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. 
I'll replace OP's $\alpha$ with $n$ (to avoid visual confusion with $a$). Also, I'll use a version of the curve such that each "half-leaf" is traced by $t$ in the range of consecutive integer multiples of $\pi/(n-1)$.
$$P = (\;\cos t + a \cos n t \;,\; \sin t + a \sin n t\;) \tag{1}$$
With this version of the curve, we observe that, for $a>0$, when adjacent loops are tangent, the point of tangency lies on the $x$-axis, and the tangent vector at that point is horizontal. (When $a<0$, the curve ---and its corresponding tangent vectors--- are rotated by an angle of $\pi/(n-1)$. In the analysis below, $a$ will ultimately appear to even powers in the key equations, so we'll get negative $a$-values "for free" if we just focus on positive ones. (That said, there's a way to re-frame the argument to incorporate non-horizontal tangents from the get-go. See the Addendum.)) Thus, a necessary condition for tangency is that the $y$-components of $P$ and $\frac{d}{dt}P$ simultaneously vanish: $$\sin t + a \sin n t = 0 = \cos t + na \cos n t \tag{2}$$ for some $\dfrac{\pi}{n-1}\leq t < \dfrac{2\pi}{n-1}$. Also, we observe that $|a|>1$ causes loops to surround the origin, precluding tangency; so we may assume $|a|\leq 1$.
Our goal is to eliminate $t$ from the system in $(2)$. One route ventures through the complex realm. Recall the expressions for cosine and sine in terms of the complex exponential: $$\cos\theta = \frac12\left(e^{i\theta}+e^{-i\theta}\right)\qquad \sin\theta = \frac1{2i}\left(e^{i\theta}-e^{-i\theta}\right) \tag{3}$$ Defining $\omega = \exp(it)$ and $\omega_n = \exp(int)$, our system becomes $$\begin{align} \omega_n(\omega^2 - 1) + \phantom{n}a \omega(\omega_n^2-1) &= 0 \\ \omega_n(\omega^2+1)+ na\omega(\omega_n^2+1) &= 0 \end{align} \tag{4}$$ which has four solutions of the form $$\omega = \frac{p+q}{\sqrt{n^2-1}} \qquad \omega_n = -\frac{p+nq}{na\sqrt{n^2-1}} \qquad\qquad \begin{array}{l} p^2 := n^2 (1-a^2) \geq 0 \\ q^2 := 1-n^2a^2\;\; \text{(possibly $<0$)} \end{array} \tag{5}$$ (taking all combinations of signs on $p$ and $q$). Since $\omega^n=\omega_n$, we can write
So, for a given $n$, candidate values of $a$ that cause tangent loops satisfy the four sign variants of $(\star)$.
Interestingly, $p^2-q^2=n^2-1$ and $p^2-n^2q^2 = n^2a^2(n^2-1)$, so that can square the equation (potentially introducing extraneous $a$ candidates) and rewrite it a bit more symmetrically as
$$\left(\frac{p+q}{p-q}\right)^{n} = -\frac{p+nq}{p-nq} \tag{6}$$
In this form, we see that the simultaneous substitutions $p\to-p$ and $q\to-q$ preserves the relation, so there are really only two sign combinations. Moreover, the substitution $p\to-p$ effectively reciprocates and swaps the signs of the fractions on each side of the equation.
Consequently, we can handle the sole sign ambiguity by squaring:
We see immediately that $a=\pm1$ and $a=\pm 1/n$ satisfy $(\star\star)$ by making either $p$ or $q$ vanish. Note that the latter case corresponds to epitrochoids with tangencies, but not quite the tangencies we seek:
So, perhaps $a=\pm1/n$ (ie, $q=0$) should be considered extraneous. Moreover, $a=\pm1$ only "works" for $n=3$.
Cross-multiplying $(\star\star)$ and expanding via the Binomial Theorem, we get some even-odd term cancellation. The result is something we can express in a convenient, radical-free (except for the factor $pq$) form:
Therefore, for a given $n$, the the values of $a$ corresponding to self-tangent epitrochoids must satisfy $(\star\star\star)$. Examples are provided below.
