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Graphing a Circle from the General Equation

Date: 05/20/2004 at 20:29:53
From: be
Subject: equation of a circle

How do you graph x^2 + y^2 = 25?  I know it's a circle, but I don't
know how to graph it.

Date: 05/21/2004 at 18:04:11
From: Doctor Ian
Subject: Re: equation of a circle

Hi Be,

One way to do it is to substitute values of x, and calculate the
corresponding values of y.  For example, if x = 0, then

  x^2 + y^2 = 25

  0^2 + y^2 = 25

        y^2 = 25

          y = 5 or -5

So the points (0,5) and (0,-5) will be on the circle.  Similarly, if
we substitute y = 0, then

  x^2 + y^2 = 25

  x^2 + 0^2 = 25

        x^2 = 25

          x = 5 or -5

So the points (5,0) and (-5,0) are on the circle. 

If you plot those four points, you'll see that the center of the
circle has to be at the origin!  So now you can just fill in the rest
of the circle using that information. 

So that's one way.  There is another way that is faster and easier, if
you understand how the equation of a circle works.  The general form
of the equation of a circle is 

  (x - h)^2 + (y - k)^2 = r^2

To graph this, you have to recognize the form; and that tells you that
the center will be at the point (h,k), and the radius will be r. 
Which is pretty easy!

So, if we write your example in general form, it would look like this:

    x^2     +   y^2     = 25
  (x - 0)^2 + (y - 0)^2 = 5^2

This tells us the center of the circle is at (0,0) and the radius is
5.  So start at the origin (0,0) and count 5 units to the right, left,
up, and down.  Those four points are all on the circle since they are
5 units from the center.  Do you notice that those are the same four
points we got earlier?

Here's another example.  Suppose the equation is:

  (x - 5)^2 + (y - 2)^2 = 49

Now the center is at (5,2) and the radius is 7 since r^2 = 49.  Again,
start at (5,2) and count 7 units in each direction, mark those points
and connect them with your circle.

One last example:

  (x + 6)^2 + (y - 2)^2 = 16

Since the general form has (x - h)^2 and this equation has (x + 6)^2,
we need to think of x + 6 as x - (-6), so it's really [x - (-6)]^2:

  [x - (-6)]^2 + (y - 2)^2 = 4^2

Now the center of the circle is at (-6,2) and the radius is 4.  Mark
the four points as above and draw the circle.

Unfortunately, the equation of the circle is not always given in
general form, but we can always use algebra to put it in general form
using a process called "completing the square".  But we'll save that
for another day!

Does that help? 

- Doctor Ian, The Math Forum 
Associated Topics:
High School Conic Sections/Circles

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