$D^*\subset\mathbb{R}^2$ is the punctured unit disc, i.e. the set $ \{ (u,v) \in \mathbb{R}^2 | 0 < u^2 + v^2 < 1 \}$.
Here's what I've got so far: two points in $D^*$ can be joined by a radial line to the circle with (Euclidean) radius $\frac{1}{2}$ centred on the origin, an arc of that circle, and another radial line. I've shown that radial lines towards the edge of the unit circle, and arcs of the circle of radius $\frac{1}{2}$, are bounded in length whatever $a$ is. I've also shown that radial lines within the circle of radius $\frac{1}{2}$ have bounded length if $a<1$, and have unbounded length otherwise. Hence I know that distances are bounded in $D^*$ with this metric if $a<1$.
Some informal reasoning to go further: I suspect that a radial line is the shortest piecewise smooth curve joining its endpoints, and so distances in $D^*$ are unbounded if $a \geq 1$. My reasoning behind this is that, because the Riemannian metric is a local scaling of the Euclidean one, with the scaling depending only on the (Euclidean) distance from the origin, any curve which is longer in the Euclidean sense must be longer in the Riemannian sense too. This is because the curve has to pass through all the same distances from the origin as the radial line and so undergoes the same stretching of distances in some places, but there is more of the curve to stretch.
I'm struggling to turn this reasoning into something concrete though. Is it possible to do so?
Note: I'm aware that it's possible to use the Euler-Lagrange equations to prove that a curve is geodesic. However, this more general theory is not approached until later in the textbook I'm working from, so I assume it's not necessary for the question. Anyway, from what I understand from glancing at that section quickly, that only means the curves are locally, rather than globally, length-minimising.