Find the Center of a Circle Using Compass and Straightedge

Date: 10/15/2003 at 08:47:40
From: Daniel
Subject: Finding the center of any circle?

I need to find the center of a circle in order for it to rotate about
this point.  It involves a specific set of procedures I cannot remember.

I actually saw this problem worked on the movie "The Man Without A 
Face".

Date: 10/15/2003 at 09:11:16
From: Doctor Edwin
Subject: Re: Finding the center of any circle?

Hi, Daniel.

It sounds like you're talking about finding the center of an actual,
physical circle.  If you have a compass and straightedge, you can find
the center of the circle.

The idea is to make two lines that are perpendicular to the circle 
(when I say circle here, I'm referring to the outside edge of the 
circle, not the whole disk).

All lines that are perpendicular to the circle run through the center. 

So, how do you make a line perpendicular to a circle?  This will be a 
little difficult to describe without pictures, but here it goes.  Take 
a point on the circle.  Let's call it "b".  Take your compass and set 
it to some size smaller than the radius of the circle and then use it 
to mark two points, a and c, one on either side b the same distance 
apart.  You now have a circle with three points, a, b, c, in that 
order, evenly spaced. 

Now place the point of your compass on point a and use the other end 
to trace a little arc inside the circle, between point b and where 
you think the center is.  Do the same for point c.  You should now
have a little curved X inside the circle.  Now take your straight edge
and draw a line from point b, through the center of the x, and
straight on past where you think the center of the circle is. 

You now have one line that goes through the center of the circle.

Now pick a new point b somewhere on the circle and repeat the 
process.  Now you have two lines that intersect at the center of the 
circle.

Did that make sense?  Write back if any of this was confusing, or 
doesn't apply to your situation, and I'll try to help.

- Doctor Edwin, The Math Forum
  http://mathforum.org/dr.math/