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 = 4 = 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=9=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 = 4 = 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 =  4  =  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 =  4  =  2

(b)  The graph is                                              

  
  
HOMEWORK-For each problem, (a) find the center and radius of the circle and (b) Graph the circle.
 
1.    (x-2)+  (y+1) = 4.
2.    (x-3)+  (y-2)  = 9 
3.    x+  y- 6x  - 10y  +  30  =  0. 
4.    x+  y- 6x  +  4y  +  9  =  0.  
5.    x+  y- 10x  =  0. 
6.    x+  y  =  8. 
7.    x+  y2   =  1.
8.    4x+ 4 y2   =  9.