Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm developing a mathematical formula for a programming interface. Please see following image for description: enter image description here

As you will see in the diagram, I have variables that will be set by my software( a, d and x), but I need a function to calculate the radius of the big circle ( r ). I can't think of any solution and I don t really need an exact one, approximation will work (rounded up even better) if there isn't any solution.

Any help will be much appreciated.

share|cite|improve this question
Is the black circle supposed to pass through the center of each small circle? – copper.hat Apr 16 '12 at 5:10
yes, the center of the small circles is for sure on the perimeter of the big circle. – denislexic Apr 16 '12 at 5:30

Assuming that the big circle passes through the center of each small one, and each small circle just touches its neighbor, you can use:

$$ r = \frac{\frac{d}{2}}{\sin(\frac{a}{2x})}$$

(Assuming the angle $a$ is in radians.)

To understand the formula, draw a radial line from the center of the big circle through the center of one of the small ones and another that just touches the top of the small circle. Draw a line from the touching point to the center of the small circle, this makes a right angle, the rest is basic trigonometry and division.

share|cite|improve this answer
If computing $\sin$ is a big deal, you can approximate it for small values of $a$ by $\sin(\frac{a}{2x}) \approx \frac{a}{2x}$, to get $r \approx \frac{d x}{a}$, as @pedja has above (except with $a$ in degrees, not radians). – copper.hat Apr 16 '12 at 5:30

$$x \cdot d \approx \frac{r \cdot \pi \cdot a}{180^{\circ}}$$

Hence :

$$r \approx \frac{x\cdot d \cdot 180^{\circ}}{a \cdot \pi}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.