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Centre C(a,b) Radius r 1. Radius r Centre C(-g,-f) 2. x 2 + y 2 + 2gx + 2fy + c = 0 HigherUnit 2 Outcome 4 Circle Wednesday, 07 January 2009
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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
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