### A problem with a chord: find the radius

Today I had the opportunity to solve a real math problem involving a circle and a chord of known length in it. I had to find the radius. It wasn't a textbook problem or a puzzle on some website, but a math problem I needed to solve for my own needs.

For a tiny while I thought I could find the answer online, but I didn't, so I'm writing it out in case someone else needs it -- they should be able to find this solution by searching the Internet.

I wanted to make a kind of "moon-sliver shapes" in CorelDraw, to use as watermarks in my new books. I have the height and the width of the "sliver".

Here's the problem mathematically:
I have a chord of a circle, 17 mm in length in my example, and the other distance marked in the image is 5 mm. I need to find the radius of the circle, AND the angle measure of the arc of the circle that makes the sliver's rounded part.

At first, like I said, I searched around if there was some theorem or formula that would tell me what I needed directly. I didn't find any, but I did realize that I can use this theorem to solve my problem:

If two chords intersect, the product of the segments of one chord equals the product of the segments of the other chord (see proof).

My problem looks like this:

My chord intersects the diameter of the circle, which is a chord too. The two parts of the first cord are 8.5 and 8.5, and the two parts of the other are 5 and d − 5. Thus I get the equation

5(d − 5) = 8.52

From this it is quick to solve that d = 19.45. Then, the radius is of course half that, or 9.725.

On to the second part of my problem: to find the angle measure.

In this picture, I have a right triangle. I can therefore find the unknown angle Î± by using simple trigonometry, in this case the tangent.

The length 4.725 comes from the fact that the radius is 9.725, and then I subtract 5 from that.

The equation is tan Î± = 8.5/4.725, from which Î± ≈ 60.931°. The actual angle I want is double that, or about 121.862°.

I had to repeat this calculation several times for slivers of different "heights." (Or, actually I let Excel calculate the rest. Oh, how I love Excel!)

Here's one of the final pictures I made for my book: