Archive for the Equation Of Circle Category

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

Advertisements

Equation of circle

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

STANDARD EQUATION OF A CIRCLE 

 

The equation of a circle with the centre at (a, b) and radius r, is given by (x – a)2 + (y – b)2 = r2 .If the centre of the circle is at the origin and the radius is r, then equation of circle is x2 + y2 = r2.

circle

EQUATION OF CIRCLE IN DIFFERENT CONDITIONS 

 

Condition        Equation  
 (i)         Touches both the  axes with  centre (a, a) and  radius a  (x-a)2  + (y-a)2  = a2 circle
 (ii)        Touches  x-axis only with centre             (a, a)  and  radius  a  (x-α)2 + (y-a)2  = a2 circle
 (iii)       Touches y-axis only with centre (a, b)              and radius  a  (x-a)2 + (y-β)2 = a2 circle
 (iv)      Passes through the origin with centre             (α/2, b/2) and radius √(α2 + b2/4) x2 +a y2 – αx – by  = 0 circle

 

by,

Daphne George

Equation Of Circle

Posted in Equation Of Circle with tags on January 10, 2010 by msrica

This article is taken from the notes for my Math 107 web page. It’s purpose is to explain how to find equations for circles and also how to use this knowledge to provide a shortcut for graphing such equations. In order to do this you need to know two things.

  1. The distance formula
  2. The definition of a circle

For the distance formula see my notes.

What is a circle?

ãA circle is a set of all points equidistant from a point.

That is the standard geometry definition for a circle, and it is this kind of definition that allows us to write down equations for things like circles. Since we have a definition in terms of distances, we can use the distance formula to come up with an equation. Remember that the equation for a geometrical figure is an equation that all of the points in it have to satisfy, a kind of entrance test to get on the graph.

The point we want to make our points equidistant from is called the center and the distance is called the radius. (Think about how you draw a circle with a compass.) So let’s find the equation of the circle with radius r and center (h,k). I am using (h,k) for the coordinates of the circle rather than (x,y) because I have to reserve (x,y) for the points on the circle.

By the distance formula, the distance between (x,y) and (h,k) is given by this expression. By our definition of a circle, in order for (x,y) to pass the entrance test, this must be r, so we get this. This looks nicer if we square both sides, so we get this. for the standard equation of a circle with center (h,k) and radius r.

This means that if you were asked to write down the equation for a circle with a given center and radius you should be able to do this by just putting the numbers into this formula. You would substitute the x coordinate of the center for h and the y coordinate of the center for k and the radius for r.

Example: Find the equation of the circle with center (2,3) and radius 5.

Solution: Just put the 2 where the h is and the 3 where the k is and the 5 where the r is to get the equation.

(x-2)2+(y-3)2=25

This means that if someone were to plot all of the points that satisfy this equation, they would get exactly the circle that we want them to get, namely the one that has a center at (2,3) and a radius of 5.

Knowing about this standard equation for a circle can also be useful for making it easier to graph equations, because now whenever you see an equation like this, you don’t need to plot any points, you just need to recognize it is of this form and draw the circle. Before when you graphed equations you could only do it be just substituting x’s and y’s and finding points.  If you have an equation of this kind of form and are asked to graph it, you need to find the center and the radius of the circle, so you need to recognize what the h, k, and r are. The right side of the equation will be r2, so to find r, you have to take its square root. You need to put the left side into the form (x-h)2+(y-k)2 in order to determine h and k. One problem you might have is that you might have addition instead of subtraction, so to deal with this you would have to see addition as subtracting a negative, which is kind of backwards of the way we normally think about these things. My suggestion for an approach so that you don’t have to remember too many things like that is just always find x and y so that what is in the parentheses is 0, and that will be your h and k, the coordinates of the center. Then once you have found the center and the radius of the circle, you can graph it by plotting the center and then drawing a circle of radius r around that center. A good way to do this is to plot four points, r to the right, r to the left, r up, and r down from the center and then just sketch a circle through those points. It doesn’t have to be perfect, just try your best. The main thing is that you show you teacher that you know what the center and radius is and what that means.

But since math instructors are all are mean and wicked, there is one more problem I need to tell how to deal with. The problem is, somebody just might have taken a nice equation for a circle like

(x+1)2+(y-2)2=9,

and expanded everything out and rearranged things like this

x2+2x+1+y2-4y+4=9
x2+y2+2x-4y-5=0,

and then presented you with the equation

x2+y2+2x-4y-5=0

to graph. The problem is that it is a lot easier to expand that out than to put it back to where it came from. But there is a way to do it and it is to use the method of completing the squares. So here is what you do.

