Circle Caustics for a Radiant Point inside the Circle

Question

I'm currently learning about caustics and envelopes and I'm tackling the classic coffee cup caustic. I understand the proof for the cardioid and nephroid. To my understanding, they occur for when the radiant point lies on the circumference and when light rays are parallel respectively. I'm aware that to solve for the envelope I must solve the equation for the family of reflected rays and its partial derivative simultaneously. Then, I came across the Wolfram MathWorld website that documented more shapes for when the radiant point a finite distance away within a unit circle.

The resulting equation is such:

I've started the derivation with the diagram below, but I just have no idea how to define the family of reflected lines to include μ. I would appreciate some pointers for starting this derivation!

Answer 1

I managed to get the same result as on Wolfram with the following steps:

1. The "cup" is a unit circle centred at $O$, which we'll take as the origin. Define points $L(\mu,0)$ (the light source), $P(\cos t,\sin t)$ (the point a light ray hits the coffee cup) and let $Q$ be the intersection of the reflected ray with the (extended) line $LO$.

2. The coordinates of $L$ and $P$ allow you to work out $\tan \angle OLP$

3. Using the fact that $\angle OPQ = \angle OPL$ (as it's a reflection), with a bit of angle chasing you can find $\angle OQP = \pi-2t+ \angle OLP$

4. This gives you the gradient of $PQ$ (the reflected ray) and since you know it goes through $P$, you can get the equation for that line. After a bit of trig, I found its equation to be $F(x,y,t)=0$, where $F$ is

$$F(x,y,t) = (\sin{t}-2\mu \sin{2t})x - (\cos{t}-2\mu \cos{2t}) + \mu \cos{t}y$$

5. We want the envelope of these rays as $t$ varies; as usual, this is found by solving $$F(x,y,t)=0,\quad\frac{\partial F}{\partial t} = 0$$

simultaneously for $x$ and $y$. This gives the result.

I think the trickiest step is (4); it's worth taking time to manipulate the trig functions so the next step is easier. As always, Wolfram|Alpha can be pretty helpful for some of those manipulations!