What's the cause of these limacon figures? 


I made some photographs of the reflection of a light source in a cylindrical shaped cup. I understand how you can construct limacons by letting two one-dimensional shapes (of which one is tied to a point) roll over each other, like the form of a hanging chain is created by letting a parabola roll over a straight line, while the focus point of the parabola traces out the form of the hanging chain.
But how are the forms in the photographs connected to two one-dimensional shapes rolling over each other?
 A: Both cardioids and nephroids tend to appear in coffee cups. To explain why, let me use a geometric argument. 
First, the cardioid:

In the diagram above, the large circle represents the bottom of the cup. The two smaller circles with radius one third of the radius of the larger circle are used to construct the cardioid. Specifically, point $C$ is traced as the outer small circle rolls (without slipping) around the small circle centered at point $A$. Initially, point $C$ started at point $H$. Because the outer small circle rolls without slipping, arc $HG$ and $CG$ are equivalent. Let the small outer circle intersect the big circle at point $D$. If a ray of light from point $E$ hits the cup at point $D$, it will reflect so that the angles that the rays of light form with the tangent at $D$ will be equal. I will attempt to show that this reflected ray intersects $C$ and is tangent to the cardioid. To see why the ray intersects point $C$, note that the angle that segment $ED$ makes with the tangent is one half arc $ED$. Next, note that the angle that ray $DC$ makes with the tangent is 90 degrees minus angle $CDG$. Now, angle $CDG$ is equal to one half arc CG. The measure of arc $CG$ is equal to the measure of arc $GH$, is equal to the measure of arc $DF$, is equal to 180 degrees minus the arc $DE$. This implies that the angle that ray $DC$ makes with the tangent is equal to one half arc $DE$, which is equal to the angle that segment $DE$ makes with the tangent. This means that ray $DC$ must be the path that the reflected ray of light takes. To prove that this ray is tangent with the cardioid, note that $G$ is an instant center of rotation, making the tangent to the trajectory of $C$ (which is the cardioid) perpendicular to $CG$. Finally, note that segment $DC$ is perpendicular to segment $GC$ because angle $DCG$ inscribes a semicircle. All of this implies that if light hits a coffee cup so that light is dispersed from one point at the bottom of the cup, the reflected rays will be tangent to a cardioid. This is why you see a cardioid at the bottom of your cup.
A very similar argument can be made for a nephroid, but the light source has to be infinitely far away. 
A: 
Light rays from a source at $\:\boldsymbol{-}\infty\:$ of the axis $\:y\:$ (red vertical lines) incident on a circular mirror and reflect (blue rays). The envelope of the reflected blue rays is the cardioid (from the Greek word $\kappa\alpha\rho\delta\dot{\iota}\alpha =\:$heart).We can find the parametric equation of the cardioid after a nice application of the Legendre Transformation. For the cardioid of above Figure this equation is
$$
\dfrac{y}{R}=\dfrac{1}{2}\left[ 1+2\, \left(\dfrac{x}{R}\right)^{\tfrac{2}{3} } \right]\sqrt{1-\left(\dfrac{x}{R}\right)^{\tfrac{2}{3}} }
$$
where $\:R\:$ the radius of the circular mirror. 

The following 3 Figures show gradually the formation of the cardioid as envelope of the first reflected rays (blue color) from rays emitted by point $\:\mathrm{P}\:$ of the circular mirror (red color).