Get all the x’s together and all the y’s together and add something to both sides to get the constant on the right side of the equation. Leave space so that there is room to complete the square.

x2+2x         +y2-4y        =5

Then complete the square to find something to add to the x’s part and something to add to the y’s part so that they are both perfect squares. Remember to complete the square you take half of the coefficient and the square it, so for the x’s you get 2/2=1, 12=1, so you add 1, and for the y’s 4/2=2, 22=4, so you add 4. (You don’t have to worry about the minus on the 4 because you are going to square, so it will go away anyway.)

x2+2x +1+y2-4y +4=5+1+4

Then write the left side as squares and add up the right side and you get

(x+1)2+(y-2)2=9,

and now you can find the center and radius and graph it. Here the center would be (-1,2) and the radius would be 3. So to graph it, all you need to do is find the point (-1,2) and then plot the points 3 up, 3 down, 3 to the right, and 3 to the left of it, and draw the circle through them.

by,

Erica Olivia Henry

Equation of Circle

Posted in Equation Of Circle on December 27, 2009 by hj306
Equation of Circles 

Let’s review what we already know about circles.

Definition:  A circle is a locus (set) of points in a plane equidistant from a fixed point.

Circle whose center is at the origin Circle whose center is at (h,k)
(This will be referred to as the “center-radius form”.
It may also be referred to as “standard form”.)
Equation:  
Example:  Circle with center (0,0), radius 4
               
Graph:
 
Equation:  
Example:  Circle with center (2,-5), radius 3
                
Graph:
  

Now, if we “multiply out” the above example we will get:

  When multiplied out, we obtain the
“general form” of the equation of a circle.  Notice that in this form we can clearly see that the equation of a circle has both x2 and y2 terms and these terms have the same coefficient (usually 1). 
   

When the equation of a circle appears in “general form”, it is often beneficial to convert the equation to “center-radius” form to easily read the center coordinates and the radius for graphing.

Examples:

1.  Convert   into center-radius form.

This conversion requires use of the technique of completing the square.

We will be creating two perfect square trinomials within the equation.
  • Start by grouping the x related terms together and the y related terms together.  Move any numerical constants (plain numbers) to the other side.
• Get ready to insert the needed values for creating the perfect square trinomials.  Remember to balance both sides of the equation.
• Find each missing value by taking half of the “middle term” and squaring.  This value will always be positive as a result of the squaring process.
• Rewrite in factored form.
You can now read that the center of the circle is at (2, 3) and the radius is .

 

2How do the coordinates of the center of a circle relate to C and D when the equation of the circle is in the general form
                                      ?

Let’s make some observations.  Re-examine our previous equations in general form and center-radius form.  Do you see a relationship between the center coordinates and C and D?

General form Center-radius form
C = -4,  D = 10 Center (2, -5)
C = -4,  D = -6 Center (2, 3)
 

It appears that the values of C and D are (-2) times the coordinates of the center respectively.  Why is this occurring?
When is expanded, becomes , where the center term’s coefficient doubles the value of -2.  Remember that while the equation deals with , the actual x-coordinate of the center of this circle is +2.

(Read more about these relationships at the Resource Page.)

3.  Write the equation of a circle whose diameter has endpoints (4, -1) and (-6, 7).

 

  Find the center by using the midpoint formula.
  
Find the radius by using the distance formula.
Points (-6,7) and (-1,3) were used here.  (d = distance, or radius)
  
                 Equation:  

 

 

4.  Write the equation for the circle shown below if it is shifted 3 units to the right and 4 units up.

 

 
A shift of 3 units to the right and 4 units up places the center at the point (3, 4).  The radius of the circle can be seen from the graph to be 5.Equation:   

 


 

5.  Convert into center-radius form.

 

  Whoa!!!  This equation looks different.  Are we sure this is a circle???In this equation, both the x and y terms appear in squared form and their coefficients (the numbers in front of them) are the same.  Yes, we have a circle here!  We will, however, have to deal with the coefficients of 2 before we can complete the square.
 Center:  (-3/2, 2)        Radius:  1/2 •  group the terms•  divide through by 2
•  get ready to create perfect squares

•  take half of the “middle term” and square it
•  factor and write in center-radius form

 

 


   

 

by,

Heperizan Junaidi