The reader will observe that $a$ occurs to even powers, so both positive and negative values satisfy $(\star\star\star)$. One of these values yields a curve with a horizontal tangency on the $x$-axis; the other corresponds to a "twin" curve, identical to the first but rotated by $\pi/(n-1)$, so the same loop tangencies are present.
Also, the algebra does recognize "non/adjacent" leaves in the curve. For example, $n=6$ case shows two self-tangent epitrochoids, but the latter's tangencies skip a leaf.
Interestingly, while extraneous (real) values of $a$ don't correspond to curves with self-tangencies, the examples suggest that "twin pairs" are tangent (with possible leaf-skipping). (See the Addendum.)
$n=3: pq( 27 a^4 - 18 a^2 - 1 ) = 0$
$a = \pm 1$ (same curve):
Twins:
$n=4: pq( 32 a^2 - 27 ) (2048 a^4 - 896 a^2 -27 ) = 0$
$a = \pm \sqrt{27/32}$:
Twins:
$n=5: pq( 25 a^2 - 16 ) (3125 a^6 - 3750 a^4+ 825 a^2 + 16)= 0$
$a=\pm 4/5$:
Twins:
$n=6: pq (6912 a^4 - 9792 a^2 + 3125) (1492992 a^6 - 1430784 a^4 + 234792 a^2 + 3125) = 0$
$a = \pm 0.6968\ldots$, and $a = \pm 0.9649\ldots$ (self-tangent, but non-adjacent leaves):
Twins:
Addendum. Why does $(\star\star\star)$ cover more tangent scenarios than originally intended? Symmetry!
For a given $n$, the epitrochoid has $(n-1)$-fold rotational symmetry. If the curve is to exhibit leaf-to-leaf tangency (with itself or its rotated "twin"), then the tangent vectors at the points-of-tangency will be symmetrically arranged. Consequently, those vectors must each make an angle of, say, $\phi_k := k\pi/(n-1)$ with the horizontal. (For self-tangent epitrochoids, we'd use only-even $k$ or only-odd $k$; for tangent twins, we'd use all $k$. In any case, the $\phi_k$ cover all angles we could possibly need.) Moreover, the tangent vectors are direction vectors for the points-of-tangency.
So, we needn't restrict attention to a horizontal tangent vector at a point on the $x$-axis. Rather, we can consider an angle-$\phi_k$ tangent vector at a point with direction angle $\phi_k$. That is, we replace $(2)$ with $$\frac{\sin t+a \sin nt}{\cos t+a \cos n t} = \frac{\sin \phi_k}{\cos\phi_k}= \frac{\phantom{-}\cos t+na \cos nt}{-\sin t-na \sin n t} \tag{A.1}$$ Cross-multiplying across the left inequality, and then also across the right one, we find $$\begin{align} \sin(t-\phi_k) + \phantom{n}a \sin(nt-\phi_k) &= 0 \\ \cos(t-\phi_k) + na\cos(nt-\phi_k) &= 0 \end{align} \tag{A.2}$$ Now we note the following about $\phi_k$: $$n\phi_k = \frac{nk\pi}{n-1} = \frac{(n-1+1)k\pi}{n-1}=k\pi+\phi_k\quad\to\quad nt-\phi_k=n(t-\phi_k)+k\pi \tag{A.3}$$ Since $\sin(\theta+k\pi)=\pm\sin\theta$ and $\cos(\theta+k\pi)=\pm\cos\theta$, we can re-write $(A.2)$ as $$\sin(t-\phi_k)\pm a\sin(n(t-\phi_k)) = 0 = \cos(t-\phi_k)\pm na\sin(n(t-\phi_k)) \tag{A.4}$$ This is just $(2)$ with $t\to(t-\phi_k)$ and a now-ambiguously-signed $a$. The former requires adjusting our earlier analysis by defining $\omega:=\exp(i(t-\phi_k))$ and $\omega_n:=\exp(in(t-\phi_k))$, but how we use these quantities is unchanged. The ambiguous sign on $a$ becomes a non-issue once all occurrences of $a$ appear to even powers. Consequently, even this more-general framing of the problem ultimately leads to the same solution, $(\star\star\star)$. $\square$