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:

Midpoint of AB =

Gradient of AB =

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

213

,25

285

,2

32

313

2358

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 (1)y = -2x – 4 (2)

To solve simultaneously I will add (1) and (2) to get:

2y = 4y = 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.

Coordinate Geometry – The Circle

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.

A

C

B

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.