1. 18: Circles, Tangents and Chords
“Teach A Level Maths” Vol. 1: AS Core Modules
18: Circles, Tangents and Chords
© Christine Crisp

2. Module C1 Module C2 AQA Edexcel MEI/OCR OCR
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

3. For a circle, the radius is a normal.
Tangents to Circles
Some properties of circles may be needed in solving problems. This is the 1st one
The tangent to a circle is perpendicular to the radius at its point of contact
A line which is perpendicular to a tangent to any curve is called a normal. x
radius
For a circle, the radius is a normal.
tangent

4. Diagrams are very useful when solving problems involving circles
Tangents to Circles
Diagrams are very useful when solving problems involving circles
e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3)
Method: The equation of any straight line is x
gradient
We need m, the gradient of the tangent.
(5, 7) x
gradient
Find using
(2, 3)
The tangent to a circle is perpendicular to the radius at its point of contact
tangent
Find m using
Substitute for x, y, and m in to find c.

5. Substitute the point that is on the tangent, (5, 7):
e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3)
Solution: x
(2, 3)
(5, 7)
tangent
gradient
Substitute the point that is on the tangent, (5, 7): or

6. Use 1 tangent and join the radius.
e.g.2 The centre of a circle is at the point C (-1, 2). The radius is 3. Find the length of the tangents from the point P ( 3, 0).
Method: Sketch!
tangent
Use 1 tangent and join the radius.
The required length is AP. x
C (-1, 2)
Find CP and use Pythagoras’ theorem for triangle CPA 3
Solution:
P (3,0) x A

7. Exercises
Solutions are on the next 2 slides
1. Find the equation of the tangent at the point A(3, -2) on the circle
Ans:
2. Find the equation of the tangent at the point A(7, 6) on the circle
Ans:

8. A(3, -2) on the circle Find the equation of the tangent at the point
Solution: Centre is (0, 0). Sketch!
Gradient of radius,
gradient x
(0, 0)
(3, -2)
gradient m
Gradient of tangent,
Equation of tangent is or

9. 2. Find the equation of the tangent at the point A(7, 6) on the circle
gradient
(4 , 2)
(7, 6) x
tangent
Solution: Centre is (4, 2).
Gradient of radius,
Gradient of tangent, or

10. Another useful property of circle is the following:
Chords of Circles
Another useful property of circle is the following:
The perpendicular from the centre to a chord bisects the chord x
chord

11. e.g. A circle has equation
The point M (4, 3) is the mid-point of a chord. Find the equation of this chord.
e.g. A circle has equation
Method: We need m and c in x
Complete the square to find the centre
Find the gradient of the radius
Find the gradient of the chord
chord
Substitute the coordinates of M into to find c.

12. C e.g. A circle has equation
The point M (4, 3) is the mid-point of a chord. Find the equation of this chord.
e.g. A circle has equation x
chord
Solution: C
Centre C is
Tip to save time: Could you have got the centre without completing the square?

13. C Exercise A circle has equation
(a) Find the coordinates of the centre, C.
(b) Find the equation of the chord with mid-point (2, 6).
Solution: (a)
(b) x
chord
Centre is ( 1, 5 ) C
Equation of chord is
on the chord
Equation of chord is

14. The 3rd property of circles that is useful is:
Semicircles
The 3rd property of circles that is useful is:
The angle in a semicircle is a right angle P x B Q
diameter A

15. Hence and P is on the circle.
e.g. A circle has diameter AB where A is ( -1, 1) and B is (3, 3). Show that the point P (0, 0) lies on the circle.
Method: If P lies on the circle the lines AP and BP will be perpendicular. x
B(3, 3)
diameter
Solution:
Gradient of AP:
A(-1, 1)
Gradient of BP:
P(0, 0)
So,
Hence and P is on the circle.

16. Since AC and BC are perpendicular, C lies on the circle diameter AB.
Exercise
A, B and C are the points (3, 5), ( -2, 4) and (1, 2) respectively. Show that C lies on the circle with diameter AB.
B(-2, 4)
diameter
C(1, 2)
A(3, 5)
Solution:
Gradient of AC x
Gradient of BC
Since AC and BC are perpendicular, C lies on the circle diameter AB.

18. The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.
For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

19. A line perpendicular to a tangent to any curve is called a normal
A line perpendicular to a tangent to any curve is called a normal. The radius of a circle is therefore a normal.
Properties of Circles
The perpendicular from the centre to a chord bisects the chord
The tangent to a circle is perpendicular to the radius at its point of contact
The angle in a semicircle is a right angle
Diagrams are nearly always needed when solving problems involving circles.

20. Substitute the point that is on the tangent, (5, 7):
e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3)
Solution:
Substitute the point that is on the tangent, (5, 7): x
(2, 3)
(5, 7)
tangent
gradient or

21. C e.g. A circle has equationx
chord
Centre C is C
Solution:
The point M (4, 3) is the mid-point of a chord. Find the equation of this chord.
e.g. A circle has equation

22. Hence and P is on the circle. Gradient of AP:x
e.g. A circle has diameter AB where A is ( -1, 1) and B is (3, 3). Show that the point P (0, 0) lies on the circle.
diameter
A(-1, 1)
B(3, 3)
Method: If P lies on the circle the lines AP and BP will be perpendicular.
Solution:
P(0, 0)
Hence and P is on the circle.
Gradient of AP:
Gradient of BP:
So,