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m f = y 1 - y 0 x 1 - x 0

The tangent (line g ) is perpendicular to this line. Therefore,

m f × m g = - 1

So,

m g = - 1 m f

Now, we know that the tangent passes through ( x 1 , y 1 ) so the equation is given by:

y - y 1 = m ( x - x 1 ) y - y 1 = - 1 m f ( x - x 1 ) y - y 1 = - 1 y 1 - y 0 x 1 - x 0 ( x - x 1 ) y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 )

For example, find the equation of the tangent to the circle at point ( 1 , 1 ) . The centre of the circle is at ( 0 , 0 ) . The equation of the circle is x 2 + y 2 = 2 .

Use

y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 )

with ( x 0 , y 0 ) = ( 0 , 0 ) and ( x 1 , y 1 ) = ( 1 , 1 ) .

y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 ) y - 1 = - 1 - 0 1 - 0 ( x - 1 ) y - 1 = - 1 1 ( x - 1 ) y = - ( x - 1 ) + 1 y = - x + 1 + 1 y = - x + 2

Co-ordinate geometry

  1. Find the equation of the cicle:
    1. with centre ( 0 ; 5 ) and radius 5
    2. with centre ( 2 ; 0 ) and radius 4
    3. with centre ( 5 ; 7 ) and radius 18
    4. with centre ( - 2 ; 0 ) and radius 6
    5. with centre ( - 5 ; - 3 ) and radius 3
    1. Find the equation of the circle with centre ( 2 ; 1 ) which passes through ( 4 ; 1 ) .
    2. Where does it cut the line y = x + 1 ?
    3. Draw a sketch to illustrate your answers.
    1. Find the equation of the circle with center ( - 3 ; - 2 ) which passes through ( 1 ; - 4 ) .
    2. Find the equation of the circle with center ( 3 ; 1 ) which passes through ( 2 ; 5 ) .
    3. Find the point where these two circles cut each other.
  2. Find the center and radius of the following circles:
    1. ( x - 9 ) 2 + ( y - 6 ) 2 = 36
    2. ( x - 2 ) 2 + ( y - 9 ) 2 = 1
    3. ( x + 5 ) 2 + ( y + 7 ) 2 = 12
    4. ( x + 4 ) 2 + ( y + 4 ) 2 = 23
    5. 3 ( x - 2 ) 2 + 3 ( y + 3 ) 2 = 12
    6. x 2 - 3 x + 9 = y 2 + 5 y + 25 = 17
  3. Find the x - and y - intercepts of the following graphs and draw a sketch to illustrate your answer:
    1. ( x + 7 ) 2 + ( y - 2 ) 2 = 8
    2. x 2 + ( y - 6 ) 2 = 100
    3. ( x + 4 ) 2 + y 2 = 16
    4. ( x - 5 ) 2 + ( y + 1 ) 2 = 25
  4. Find the center and radius of the following circles:
    1. x 2 + 6 x + y 2 - 12 y = - 20
    2. x 2 + 4 x + y 2 - 8 y = 0
    3. x 2 + y 2 + 8 y = 7
    4. x 2 - 6 x + y 2 = 16
    5. x 2 - 5 x + y 2 + 3 y = - 3 4
    6. x 2 - 6 n x + y 2 + 10 n y = 9 n 2
  5. Find the equations to the tangent to the circle:
    1. x 2 + y 2 = 17 at the point ( 1 ; 4 )
    2. x 2 + y 2 = 25 at the point ( 3 ; 4 )
    3. ( x + 1 ) 2 + ( y - 2 ) 2 = 25 at the point ( 3 ; 5 )
    4. ( x - 2 ) 2 + ( y - 1 ) 2 = 13 at the point ( 5 ; 3 )

Transformations

Rotation of a point about an angle θ

First we will find a formula for the co-ordinates of P after a rotation of θ .

We need to know something about polar co-ordinates and compound angles before we start.

Polar co-ordinates

Notice that : sin α = y r y = r sin α
and cos α = x r x = r cos α
so P can be expressed in two ways:
  1. P ( x ; y ) rectangular co-ordinates
  2. P ( r cos α ; r sin α ) polar co-ordinates.

Compound angles

(See derivation of formulae in Ch. 12)

sin ( α + β ) = sin α cos β + sin β cos α cos ( α + β ) = cos α cos β - sin α sin β

Now consider P ' After a rotation of θ

P ( x ; y ) = P ( r cos α ; r sin α ) P ' ( r cos ( α + θ ) ; r sin ( α + θ ) )
Expand the co-ordinates of P '
x - co-ordinate of P ' = r cos ( α + θ ) = r cos α cos θ - sin α sin θ = r cos α cos θ - r sin α sin θ = x cos θ - y sin θ
y - co-ordinate of P' = r sin ( α + θ ) = r sin α cos θ + sin θ cos α = r sin α cos θ + r cos α sin θ = y cos θ + x sin θ

which gives the formula P ' = ( x cos θ - y sin θ ; y cos θ + x sin θ ) .

