It’s Maths!!

Posted in General with tags , , on January 20, 2010 by Admin

Maths Forum

Not only talks about Conic section(Circle) but also

other math topics.

KLIK to view;)

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Here’s another link which I think helps a lot for those who had troubles with Conic Section

Click HERE or HERE

or maybe HERE.

The Tab Tutor program sure is useful;)

 

moderator,

Nor Hidayah Binti Kamin

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Inscribed Angles

Posted in The Angle of Circle with tags , on January 20, 2010 by dramakcjt

by,

Kong Ngiik Hiong

angle of circle

Posted in The Angle of Circle with tags on January 20, 2010 by dramakcjt

Inscribed Angles Conjectures



Explanation:

An inscribed angle is an angle formed by two chords in a circle which have a common endpoint. This common endpoint forms the vertex of the inscribed angle. The other two endpoints define what we call an intercepted arc on the circle. The intercepted arc might be thought of as the part of the circle which is “inside” the inscribed angle. (See the pink part of the circle in the picture above.)

A central angle is any angle whose vertex is located at the center of a circle. A central angle necessarily passes through two points on the circle, which in turn divide the circle into two arcs: a major arc and a minor arc. The minor arc is the smaller of the two arcs, while the major arc is the bigger. We define the arc angle to be the measure of the central angle which intercepts it.

The Inscribed Angle Conjecture I gives the relationship between the measures of an inscribed angle and the intercepted arc angle. It says that the measure of the intercepted arc is twice that of the inscribed angle.

The precise statements of the conjectures are given below. Each conjecture has a linked Sketch Pad demonstration to illustrate its truth (proof by Geometer’s Sketch Pad!). The linked activities sheet also include directions for further “hands on” investigations involving these conjectures, as well as geometric problems which utilize their results.


The precise statement of the conjectures:

Conjecture (Inscribed Angles Conjecture I ): In a circle, the measure of an inscribed angle is half the measure of the central angle with the same intercepted arc..


Corollary (Inscribed Angles Conjecture II ): In a circle, two inscribed angles with the same intercepted arc are congruent.

Proof: The measure of each inscribed angle is exactly half the measure of its intercepted arc. Since they have the same intercepted arc, they have the same measure.


Corollary (Inscribed Angles Conjecture III ): Any angle inscribed in a semi-circle is a right angle.

Proof: The intercepted arc for an angle inscribed in a semi-circle is 180 degrees. Therefore the measure of the angle must be half of 180, or 90 degrees. In other words, the angle is a right angle.

by,

Kong Ngiik Hiong

Picture of circle applications

Posted in Gallery with tags , , , on January 20, 2010 by ctsuraidah

Post by,

ctsuraidah

Hello Again…

Posted in General on January 16, 2010 by Admin

It’s been such a long time since this blog is created.And whadaa…everyone is participating I can say.

Thanx a lot for the coorperation my dear tutorialmate.And if you are facing any problem related to posting and such,just give me a call or text me;)

Here are the list of those who already post:

1.NOR HIDAYAH BINTI KAMIN

2.ELENA ROZELLA JARID

3.SITTI SURAIDAH BINTI ANAS

4.BONG SING KOCK

5.KONG NGIIK HIONG

6.WONG SWAN SWAN

7.DELTA JENNETY DENIL

8.MARK RYAN

9.SEAH SHU YEN

10.HEPERIZAN JUNAIDI

11.GLORIA TERRENCE FIRAON

12.SAWAI A/K JANTAN

13.ERICA OLIVIA HENRY

14.GLORIA HAREN SAGING

15.DAPHNE GEORGE

16.MELSON MANGGIS

17.CAROLYNE DAIRIS

thanx again for those who has posted.and double thanx to those who posted 2 post(what a hardworking person you areO_0)

I hope the rest will post soon.

Thank you.

Moderator,

NOR HIDAYAH KAMIN

Pictures of Application of Circle

Posted in Application of Circle on January 12, 2010 by Admin

 

 

 

 

by,

Carolyne Dairis

Uploaded by,

Admin

 

Equation of circle

Posted in Equation Of Circle on January 10, 2010 by daphne27

EQUATION OF A CIRCLE      

 
The equation of a circle comes in two forms: 
  1) The standard form:  (x – h)+  (y-k)2   =   r2
  2) The general form  :  x+  y+  Dx  +  Ey  +  F   =   0,   where D, E, F are constants.
       
If the equation of a circle is in the standard form, we can easily identify the center of the circle, (h, k), and the radius, r .  Note: The radius, r, is always positive.
     
Example 1:   (x-2)2 + (y-3)2 = 4.    (a)  Find the center and radius of the circle.  (b) Graph the circle. 
     
Note:    A common mistake is to take h= -2 and K= -3.   In an equation, if the sign preceding h and k , ( h, k)  are  negative, then h and k are positive.  That is, h= 2 and k= 3.
    
