HigherUnit 2 Outcome 4 Circle x 2 + y 2 + 2gx + 2fy + c = 0 The General equation of a circle...

Higher Unit 2 Outcome 4Circle

x 2 + y 2 + 2gx + 2fy + c = 0

The General equation of a circle

Wednesday, 07 January 2009

Higher Circle Unit 2 Outcome 4

x 2 + y 2 + 2gx + 2fy + c = 0 The equation of a circle is

(x – 2)2 + (y – 3)2 = 25Write the equation without brackets

(x – 2) (x – 2) + (y – 3) (y – 3) = 25 x2 - 4x + 4 + y2 - 6y + 9 = 25

x2 - 4x + y2 - 6y + 13 - 25 = 0

x2 + y2 - 4x - 6y - 12 = 0

x2 + y2 – 2ax – 2by +a2 +b2 – r2 = 0

(x – a) (x – a) + (y – b) (y – b) = r2

(x – a) 2 + (y – b) 2 = r2

x2 – 2ax + a2 + y2 - 2by + b2 = r2

As a , b and r are constants (numbers) then these can be collected together as one term, c

x2 + y2 – 2ax – 2by + c = 0

In the same way we can

This is the general form This is the general form of the equation of a circleof the equation of a circle

Wednesday, 07 January 2009

22222 )2()2( rbybyaxax

2))(())(( rbybyaxax

22222 22 rbaybxayx

022 22222 rbabyaxyx

222 )()( rbyax Centre C(a,b) Radius r1.

Radius r02222 cfygxyx Centre C(-g,-f) cfg 222.

222 rba c -b, f a,- g Let

cfgrcfgr

rfgcrf)((-g) c

rba c

22

222

222

222

222

x 2 + y 2 + 2gx + 2fy + c = 0 Higher Unit 2 Outcome 4Circle

Wednesday, 07 January 2009

Finding the centre and the radiusFinding the centre and the radius

Given the equation of a circle, we can find the coordinates of its centre and the length of its radius. For example:

Find the centre and the radius of a circle with the equation (x – 2)2 + (y + 7)2 = 64

By comparing this to the general form of the equation of a circle of radius r centred on the point (a, b):

(x – a)2 + (y – b)2 = r2

We can deduce that for the circle with equation

(x – 2)2 + (y + 7)2 = 64

The centre is at the point (2, –7) and the radius is 8.

Wednesday, 07 January 2009

Finding the centre and the radiusFinding the centre and the radius

When the equation of a circle is given in the form

Find the centre and the radius of a circle with the equation x2 + y2 + 4x – 6y + 9 = 0

Start by rearranging the equation so that the x terms and the y terms are together:

x2 + 4x + y2 – 6y + 9 = 0

x2 + y2 – 2ax – 2by + c = 0

we can use the method of completing the square to write it in the form

(x – a)2 + (y – b)2 = r2

For example:

Wednesday, 07 January 2009

Finding the centre and the radiusFinding the centre and the radius

We can complete the square for the x terms and then for the y terms as follows:

The equation of the circle can now be written as:

x2 + 4x = (x + 2)2 – 4

y2 – 6y = (y – 3)2 – 9

(x + 2)2 – 4 + (y – 3)2 – 9 + 9 = 0

(x + 2)2 + (y – 3)2 = 4

(x + 2)2 + (y – 3)2 = 22

The centre is at the point (–2, 3) and the radius is 2.

x2 + 4x + y2 – 6y + 9 = 0

Wednesday, 07 January 2009

Higher Circle Unit 2 Outcome 4

x 2 + y 2 + 2gx + 2fy + c = 0 Alternative approach

Rearrange to get in the general form

x2 + 4x + y2 – 6y + 9 = 0

x2 + y2 + 4x – 6y + 9 = 0

2g = 4 2f = -6 c = 9

x 2 + y 2 + 2gx + 2fy + c = 0

g = 2 f = -3 c = 9

As before It therefore follows that

The centre is at the point (–2, 3) and the radius is 2.

(x + 2)2 + (y – 3)2 = 22

C is sum of all the constants

Wednesday, 07 January 2009

r2 = g2 +f2 - c

r2 = 22 + - 32 - 9

Centre (-g, -f)

Higher Circle Unit 2 Outcome 4

x 2 + y 2 + 2gx + 2fy + c = 0

Wednesday, 06 January 2009

Show that the equation x2 + y2 - 6x + 2y - 71 = 0represents a circle and find the centre and radius.

x2 + y2 - 6x + 2y - 71 = 02g = -6 2f = 2 c = -71

g = -3 f = 1 c = -71

(x + 3)2 + (y – 1)2 = 92

r2 = g2 + f2 -c

r2 = 9 + 1 - -71

r2 = 81

This is now in the form (x-a)2 + (y-b)2 = r2

So represents a circle with centre (3,-1) and radius = 9

Centre (-g, -f)

Higher Circle Unit 2 Outcome 4

x 2 + y 2 + 2gx + 2fy + c = 0

Wednesday, 06 January 2009

Show that the equation x2 + y2 + 6x - 2y - 15 = 0represents a circle and find the centre and radius.

x2 + y2 + 6x - 2y - 15 = 02g = 6 2f = -2 c = -15

g = 3 f = -1 c = -15

(x - 3)2 + (y + 1)2 = 52

r2 = g2 + f2 -c

r2 = 9 + 1 - -15

r2 = 25

This is now in the form (x-a)2 + (y-b)2 = r2

So represents a circle with centre (-3,1) and radius = 5

Centre (-g, -f)

Higher Circle Unit 2 Outcome 4

x 2 + y 2 + 2gx + 2fy + c = 0

Wednesday, 06 January 2009

Show that the equation x2 + y2 - 4x - 6y + 9 = 0represents a circle and find the centre and radius.

x2 + y2 - 4x - 6y + 9 = 02g = -4 2f = -6 c = 9

g = -2 f = -3 c = 9

(x + 2)2 + (y + 3)2 = 22

r2 = g2 + f2 -c

r2 = 4 + 9 - 9

r2 = 4

This is now in the form (x-a)2 + (y-b)2 = r2

So represents a circle with centre (2,3) and radius = 2

Centre (-g, -f)

Higher Circle Unit 2 Outcome 4

x 2 + y 2 + 2gx + 2fy + c = 0

Wednesday, 06 January 2009

Show that the equation x2 + y2 + 2x + 8y + 1 = 0represents a circle and find the centre and radius.

x2 + y2 + 2x + 8y + 1 = 02g = 2 2f = 8 c = 1

g = 1 f = 4 c = 1

(x - 1)2 + (y - 4)2 = 42

r2 = g2 + f2 - c

r2 = 1 + 16 -1

r2 = 16

This is now in the form (x-a)2 + (y-b)2 = r2

So represents a circle with centre (-1,-4) and radius = 4

Centre (-g, -f)

Higher Unit 2 Outcome 4Circle

Wednesday, 06 January 2009

Page 172

To build skills Complete

Exercise 3A Q 1, Q2,

(x – a)2 + (y – b)2 = r2 Centre C (a,b)(a,b) and radius rr

Higher Unit 2 Outcome 4Circle

Tuesday, 06 January 2009

What What do do you you

see ?see ?