Post on 07-Jan-2016
description
Higher Maths 2 4 Circles
1
Distance Between Two Points
2Higher Maths 2 4 Circles
The Distance Formula
d = ( y2 –
y1)²+
( x2 –
x1)²√
B ( x2 ,
y2 )
A( x1 ,
y1 )
y2 – y1
x2 – x1
ExampleCalculate the distance between (-2,9) and (4,-3).
d = +
6²√ 12²
= 180
√ = 5√6
Where required, write answers as a surd in its simplest form.
Points on a Circle
3Higher Maths 2 4 Circles
ExamplePlot the following points
and find a rule connecting x
and y.( 5 , 0 ) ( 4 , 3 ) ( 3 , 4 ) ( 0 , 5 )
(-3 , 4 ) (-4 , 3 ) (-5 , 0 ) (-4 ,-3 )
(-3 ,-4 ) ( 0 ,-5 ) ( 3 ,-4 ) ( 4 ,-3 )
All points lie on a circle with radius
5 units and centre at the origin.
x ² + y ² = 25
x ² + y ² =
r ²For any point on the
circle,
For any
radius...
The Equation of a Circle with centre at the Origin
4Higher Maths 2 4 Circles
x ² + y ² = r
²
For any circle with
radius r and centre the
origin,
The ‘Origin’
is thepoint (0,0)
origin
ExampleShow that the point (-3 , )lies on the circle with equation
7
x ² + y ² = 16
x ² + y
²= (-3)² + ( )²7
= 9 + 7
= 16
Substitute point into equation:
The point lies on the circle.
The Equation of a Circle with centre
( a , b )
5Higher Maths 2 4 Circles
( x – a ) ² + ( y – b ) ² =
r ²
For any circle with radius
r and centre at the point
( a , b ) ...
Not all circles are centered at the origin. ( a , b )
r
ExampleWrite the equation of the
circle with centre ( 3 ,-
5 )and radius 2 3 .
( x – a ) ² + ( y – b ) ² =
r ²( x – 3 ) ² + ( y – (-5) ) ² = (
) ²2 3
( x – 3 ) ² + ( y + 5 ) ² = 12
6Higher Maths 2 4 Circles
The General Equation of a Circle
( x + g )2 + ( y + f )2 =
r 2
( x
2 + 2g x + g
2 ) + ( y
2 + 2fy + f 2 )
= r 2
x
2 + y
2 + 2g x + 2f y + g
2 + f 2 – r
2 = 0
x
2 + y
2 + 2g x + 2f y + c = 0
c = g
2 + f 2 – r
2
r
2 = g
2 + f 2 – c
r = g
2 + f 2 – c
Try expanding the equation of a circle with centre ( -g , - f ) .
General Equation of a Circle with
center ( -g , - f )and radiusr = g
2 + f 2 – c
this is just a number...
7Higher Maths 2 4 Circles
Circles and Straight LinesA line and a circle can have two, one or no points of intersection.
rA line which intersects a circle at only one point is at 90° to the radius and is is called a tangent.
two pointsof
intersection
one pointof
intersection
no pointsof
intersection
8Higher Maths 2 4 Circles
Intersection of a Line and a Circle
ExampleFind the
intersection of
the circle
and the line
2 x – y = 0
x
2 + (2 x)2 = 45
x
2 + 4 x
2 = 45
5 x
2 = 45
x
2 = 9
x = 3 or -3y = 2 x
x
2 + y2 = 45
Substitute into y =
2 x :
How to find the points of intersection between a line
and a circle:
• rearrange the equation of the line into the form y =
m x + c • substitute y = m x + c into the equation of the circle
• solve the quadratic for x and substitute into m x + c
to find y
y = 6 or -6
Points of intersection are
(3,6) and (-3,-6).
9Higher Maths 2 4 Circles
Intersection of a Line and a Circle (continued)Example 2Find where the line 2 x – y + 8 = 0
intersects the circle x
2 + y
2 + 4 x + 2 y
– 20 = 0x
2 + (2 x + 8)2 + 4 x + 2 (2 x + 8) – 20 = 0
x
2 + 4 x
2 + 32 x + 64 + 4 x + 4 x + 16 – 20 = 0
5 x
2 + 40 x + 60 = 0
5( x
2 + 8 x + 12 ) = 0
5( x + 2)( x + 6) = 0
x = -2 or -6
Substituting into y = 2
x + 8 points of
intersection as
(-2,4) and (-6,-4).
Factorise and solve
10
Higher Maths 2 4 Circles
The Discriminant and Tangents x = -b b
2 – (4
ac )
±
2 a
b 2 – (4 ac )
Discriminant
The discriminant can be
used to show that a line is a
tangent: • substitute into the circle
equation
• rearrange to form a quadratic equation
• evaluate the discriminant
y = m x + c
b 2 – (4 ac ) > 0 Two points of
intersectionb
2 – (4 ac ) = 0 The line is a tangent
b 2 – (4 ac ) < 0 No points of
intersection
r
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Higher Maths 2 4 Circles
Circles and Tangents
Show that the line 3 x + y = -10 is a
tangent to the circle x
2 + y
2 – 8 x + 4 y
– 20 = 0
Example
x
2 + (-3 x – 10)2 – 8 x + 4 (-3 x – 10) – 20 = 0
x
2 + 9 x
2 + 60 x + 100 – 8 x – 12 x – 40 – 20 = 0
10 x
2 + 40 x + 40 = 0
b 2 – (4 ac )= 40
2 – ( 4 × 10 × 40 )
= 0
= 1600 – 1600
The line is a tangent to the circle sinceb 2 – (4 ac ) = 0
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Higher Maths 2 4 Circles
Equation of TangentsTo find the equation ofa tangent to a circle:
• Find the center of the circle
and the
point where the tangent
intersects• Calculate the gradient of the
radius using the gradient formula
• Write down the gradient of the
tangent• Substitute the gradient of the
tangent
and the point of intersection
into
y – b = m ( x – a )
Straight Line Equation
y – b = m ( x – a ) m tangent =
–1m
radius
x2 – x1 y2 – y1
m radius
=
r