Is there a way to parametrize a line and a circle in the complex plane [by $z = z(t)$], to show that under the inversion function $f(z) = 1/z$, a line is mapped either to a line or a circle, and a circle is mapped either to a line or a circle?
This is simple in the case of translations or dilations: For example, parametrizing a circle by $z(t) = z_0 + Re^{it}$, under the dilation $f(z) = az$, $f[z(t)] = az_0 + aRe^{it}$, which is still the parametrization of a circle.
Yes, but it won't be easy, because a line parameterized by $$ L(t) = a + bt $$ will map to a circle that's parameterized by $$ C(s) = z_0 + Re^{is} $$ where $s$ and $t$ are related by some relatively complicated function. In particular, the relationship between $t$ and $s$ is certainly not linear, for the line is traversed once as $t$ ranges over the real line, while the circle is traversed infinitely often as $t$ ranges over the real line. So that map that expresses $s$ as a function of $t$ must take the whole real line to something like the interval $-\pi < s < \pi$.
In short: it's possible, but possibly not worth the effort. :(
Alternative short summary: While inversion takes lines and circles to lines and circles, it does not carry standard parameterizations of lines to standard parameterizations of circles.
Here are the details for one particular line:
$$ L(t) = 2 + it $$ where $t \in \mathbb R$.
Then after transformation, we get the point $$ M(t) = \frac{1}{2+it} = \frac{2 - it}{4 + t^2} $$ This is a point of the circle whose center is at $z = \frac{1}{4} + 0i$ and whose radius is $\frac{1}{4}$. Subtracting $\frac{1}{4}$ from $M$ gives $$ v(t) = \frac{4- t^2 - 4ti}{4(4+t^2)} $$ as the vector from the circle-center to the point on the circle. It's not too hard to see that the squared length of $v$ is $1/4$, by direct computation and simplification.
Now $v$ points in some direction $\theta$, and since the length of $v$ is $1/2$, we know that the real and imaginary parts of $2v$ are just $\cos \theta$ and $\sin \theta$. so $$ \theta = atan2(-4t, 4-t^2) $$ and we can say that the point $\frac{1}{4} + e^{i\theta}$, under inversion, corresponds to the point $2 + ti$ on the line at $x = 2$.
Going the other way might be simpler.
Let's look at $$ C(s) = \frac{1}{4} (1 + e^{is}) $$ which describes a circle of radius $1/4$ around the point $z_0 = 1/4$.
Under inversion, we get $$ D(s) = \frac{1}{C(s)} = \frac{1}{\frac{1}{4} (1 + e^{is}))} = \frac{4}{1+ e^{is}} = \frac{4}{(1 + \cos s) + i \sin s}. $$ Rationalizing the denominator, we get \begin{align} D(s) &= \frac{4}{(1 + \cos s) + i \sin s}\\ &= \frac{4 \left( (1 + \cos s) - i \sin s \right)}{(1 + \cos s)^2 + \sin^2 s}\\ &= \frac{4 \left( (1 + \cos s) - i \sin s \right)}{1 + 2 \cos s + \cos^2 s + \sin^2 s}\\ &= \frac{4 \left( (1 + \cos s) - i \sin s \right)}{2 + 2 \cos s}\\ &= \frac{2 \left( (1 + \cos s) - i \sin s \right)}{1 + \cos s}\\ &= \frac{2 (1 + \cos s)}{1 + \cos s} - i \frac{\sin s }{1 + \cos s}\\ \end{align} &= 2 - i \frac{\sin s }{1 + \cos s} \end{align} which has the form $2 + it$ where $$ t = -\frac{\sin s }{1 + \cos s}. $$
Hence that circle is transformed into the line $\Re( z )= 2$.
As I said: probably more complicated than it's worth.
By the way, the corresponding transformation for the line $\Re(z) = 1$ turns out to be essentially the "tan(t/2)" substitution used in calculus to turn any rational function of sines and cosines into a rational function in a single real variable. You might want to write it out and see if you can figure out why they're related.