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 ?
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 ?
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.)
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.