An asymptote is going to mean that $r\rightarrow\infty$, which happens when the denominator approaches zero. Vertical or horizontal means that $\theta$ is a multiple of $\pi/2$ (vertical for even, horizontal for odd multiples). The only asymptote would therefore be when $\theta\rightarrow\pi$, which would be a horizontal one above (and parallel to) the negative $x$ axis.
Now as $\theta\rightarrow\pi$,
$$
y = r\sin\theta
= \frac{\theta\sin\theta}{\pi-\theta}
= \frac{\theta\sin(\pi-\theta)}{\pi-\theta}
\rightarrow \theta
\rightarrow \pi
$$
so the horizontal asymptote is $y=\pi$.
For visualization, we can reason that as $\theta$ increases from $0$ to $\pi$, $r$ increases from $0$ to $\infty$, crossing $1$ at the positive $y$ axis. Here are some plots for $\theta\in[0,.6\pi]$ and $[0,.99\pi]$ made with sage (online):
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.6*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.6*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.99*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.99*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
Horizontal and vertical tangents can be found by finding the roots of
$\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ respectively (assuming
that they are never simultaneously zero, in which case we'd have to
resort to higher order derivatives). A simple way to find these roots
is to observe that
$$r = \left(\frac{\pi}{\theta}-1\right)^{-1}
\implies \frac{dr}{d\theta}
= -\left(\frac{\pi}{\theta}-1\right)^{-2} \cdot
\left(-\pi\theta^{-2}\right)
= \frac{\pi}{(\theta-\pi)^2}
$$
so that
$$
0 = \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta
\iff
\tan\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)}
$$
and
$$
0 = \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta
\iff
-\cot\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)}
$$
which show us that they never vanish simultaneously.
From the shape of $\frac{\pi}{\theta(\pi-\theta)}$,
which is symmetric and positive on $(0,\pi)$
with global minimum at $\frac{\pi}{2}$
and vertical asymptotes at the endpoints,
we see that it will meet $\tan\theta$
at exactly one point $\theta_0\in(0,\frac{\pi}{2})$;
thus, there is one unique vertical tangent to our curve.
From the graph above, we might estimate
$$
\tan\theta_0\approx\frac{.4}{.25}=1.6
\implies
\theta_0
\approx \frac{\pi}{2}-\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}-\frac{5}{4}\right)}
\approx .8609
$$
where we used the quadratic equation
to deduce $\theta$ from $\tan\theta$.
However, the actual solution seems to be
closer to $0.97803904765198235$:
t=var('t'); find_root(t*(pi-t)*tan(t)==pi, 0, pi/2)
For horizontal tangents, the same function
$\frac{\pi}{\theta(\pi-\theta)}$ must now meet
$-\cot\theta$, which is equivalent to the equation
$$\tan\theta=\theta\left(1-\frac{\pi}{\theta}\right) $$
which on $[0,\pi]$ has only solutions at the two endpoints,
i.e., on our curve, at the origin and
at the horizontal asymptote found above.
$(\cos t,\sin t)$ is the constant speed parametrization of the unit circle. Adding to the coordinates shifts its center, while multiplying these terms scales the circle and hence changes its radius. So as you already recognized, $(1+\cos t,â3+\sin t)$ is a circle or radius $1$ with center $(1,-3)$. The other, $(2+4\cos s,kâ4\sin s)$, is a circle with center $(2,k)$ and radius $4$.
The fact that you have $-4\sin s$ instead of $+4\sin s$ means that the parametrization is the other way round from the conventional one: you start at the righternmost point, but then go down (negative $y$ direction) instead of up. Doesn't change which circles this describes, only what parameter $s$ corresponds to which point on that circle.
So you have two circles, and you want to know how to choose the $y$ coordinate for the center of the second in such a way that the two circles have exactly one point in common, i.e. they touch. There are two ways the circles might touch: they might touch on the outside, or the smaller one might touch the larger one from the inside. In the former case, the distance between the centers has to be the sum of the radii, in the latter case it's the difference of the radii. So you have touching circles for
$$ \sqrt{(2-1)^2+(k+3)^2}\in\{4+1,4-1\}=\{5,3\} $$
To solve this, square the equation and you get
\begin{align*}
(2-1)^2+(k+3)^2 &= 5^2 & (2-1)^2+(k+3)^2 &= 3^2 \\
k^2 + 6k + 9 &= 25 - 1 & k^2 + 6k + 9 &= 9-1 \\
k^2 + 6k - 15 &= 0 & k^2 + 6k + 1 &= 0 \\
k_{1,2} = \frac{-6\pm\sqrt{36+60}}{2} &= -3\pm2\sqrt6 &
k_{3,4} = \frac{-6\pm\sqrt{36-4}}{2} &= -3\pm2\sqrt2 &
\end{align*}
Best Answer
Consider the parametrization of the circle $(x_c(\theta), y_c(\theta))=(r\cos \theta,r\sin \theta)$. It's derivative $(x_c'(\theta), y_c'(\theta))=(-r\sin \theta,r\cos \theta)$ gives the direction of the tangent line at $\theta$. For every $\theta$, this vector is perpendicular to the tangent of the involute: $$ (x_c'(\theta),y_c'(\theta))\cdot(x'(\theta),y'(\theta)) = (-r\sin\theta)(r\theta \cos\theta) + (r\cos\theta)(r\theta \sin \theta) = 0. $$ This proves the statement.
Alternatively, it suffices to note that for every $\theta$ the vector $(\cos \theta,\sin \theta)$ is normal to the circle. And since $(x'(\theta),y'(\theta))$ is just a multiple of the normal, the statement follows.