Calculating the Radius from a Chord

Date: 08/18/98 at 23:49:52
From: Katie Auer
Subject: Radius of a circle

Dr. Math,

If I know the chord length and chord height, is it possible to 
determine the radius of the circle? Is there a formula?

Thank you,
Katie

Date: 08/20/98 at 12:02:31
From: Doctor Peterson
Subject: Re: Radius of a circle

Hi, Katie. I assume that by "chord height" you mean the distance from 
the chord to the middle of the arc it cuts off. If so, there's a 
simple formula, and I'll even show you how to find it, if you can do a 
little algebra.

Here's my picture:

              *******
         *       |h      *
       *    d    |    d    *
      *----------+----------*
     *           |       /   *
    *         r-h|    /r      *
    *            | /          *
    *------------+------------*

I've called the length of the chord 2d, to simplify the calculations, 
so half the chord is d. The distance from the chord to the arc (your 
"chord height") is h, and the unknown radius of the circle is r. Then 
the distance from the center to the chord is (r-h). Incidentally, the 
technical term for h is the "sagitta" of the chord, and r-h is the 
"apothem." You don't hear those words used much. "Sagitta" is Latin 
for arrow. If you think of the arc as the bow and the chord as the 
string, you can see why.

Look at the right triangle with sides d, r-h, and r. Using the 
Pythagorean Theorem, we can say that:

    d^2 + (r-h)^2 = r^2         ("^2" means squared)

which expands to:

    d^2 + r^2 - 2rh + h^2 = r^2

and by subtracting r^2 from both sides we get:

    d^2 - 2rh + h^2 = 0

Now we can add 2rh to both sides:

    d^2 + h^2 = 2rh

and divide both sides by 2h to get:

        d^2 + h^2
    r = ---------
           2h

So the radius is just the sum of the squares of the height and half the 
length, divided by twice the height.

If you look back at the picture, d^2 + h^2 is just the square of the 
straight-line distance from one end of the chord to the middle. There's 
a simple geometric way to get the formula in that form, if you know 
enough about similar triangles.

- Doctor Peterson, The Math Forum
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