# Seemingly easy circle geometry can't figure it out.

1. Jun 3, 2010

### tectactoe

In the image

[PLAIN]http://img411.imageshack.us/img411/6857/circle.png [Broken]

If given the diameter of the circle, as well as the required distance for AD, is it possible to find out what BC must be? It seems like this should be possible given the information, but I can't seem to find out how. Is there an equation I'm missing or something, or do I need more info?

EDIT: BC is parallel to the horizon.

Thanks...

Last edited by a moderator: May 4, 2017
2. Jun 3, 2010

### Mentallic

From the info given, no, but by the looks of it I think it would be assuming AD is perpendicular to BC and if DA is extended it passes through the centre of the circle.

If this is the case, think about create a right-angled triangle with one of the sides being AC.

3. Jun 3, 2010

### tectactoe

Yes you are correct. That had crossed my mind, however that would give me length AC and that right angle, but to solve for side AD, I'd still need either the hypotenuse or another angle, and I don't see a way to figure those out...

I feel like I'm in sixth grade all over again. >_> LOL, thanks for the help - any more suggestions?

4. Jun 3, 2010

### Mentallic

Actually what you're missing a year 6 kid will probably know haha :tongue:

You know the length of the diameter, well then what's half that?

5. Jun 3, 2010

### tectactoe

Well that would give me the radius of course... but how does that help me here?

6. Jun 3, 2010

### Mentallic

Circle geometry gets to a point where all the lines necessary to find the answer aren't drawn up for you, you have to draw them yourself. Use a little imagination!
Start drawing some radii, I'm sure you'll figure out where I'm trying to lead you.

7. Jun 4, 2010

### HallsofIvy

That's the inverse of a rather famous problem, known as the "bow and arrow" that goes back to Babylonian times.

To solve the "bow and arrow" problem, draw the radii to point B and the line from the center of the circle to A. That gives you a right triangle. Since you are given the diameter of the circle, you know the hypotenuse of that right triangle: half the diameter. Since you are given the length of BC, you know the length of one leg of that triangle: half of BC.

Use the Pythagorean theorem to find the length of the other leg of that right triangle: the distance fromthe center of the circle to A.

Finally, the distance from A to D is the radius of the circle minus the distance from the center to A.

To solve this problem, use that reasoning, leaving BC as a variable. Call the known diameter "d", the known length of AD, "a", and the length of bc "x". Then the distance from the center of the circle to A is $\sqrt{d^2/4- x^2/4}= \left(\sqrt{d^2- x^2}\right)/2$. Subtracting that from r gives a: $a= r- \sqrt{d^2- x^2}\right)/2$.

Solve for x.

Last edited by a moderator: Jun 4, 2010
8. Jun 4, 2010

### tectactoe

Brilliant. Gosh, when you see someone else explain it, it seems so easy, right? Gahhh..

Thank you, this will help a ton.

9. Jun 4, 2010

### sullivanjayle

BC must be? It seems like this should be possible given the information, but I can't seem to find out how.

10. Jun 4, 2010

### luma

call the centre of the circle X
r = diameter / 2
if you have AD, then you have XD, XB = r and from pythagoras we have XD^2 + (AB/2)^2 = r^2

solve that and you have AB