Just like @alephzero already mentioned, the most general way of describing motion with polar coordinates is to write $r$ and $\theta$ as a function of time, i.e. $r(t)$ and $\theta(t)$. When deriving the path equation, we start from the acceleration formulas in polar coordinates and the fact that the body is in a central force field (inverse square in the case of orbital motion due to gravity), such that:
$$ a_{\theta} = r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt} = 0$$
$$ a_r = \frac{d^2r}{dt^2}+r(\frac{d\theta}{dt})^2 = F(r) $$
Our initial goal is to find two functions $r(t)$ and $\theta(t)$, that satisfy the above equations. However, this is hard to carry out analytically. Instead we make the assumption that $r$ is dependent on the variable $\theta$. We also know that the reduced angular momentum $L'$ is constant, because we have a central force field.
If $r$ is dependent on $\theta$, and the reduced angular momentum is zero, i.e. $r^2\frac{d\theta}{dt} = 0$, this implies that either $r(t) = 0 $ or $\frac{d\theta}{dt}=0$. In both cases the above equations for $a_r$ and $a_{\theta}$ vanish, so that we don't even have an equation to start with, we can't even speak of motion actually.
If $r$ is dependent on $\theta$, and $L'\not=0$ we can, after some manipulation of the above equations, eliminate the variable $t$, such that we try to find $r$ in function of the angle $\theta$, without wanting to know where the particle is at what time $t$.
The final result is the path equation, which describes the path the object takes. The path equation makes the assumption that $r$ is dependent on $\theta$ and the reduced angular momentum $L'\not=0$. The radial trajectory is the case where the angular momentum is zero, and thus no solution of the path equation.
