Epitrochoids and adjacent loop touching Consider the pair of parametric equations which describe a (simplified) epitrochoid:
\begin{align}
x(t) &= \cos (t) - a \cos (\alpha t)\\
y(t) &= \sin (t) - a \sin (\alpha t).
\end{align}
Here $a \in \mathbb{R}$ while $\alpha$ is a positive integer such that $\alpha \geqslant 3$ (so that the epitrochoid will have two or more loops).
Selecting a given $\alpha$, how is the value for $a$ found such that adjacent loops just touch one another? When the loops touch one another such points on the curve are self-tangential.
It seems these self-tangential points can be found by setting $y (t) = 0$ and $\frac{dy}{dt} = 0$, but I do not understand why this ought to be the case.
As an example, when $\alpha = 5$, adjacent loops in the corresponding epitrochoid (which has 4 loops) touch when $a = \pm \frac{4}{5}$.
Edit
I have added two curves to show what I exactly mean for my example when $\alpha = 5$.  For the curve on the left $a= \frac{3}{5}$ and none of the loops touch their adjacent loops. For the curve on the right $a = \frac{4}{5}$ and each of the loops just touch their adjacent loops. 
 A: There will be two solutions of $a$ for a given $\alpha$, although one is simply the negative of the other. For now, I will assume that $a < 0$. Because the epitrochoid is symmetric about the $x$-axis, there will be a tangent point at $y=0$ with $x > 0$.
$$\sin(t) - a \sin(\alpha t) = 0$$ must be true there. In the interval $0 \le t < 2\pi$, we want exactly $6$ solutions of $t$ for $\alpha \ge 4$: $1$ is for $t = 0$, $2$ are for the first loop and the last loop. If $\alpha$ is odd, the last $3$ are for the two middle loops on the left side touching the $x$-axis and the "outwards left" loop crossing the axis. If $\alpha$ is even, the last $3$ are for the middle inner loop on the left side of the $y$-axis crossing the axis and then looping back around twice (one point counts for two since there are different values of $t$ that could produce this point).
What this means in terms of the math is that the first relative minimum of $\sin(t) - a \sin(\alpha t)$ for $t > 0$ must have a value of $0$.

In other words, this point in the picture must be on the $x$-axis. Using the derivative of $\sin(t) - a\sin(\alpha t)$, I find that $$\cos(t) = a \alpha \cos(\alpha t)$$ must be true at the extrema. Then, if I take the second solution of that for $t > 0$, and then plug that into $\sin(t) - a \sin(\alpha t)$, the result should be $0$. Since it is known that $\cos(t) = a \alpha \cos(\alpha t)$, I did a bunch of algebraic and trig manipulation to get that $$a\cos\left(\alpha\arccos\left(\alpha\sqrt{\frac{a^{2}-1}{1-\alpha^{2}}}\right)\right)-\sqrt{\frac{a^{2}-1}{1-\alpha^{2}}} = 0 \tag 1$$ must be true for the loops to be tangent with $-1 \le a < 0$. This will have multiple solutions in $a$ for higher values of $\alpha$, but seems that the right root is the solution that is second-closest to $0$ (although I don't have a proof for this). Since $\cos(\alpha \arccos(x))$ can be rewritten as a polynomial in $x$, $a$ will be the root of a polynomial in $\alpha$, and as such, $a$ will be an algebraic number.
Using this, I found $a$ for small values of $\alpha$. For $\alpha = 3, a = \pm 1$. For $\alpha = 4, a = \pm \frac{3\sqrt{\frac{3}{2}}}{4}$. For $\alpha = 5, a = \pm \frac{4}{5}$. For $\alpha = 6, a = \pm \frac{\sqrt{102-7\sqrt{21}}}{12}$. For $\alpha = 7, a = \pm \frac{-1+2\sqrt{7}}{7}$. For higher values of $\alpha$, the expression for $a$ becomes increasingly complex (or might not even exist in closed form).
Edit: As $\alpha$ goes to $\infty$, $a\alpha$ approaches the root of $1 + x\cos(\sqrt{-1+x^2})$ around $x = 4.6$. This can be seen by plugging in $a = \frac{c}{\alpha}$, and then finding for what value of $c$ the LHS side of $(1)$ multiplied by $\alpha$ will go to $0$ as $\alpha \to \infty$. More specifically, I found the value of $c$ such that $$\lim_{\alpha \to \infty} \left( -c \cos\left( \alpha \arccos\left( \sqrt{\frac{c^2 - \alpha^2}{1-\alpha^2}} \right) \right) - \sqrt{\frac{c^2 - \alpha^2}{1-\alpha^2}} \right) = -1-c \cos(\sqrt{-1+c^2}) = 0$$
