1. Circles :
Introduction

2. C2: Circle Geometry Introduction
KUS objectives
BAT Find midpoints and distances
BAT Solve problems with circle geometry using midpoints and distances
Starter:
You can find the mid-point of a line by using the following formula:
𝑥 1 + 𝑥 2 2 , 𝑦 1 + 𝑦 i.e. find the average of the x and y coordinates
a) Find the midpoint of this pair of points: 2, 3 𝑎𝑛𝑑 6, 9
b) Find the midpoint of this pair of points: 2𝑎, −4𝑏 𝑎𝑛𝑑 7𝑎, 8𝑏
c) Find the midpoint of this pair of points: 4, 2 𝑎𝑛𝑑 −4, 3 2

3. Notes
AB is a diameter of a circle, where A and B are the coordinates (-3,8) and (5,4) respectively. Find the coordinates of the centre of the circle.
(-3,8) A
(1,6)
(5,4) B
The centre of the circle is at the midpoint of its diameter…

4. Notes Reminder: finding lengths
(8,-5) Q P
(-4,1)
PQ is a line. Given that P is (8,-5), and Q is (-4, 1) find the length of PQ
What can we use to find distances?
𝑑= (8−−4) 2 + (−5−1) 2
= =6 5

5. WB 1 PQ is a diameter of a circle, centre (2,-2)
WB 1 PQ is a diameter of a circle, centre (2,-2). Given that P is (8,-5),
a) find the coordinates of Q.
b) Find the radius of the circle
Write the unknown coordinate in terms of x and y, and fill in the formula for the mid-point as before…
(x,y)
(8,-5) Q P
(2,-2)
𝑥+8 2 , 𝑦−5 2
(-4,1)
Set the x-part equal to 2 and the y-part equal to -2…
𝑥+8 2 = 2
𝑦−5 2 = -2
𝑥=−4, 𝑦= So Q is −4, 1
𝑟𝑎𝑑𝑖𝑢𝑠= (2−−4) 2 + (−2−1) 2 = 45 =3 5

6. WB2 Find the distance between the co-ordinates (4a,a) and (-3a,2a)y
(x2,y2)
(y2-y1)
(x2-x1)
(x1,y1) x

7. Centre 1, 5 Radius= (−1−3) 2 + (6−4) 2 = 20 =2 5
WB 3
The points (3, 4) and B (-1, 6) are the end points of a diameter
a) Find the coordinates of the centre of the circle
b) Find the radius of the circle
Centre , 5
Radius= (−1−3) 2 + (6−4) 2 = 20 =2 5

8. Find the radius of the cirle
WB 4
PQ is the diameter of a circle, where P and Q are (-1,3) and (6,-3) respectively.
Find the radius of the cirle
R adius = ½ 85

9. Since Pythagoras’ Theorem works, ACB must be right-angled!
WB 5 The line AB is the diameter of the circle, where A and B are (-3,21) and (7,-3) respectively. The point C (14,4) lies on the circumference of the circle.
Find the values of AB2, AC2 and BC2 and hence show that angle ACB is 90°
(-3,21) A
(14,4) C B
(7,-3)
Since Pythagoras’ Theorem works, ACB must be right-angled!

10. WB 6
The line AB is the diameter of the circle with centre C, where A and B are (-1, 4) and (5, 2) respectively. The line l passes through C and is perpendicular to AB.
Find the equation of l.
a) Find the gradient of the line AB
b) Then work out the gradient perpendicular to that
c) We also need to find the co-ordinates of the centre
d) We can then find the equation of l l
(-1,4) A C
(5,2) B

11. WB 7
The line PQ is the Chord of the circle centre (-3,5), where P and Q are (5,4) and (1,12) respectively. The line l is perpendicular to PQ and bisects it. Show that it passes through the centre of the circle.
a) Find the midpoint of PQ
b) Find the gradient of PQ, and then the perpendicular
c) We can then find the equation of line l and substitute (-3,5) into it
(1,12) Q l C P
(-3,5)
(5,4)

12. WB 8
The lines AB and CD are chords of a circle. The line y = 3x – 11 is the perpendicular bisector of AB. The line y = -x – 1 is the perpendicular bisector of CD. Find the coordinates of the circle’s centre.
THE PERPENDICULAR BISECTOR OF A CHORD GOES THROUGH THE CENTRE!
So: Set the bisectors equal to each other and solve the equation for x and y.
y = 3x - 11 D C A B
y = -x - 1

13. One thing to improve is –
KUS objectives BAT Find midpoints and distances
BAT Solve problems with circle geometry using midpoints and distances
self-assess
One thing learned is –
One thing to improve is –

14. END