So to find the co-ordinates of P ( 1 ; 3 ) after a rotation of 45 , we arrive at:

P ' = ( x cos θ - y sin θ ) ; ( y cos θ + x sin θ ) = ( 1 cos 45 - 3 sin 45 ) ; ( 3 cos 45 + 1 sin 45 ) = 1 2 - 3 2 ; 3 2 + 1 2 = 1 - 3 2 ; 3 + 1 2

Rotations

Any line O P is drawn (not necessarily in the first quadrant), making an angle of θ degrees with the x -axis. Using the co-ordinates of P and the angle α , calculate the co-ordinates of P ' , if the line O P is rotated about the origin through α degrees.

P α
1. (2, 6) 60
2. (4, 2) 30
3. (5, -1) 45
4. (-3, 2) 120
5. (-4, -1) 225
6. (2, 5) -150

Characteristics of transformations

Rigid transformations like translations, reflections, rotations and glide reflections preserve shape and size, and that enlargement preserves shape but not size.

Geometric transformations:

Draw a large 15 × 15 grid and plot A B C with A ( 2 ; 6 ) , B ( 5 ; 6 ) and C ( 5 ; 1 ) . Fill in the lines y = x and y = - x . Complete the table below , by drawing the images of A B C under the given transformations. The first one has been done for you.

Description
Transformation (translation, reflection, Co-ordinates Lengths Angles
rotation, enlargement)
A ' ( 2 ; - 6 ) A ' B ' = 3 B ^ ' = 90
( x ; y ) ( x ; - y ) reflection about the x -axis B ' ( 5 ; - 6 ) B ' C ' = 4 tan A ^ = 4 / 3
C ' ( 5 ; - 2 ) A ' C ' = 5 A ^ = 53 , C ^ = 37
( x ; y ) ( x + 1 ; y - 2 )
( x ; y ) ( - x ; y )
( x ; y ) ( - y ; x )
( x ; y ) ( - x ; - y )
( x ; y ) ( 2 x ; 2 y )
( x ; y ) ( y ; x )
( x ; y ) ( y ; x + 1 )

A transformation that leaves lengths and angles unchanged is called a rigid transformation. Which of the above transformations are rigid?

Exercises

  1. Δ A B C undergoes several transformations forming Δ A ' B ' C ' . Describe the relationship between the angles and sides of Δ A B C and Δ A ' B ' C ' (e.g., they are twice as large, the same, etc.)
    Transformation Sides Angles Area
    Reflect
    Reduce by a scale factor of 3
    Rotate by 90
    Translate 4 units right
    Enlarge by a scale factor of 2
  2. Δ D E F has E ^ = 30 , D E = 4 cm , E F = 5 cm . Δ D E F is enlarged by a scale factor of 6 to form Δ D ' E ' F ' .
    1. Solve Δ D E F
    2. Hence, solve Δ D ' E ' F '
  3. Δ X Y Z has an area of 6 cm 2 . Find the area of Δ X ' Y ' Z ' if the points have been transformed as follows:
    1. ( x , y ) ( x + 2 ; y + 3 )
    2. ( x , y ) ( y ; x )
    3. ( x , y ) ( 4 x ; y )
    4. ( x , y ) ( 3 x ; y + 2 )
    5. ( x , y ) ( - x ; - y )
    6. ( x , y ) ( x ; - y + 3 )
    7. ( x , y ) ( 4 x ; 4 y )
    8. ( x , y ) ( - 3 x ; 4 y )

Questions & Answers

anyone have book of Abdel Salam Hamdy Makhlouf book in pdf Fundamentals of Nanoparticles: Classifications, Synthesis
Naeem Reply
what happen with The nano material on The deep space.?
pedro Reply
It could change the whole space science.
puvananathan
the characteristics of nano materials can be studied by solving which equation?
sibaram Reply
plz answer fast
sibaram
synthesis of nano materials by chemical reaction taking place in aqueous solvents under high temperature and pressure is call?
sibaram
hydrothermal synthesis
ISHFAQ
how can chip be made from sand
Eke Reply
is this allso about nanoscale material
Almas
are nano particles real
Missy Reply
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
Lale Reply
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
Jyoti Reply
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
STM - Scanning Tunneling Microscope.
puvananathan
How we are making nano material?
LITNING Reply
Some times this process occur naturally. if not, nano engineers build nano materials. they have different physical and chemical properties.
puvananathan
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Source:  OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2
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