Me: General low math skilled user here (basic stuff, nothing fancy).

I'm kinda stuck on the following problem.

Given data:
- 2 concentric circles:
-- with there centers at the same spot. (origin, (0,0,0))
-- with radius r1 and r2.
- a distance d.

- finding point P? on circle 2 based on the distance d (from point->(x=r1, y=0).


The original problem was/is to reconstruct the relative cords/positions(3D) of 4 points by the distances between them. With this part I'm trying to get the relative cord of point 4 in relation to point 3. (The first part, getting cord on point 3, was mainly solved by finding the proper code that did the job for me. In case your wondering how I got this far.)




If you let P=(x,y) then $$ x^2+y^2=R_2^2 \\ (x-R_1)^2+y^2 = d^2 $$ subtract the second from the first to get $$ 2 R_1 x - R_1^2 = R_2^2 - d^2 \Rightarrow x = \frac{R_1^2+R_2^2-d^2}{2 R_1}$$ Substitute in the first to get $y$. You will get additional condition for $y^2 \ge 0$, i.e limits on $d$.

Sorry hit submit too soon. With $R_2 > R_1 > 0$ we need $$ R_2 - R_1 \le d \le R_2+R_1 $$

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  • $\begingroup$ I'm not sure yet if I can translate that so something I can make work for me. Will try ...<br> I'm aware of the needed minimum and maximum on d to get a solution (2 unless d is equal to the minimum or maximum. $\endgroup$ – MvGulik Feb 2 '14 at 2:58
  • $\begingroup$ Can you state the original problem? $\endgroup$ – user44197 Feb 2 '14 at 2:59
  • $\begingroup$ Original problem ? I don't think I can add anything to what I already wrote down that would add anything useful. Unless you can be a little more specific in your question. $\endgroup$ – MvGulik Feb 2 '14 at 3:06
  • $\begingroup$ I thought I gave the explicit formula. Once you have $x$ just use $y = \pm \sqrt{R_2 - x^2}$. Not sure why you can't make it work for you. $\endgroup$ – user44197 Feb 2 '14 at 3:08
  • $\begingroup$ Aha, clear. Lack of math skills on this end, as in reading and translating pure math statements it to something I understand. I'm more visual then math orientated. But I will try ... $\endgroup$ – MvGulik Feb 2 '14 at 3:15

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