Transcript Slide 1

Coordinate Geometry – The Circle
This week the focus is on perpendicular bisectors
and using the perpendicular bisector of two or more
chords of a circle to find the centre of the circle.
Coordinate Geometry – The Circle
CONTENTS:
Perpendicular Bisectors
Perpendicular Bisectors and Centre of a Circle
Example 1
Example 2
Angle in a Semi-Circle
Example 3
Assignment
Coordinate Geometry – The Circle
Perpendicular Bisectors
A perpendicular bisector of a line cuts the line in
half and is perpendicular to the line.
The perpendicular bisector of a chord of a circle
passes through the centre of the circle.
Coordinate Geometry – The Circle
Perpendicular Bisectors
If given the equations of the perpendicular bisectors of
two chords we can find the centre of the circle by solving
the equations simultaneously.
In the diagram below WX and YZ are chords of the circle.
N is the perpendicular bisector of WX and M is the
perpendicular bisector of YZ. Where the N and M
intersect, the point A, is the centre of the circle.
Coordinate Geometry – The Circle
Example 1:
Find the equation of the perpendicular bisector of the
chord, AB, where A is (2,5) and B is (3, 8).
Solution:
The perpendicular bisector of AB is a line. To find the
equation of a line we need a point on the line and the
gradient of the line.
The perpendicular bisector passes through the midpoint of
AB so we can find this point.
The gradient of the line is -1/m where m is the gradient
of AB. We can find the gradient of AB as we have the
points A and B.
continued on next slide
Coordinate Geometry – The Circle
Example 1:
Find the equation of the perpendicular bisector of the
chord, AB, where A is (2,5) and B is (3, 8).
Solution:
 2  3 5  8   5 13 
,
 , 
2
2

 2 2 
Midpoint of AB = 
Gradient of AB =
85 3
 3
32 1
Therefore gradient of perpendicular line = -1/3
Equation of perpendicular bisector is:
y – 13/2 = -1/3(x – 5/2)
substituting into formula
6y – 39 = -2x + 5
multiplying across by 6
2x + 6y – 44 = 0
taking everything to one side
Coordinate Geometry – The Circle
Example 2:
The lines AB and CD are chords of a circle. The line
y = 2x + 8 is the perpendicular bisector of AB. The line
y = -2x – 4 is the perpendicular bisector of CD. Find the
coordinates of the centre of the circle.
Solution:
The perpendicular bisectors will intersect at the centre of
the circle, therefore we need to find the point of
intersection as this is the centre.
We find the point of intersection of the two lines by
solving simultaneously.
continued on next slide …
Coordinate Geometry – The Circle
y = 2x + 8
y = -2x – 4
(1)
(2)
To solve simultaneously I will add (1) and (2) to get:
2y = 4
y = 2
Substitute back into (1) to find x:
2 = 2x + 8
-6 = 2x
-3 = x
So coordinates of the centre of the circle are (-3, 2).
Coordinate Geometry – The Circle
Angles in a Semi Circle
The angle in a semi-circle, at the circumference
and sitting on a diameter, is always a right angle.
Chp 4 – Coordinate Geometry
Angles in a Semi Circle
If we are given three points on the circumference of a
circle, A, B and C, we determine if the angle between
the points is a right angle by checking the gradients of
the lines AB, AC and BC joining the points.
If one pair of lines are perpendicular then there is a
right angle between them.
B
C
A
Chp 4 – Coordinate Geometry
Example 3:
The points U(-2, 8), V(7, 7) and W(-3, -1) lie on a circle.
Show that angle UVW has a right angle.
Solution:
If angle UVW is a right angle then either:
gradient of UV x gradient of VW will equal -1
or
gradient of UW x VW will equal -1
•
•
•
Gradient of UV = 7-8/7--2 = -1/9
Gradient of VW = -1-7/-3-7 = -8/-10 = 4/5
Gradient of UW = -1-8/-3--2 = -9/-1 = 9
We can see that UV x UW = -1.
Therefore there is a right angle.
Chp 4 – Coordinate Geometry
ASSIGNMENT:
This weeks assignment is Yacapaca Activity.
The deadline for assignments is 5:00pm on Monday
15th February 2010.
No late submissions will be accepted.