Find the circumference of a circle, given a chord and the length of the rest of the circle
If you know the length of a chord in a circle as well as the length of the circumference minus the segment cut off by the chord, can you find the circumference of the circle?
For example, given the figure below, assume we know $d$, the arc-length of the blue arc, as well as $b$, the length of the red chord. We would like to calculate either the angle $\theta$, the radius, $r$, or the full circumference of the circle.
The following equations can all be derived from the figure:
$\cos\,\theta=\frac{b}{2\,r}$
$d=c\,\frac{\pi-\theta}{\pi}=2\pi\,r\frac{\pi-\theta}{\pi}=2r(\pi-\theta) \quad \textrm{where c is the circumference of the circle}$
so
$r=\frac{d}{2(\pi-\theta)}=\frac{d}{2(\pi - \cos^{-1}\,\frac{b}{2\,r})}$
But I have no idea how to solve that.
$\endgroup$ 32 Answers
$\begingroup$Let the central angle be $\theta ^c$ and r the radius.
Via the info on the arc, we can get an equation in two unknowns, r and $\theta$.
From that of the chord, we can get another equation in r and $\sin \theta$.
The next step is to combine these two (assumed to be independent) simultaneous equations into one by either eliminating r (or $\theta$).
For the former, the resultant combined equation will have a factor in the form of $\dfrac {\sin \theta}{2\pi - \theta}$ (more or less). This equation is of the type called transcendental equation which may not be solvable by ordinary mathematical method. We can only get an approximated answer by using numerical methods. Eliminating $\theta$ will just end up with another transcendental equation.
$\endgroup$ $\begingroup$If the included angle of the chord is $\psi$ then the chord length is
$$ S = 2 r \sin \left( \frac{\psi}{2} \right) $$
Also the remaining circumference is $$K = (2\pi-\psi) r$$
From these two equations you are asked to find $r$ and $\psi$. Unfortunately there are no analytical solutions, because if you divide the two equations you get
$$ \frac{S}{K} = \frac{2 \sin \left( \frac{\psi}{2} \right) }{2\pi - \psi} $$
Once you have a numeric solution you get the radius from $$r = \frac{K}{2\pi-\psi}$$
Appendix I
If the angle is really small $\psi \rightarrow 0$ then you have $$\psi \approx \frac{2\pi S}{K+S}$$
or
$$\psi \approx \pi \left( \sqrt{\left( \frac{4 S}{K}+1\right)}-1 \right) $$
$\endgroup$ 2
