Higher Maths 2 4 Circles

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


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


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 )


( 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


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...


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


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).


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


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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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

12


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

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Higher Maths 2 4 Circles