I ran into this while attempting to draw points on a circle using polar coordinates with the origin offset from the center of the circle.
The objective is to find a closed form expression for the length of the segment labled $a$ as a function of the angle $\alpha.$ The radius $R$ is fixed, as is the distance $c$ from $\mathfrak{C}$ to $\mathfrak{c}$. Since we have two fixed sides and an angle for the triangle $\mathfrak{cCP},$ the value of $a$ is, in principle, determined. Using the law of cosines, the following two equations are easily established:
$$R^{2}=a^{2}+c^{2}+2ca\cos\left(\alpha\right)$$
$$a=\sqrt{c^{2}+R^{2}-2cR\cos\left(\rho\right)}$$
In the first of these we have both $a$ and $a^{2}.$ In the second, the value of $\rho$has not been given.
When I started trying to figure this out, it seemed straight-forward. But all I have managed to accomplish is to find new and interesting ways to derive the law of cosines.
How might we express a as a function of $\alpha?$
This is a response to a follow-up challenge in the answer.
This is a second attempt at the follow-up.
This is what motivated the original question. It's J.J. Thomson method of illustrating why the field strength of EM radiation falls off as the first power of the inverse of distance from the source, but Coulomb's law gives field strength as the inverse squared distance. The field line density is increased in the perturbed region. The the increase fall off as $\frac{1}{r}$.
Best Answer
By law of cosines, $$ R^2 = c^2 + a^2 -2ac \cos(\pi-\alpha) $$ so \begin{align} 0 &= a^2 -a\left(2c \cos(\pi-\alpha)\right) + (c^2 - R^2) \end{align} which is a quadratic in $a$, with solutions \begin{align} a &= \frac{\left(2c \cos(\pi-\alpha)\right) \pm \sqrt{\left(2c \cos(\pi-\alpha)\right)^2 - 4(c^2 - R^2) }}{2}\\ &= \frac{\left(-2c \cos(\alpha)\right) \pm \sqrt{\left(2c \cos(\alpha)\right)^2 - 4(c^2 - R^2) }}{2}\\ &= \left(-c \cos(\alpha)\right) \pm \sqrt{\left(c \cos(\alpha)\right)^2 - (c^2 - R^2) }\\ \end{align}
I'll let you try to work out geometrically what the second solution might correspond to. Hint: extend the red arrow.