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Equation of a Circle

Date: 05/13/2003 at 13:52:22
From: Balbino
Subject: Equation of a circle

Find the equation of a circle with the center at point (3,-4) and
radius 2.


Date: 05/14/2003 at 14:23:04
From: Doctor Dotty
Subject: Re: Equation of a circle

Hi Balbino,

Thanks for the question.

Here's a circle with centre (0,0) and radius R:

                         |
                         |
                     ,---+---.
                  ,-'   y+    `-P(x,y)
                ,'       |     / `.
              ,'         |    /    `.
             /           |  R/       \
            ;            |  /         :
            ;            | /          :
           ;             |/            :
    -------+-------------+------+------+------
           :             |      x      ;
            :            |            ;
            :            |            ;
             \           |           /
              `.         |         ,'
                `.       |       ,'
                  `-.    |    ,-'
                     `---+---'
                         |
                         |

Pythagoras' Theorem tells us that R^2 = x^2 + y^2.

If you think about it, that equation is true whatever x and y are. If 
they are in a negative quadrant, the radius is still positive. Can you 
see why?

So, that's a circle with centre (0,0). We don't want this; we want it 
at another centre.

So what if the centre is at (a,b)?

         |
        y+     ,-----.P(x,y)
         |   ,'      /`.
         |  /      R/   \
         | ;       /     :
        b+ |      +      |
         | :       (a,b) ;
         |  \           /
         |   `.       ,'
         |     '-----'
    -----+--------+--+--------
         |        a  x
         |

Now, with Pythagoras' Theorem again: 

   R^2 = (the distance from x to a)^2 + (the distance from y to b)^2

   R^2 = (x - a)^2 + (y - b)^2

That is the equation of a circle.

So for a circle of centre (-10,5) and radius 3, we get:

   3^2 = (x - -10)^2 + (y - 5)^2

     9 = (x + 10)^2 + (y - 5)^2

Some people prefer a general equation where the brackets have been 
multiplied out, so that it is in the form:

  x^2 + y^2 + 2gx + 2fy + c = 0

In our case, that can be worked out:

     9 = (x + 10)^2 + (y - 5)^2

     9 = x^2 + 20x + 100  +  y^2 - 10y + 25

     9 = x^2 + y^2 + 20x - 10y + 125

     0 = x^2 + y^2 + 20x - 10y + 116

Can you now find the equation of your circle?

Write back if I can be of any more help - on this or anything else.

- Doctor Dotty, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Conic Sections/Circles

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