x + x1) + f ( y + y1) + c = 0
Theorem
If the equation of a circle is
x2 + y2 + 2gx + 2fy + c = 0,
then the equation of a tangent to a circle
at the point P(x1,y1) is given by
xx1 + yy1+ g( x + x1) + f ( y + y1) + c = 0

8. PROOF:

10. An Alternative Method
Solution for Example 1:
x2 + y2 - 6x – 10y - 82 = 0
We have g = -3, f = -5 and c = -82
The equation of the tangent at K(-1,-5),
where g = -3, f = -5, c = -82
x1 = -1 and y1 = -5

11. xx1+ yy1+ g(x + x1) + f(y + y1) +c = 0
x(-1) + y(-5) + (-3)(x -1) +(-5)(y - 5) +(-82) = 0
-x - 5y - 3x y + 25 – 82 = 0
- 4x - 10y – 54 = 0
The equation of the tangent is:
2x + 5y + 27 = 0

12. The gradient of the normal is
Thus, the equation of the normal at the point P(-1,-5) is
P(-1,-5)
2y + 10 = 5x + 5
5x - 2y – 5 = 0

13. Find the length of the tangent from the point K(4,6) to the circle
Example 2
Find the length of the tangent from the point K(4,6) to the circle
x2 + y2 - 4x – 2y - 6 =0
K(4,6 ) d h r
C(2,1) M
Solution
x2 + y2 - 4x – 2y - 6 = 0
g = -2, f = -1 , c = -6
The centre = (2,1)

14. K(4,6 ) d h r
C(2,1) M
By using Pythagoras Theorem, the length of the tangent from a fixed point K(4,6) to the circle is

15. The Length of the Tangent to a Circle from a Fixed Point
N(a,b)
C(-g,-f) M
The length of the tangent from a fixed point N(a,b) to a circle with equation
x2 + y2 +2gx + 2fy +c = 0 is given by

16. If N(a,b), the centre of the circle is C(-g,-f) and the radius
PROOF:
If TA is the tangent line to the circle and the point M and N is two points on TA as given in the diagram.
If N(a,b), the centre of the circle is C(-g,-f) and the radius
N(a,b)
C(-g,-f) M y x T A

18. An Alternative Method
Solution for Example 2
We have a = 4, b = 6, g = -2, f = -1 and c = -6
The length of the tangent from K(4,6) to the circle is

19. NOTES:
Using the discriminant for circles problem solving:
( The circles touch each other)
( The circles intersect with
each other)
( The circles does not intersect each other)

20. Intersection of two circles
NOTES:
Intersection of two circles C1 C2 P Q
There are two intersection points.

21. Circles do not intersect with each other
(ii) C1 r1 C1 C2 r1 r2 C2 r2 OR

22. Two circles touches each other
(ii) C1 r1 C2 r2 C1 r1 OR C2 r2

23. Example 3
Show that the circles x2 + y2 – 10x – 8y + 8 = 0 and x2 + y2 – 8x – 4y + 14 = 0 do not intersect each other.
Solution

25. An Alternative Method
Solution for Example 3
g = -5, f = -4, c = 8
The centre:
The radius:

26. g = -4, f = -2, c = 14
The centre:
The radius:

27. P1 P2 r2 r1 r1
Since therefore these two circle don’t intersect each other.

28. xx1 + yy1+ g( x + x1) + f ( y + y1) + c = 0
CONCLUSION
the equation of a tangent to a circle at the point P(x1,y1) is given by
xx1 + yy1+ g( x + x1) + f ( y + y1) + c = 0
The length of the tangent from a fixed
point N(a,b) to a circle with equation
x2 + y2 + 2gx + 2fy + c = 0 is given by