On a circle, 2 parallel chords delimit a segment of which we know the area : A.
We also know the distance to the center of the circle of 1 chord : d1.

How to find d2, the distance of the other chord to the center of the circle ?

  • $\begingroup$ Do you know $r$ of the circle ? $\endgroup$ – Peđa Terzić Nov 1 '11 at 11:44

Let the circle be given by $x^2+y^2=r^2$ and assume that the area in question is to the right of the line $x=d$. Then we have to solve the equation $$2\int_d^x\sqrt{r^2-t^2}\ dt =A$$ for $x$. After expanding the integral you will obtain a transcendental equation for $x$ that can only be solved numerically. (Of course elementary geometric considerations will lead to the same equation.)

  • 1
    $\begingroup$ There could be two solutions, right? One with the center inside the segment, one with it outside? $\endgroup$ – Gerry Myerson Nov 1 '11 at 11:59
  • $\begingroup$ @Gerry Myerson: Thank you. I have edited that. $\endgroup$ – Christian Blatter Nov 1 '11 at 16:27

If we suppose that $r$ is given and that $d_2>d_1$ we may write following equations:

$a)$ case when center is outside of the segment:

$A=(\frac{r^2\theta_1}{2}-\frac{r^2\cdot \sin \theta_1}{2})-(\frac{r^2\theta_2}{2}-\frac{r^2\cdot \sin \theta_2}{2})$

$b)$ case when center is inside of the segment:

$A=r^2\cdot\pi-((\frac{r^2\theta_1}{2}-\frac{r^2\cdot \sin \theta_1}{2})+(\frac{r^2\theta_2}{2}-\frac{r^2\cdot \sin \theta_2}{2}))$

where: $\theta_1=2\arccos(\frac{d_1}{r}) , \theta_2=2\arccos(\frac{d_2}{r})$

So in both cases you have to solve numerically transcedental equation.


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