(a)  Center: (h= 2, k= 3) = ( 2, 3 )    and     radius   r=2   since    r2 =  4   =>   r = Ö4 = 2
(b)  The graph is                                            
   
Example 2:    (x+1)2 + (y-2)2 = 9.     (a)  Find the center and radius of the circle.  (b) Graph the circle.
   
Note:  To correctly identify the center of the circle we have to place the equation in the standard form:
The standard form is:   (x –    h)2   +  (y-k)2 =   r2
  (x – (-1))+  (y-2)2 = (3)2.  Now, you can identify the center correctly.
  
(a)  Center: (h= -1, k= 2) = ( -1, 2 )         and radius          r=3   since    r2= 9   >     r=Ö9=3
(b)  The graph is                                                                 
   
Example 3:        2x+  2y2  =  8.     (a) Find the center and radius of the circle.  (b) Graph the circle.
  
Note:  To correctly identify the center of the circle we have to place the equation in the standard form.
First divide the equation by 2.  The new equation is :     x2     +     y2      =   4   .
The standard form: (x – h)2 + (y – k) =   r2
  (x – 0)2 + (y – 0) = (2)2.  Now, you can identify the center correctly.
  
(a)  Center:    (h= 0, k= 0) = ( 0, 0 )         and   radius   r = 2   since   r2 =  4   =>     r = Ö4 = 2
(b)  The graph is                                                
  
If the equation is in the general form, we have to complete the square and bring the equation in the standard form.  Then, we can identify the center and radius correctly.  We learned how to complete the square when working with quadratic equations (E III).  We will review it through an example.
  
Example 4:   x+  y– 6x  +  4y  +  9  =  0.    (a)  Find the center and radius of the circle.  (b) Graph the circle.
  
Completing the square:
  • Write the equation in this form:   (x2 – 6x +   ?1  ) + (y2 + 4y +    ?2  ) = -9   + ?1  +  ?2 . In the first parenthesis, we group the x-terms and in the second the y-terms.  The constant is moved on the right hand side. The question mark, ?, is the number needed in each parenthesis to complete the square.   Note that we have to add this number to both sides of the equation.  That is why you see ?1  and ?2, added to both sides.
  • How to find the number to replace the question mark, ?1.   Take the coefficient of x and divide it by 2,  (-6/2), and then square it,  (-3)2 = 9.     ?1   is going to be replaced by the number 9.
  • How to find the number to replace the question mark, ?2.   Take the coefficient of y and divide it by 2,  (4/2), and then square it,  (2)2 = 4.     ?2    is going to be replaced by the number 4.
 
Putting steps 1-3 together we have the following:
  (x2 – 6x +   ?1  )  +  (y2 + 4y +    ?2  )  =  -9   +  ?1  +  ?2
  (x2 – 6x +    9  )   +  (y2 + 4y +     4   )  = -9   +   9   +  4
         ( x – 3 )2       +       ( y  +  2 )2        =  4
         ( x – 3 )2       +       ( y  – (-2) )2      =  4    This equation is in the standard form.
 
(a)  Center:      (h= 3, k= -2) = ( 3, -2 )     and    radius     r = 2     since    r2 =  4   =>     r =  Ö4  =  2
(b)  The graph is                                             
  
Example 5:    x+  y– 6x  +  2y  +  4  =  0.    (a)  Find the center and radius of the circle.  (b) Graph the circle.
  
Completing the square:
  • Write the equation in this form:   (x2 – 6x +   ?1  ) + (y2 + 2y +    ?2  ) = -4   + ?1  +  ?2 . In the first parenthesis, we group the x-terms and in the second the y-terms.  The constant is moved on the right hand side. The question mark, ?, is the number needed in each parenthesis to complete the square.   Note that we have to add this number to both sides of the equation.  That is why you see ?1  and ?2, added to both sides.
  • How to find the number to replace the question mark, ?1.   Take the coefficient of x and divide it by 2,  (-6/2), and then square it,  (-3)2 = 9.     ?1    is going to be replaced by the number 9.
  • How to find the number to replace the question mark, ?2.   Take the coefficient of y and divide it by 2,  (2/2), and then square it,  (1)2 = 1.       ?2   is going to be replaced by the number 1.
 
Putting steps 1-3 together we have the following:
  (x2 – 6x +   ?1  )  +  (y2 + 2y +    ?2  )  =  -4   +  ?1  +  ?2
  (x2 – 6x +    9  )   +  (y2 + 2y +     1   )  = -4   +   9   +   1
         ( x – 3 )2       +       ( y  +  1 )2        =  4
         ( x – 3 )2       +       ( y  – (-1) )2      =  4    This equation is in the standard form.
 
(a)  Center:      (h= 3, k= -1) = ( 3, -1 )     and    radius     r = 2     since    r2 =  4   =>     r =  Ö4  =  2
(b)  The graph is                                              

 

by,

Daphne George