Notation
Let $O$ be the centre of the disc, let $r$ be the radius of the disc and $d$ be the distance between point $O$ and point $A$.
Result
We can reduce the cut length of this problem whenever $\frac{d}{2r} < \alpha$ where $\alpha$ is the positive real solution to the equation $\cos (\pi x) = x$ (Cut 2) and whenever $\frac{d}{2r} > \beta$ where $\beta$ is the smallest positive real solution to the equation $\cos(2-2x) = x$ (Cut 3).
Proof
Cut 1 (Henning Makholm's solution)
We imagine a circle of radius $r$ and centre $A$ and cut along where this circle intersects our disc. Then slide the piece not containing $A$ to the shaded region
In this case, the length of the cut $C_1$ is equal to the length of the arc which subtends an angle of $2\theta$ at $O$
That is, $$ C_1 = 2r\theta . $$
Completing the isoceles triangle in the diagram by joining $A$ to the point of intersection of the circles, we find that
$$ \frac{r}{\sin \theta} = \frac{d}{\sin(\pi - 2\theta)} = \frac{d}{2\sin \theta \cos \theta}$$
In other words,
$$\theta = \arccos \left(\frac{d}{2r} \right),$$
and so
$$C_1 = 2r \arccos \left(\frac{d}{2r} \right).$$
Cut 2 (The circle cut)
This time we cut along the circle which has diameter $OA$, then rotate this circle through an angle of $\pi$ so that $A$ now coincides with the centre of the disc. 
Clearly, in this case, the length of the cut is
$$C_2 = \pi d$$.
NB We may want to be careful here and say that the points $O$ and $A$ should lie inside the circle. In that case, the length of our cut is $$C_2 = \pi d + \epsilon$$ where $\epsilon$ is an arbitrarily small positive quantity. This does not affect the result relating to the relationship between Cut 2 and Cut 1.
Then we have
$$C_2 < C_1$$ $$\Leftrightarrow \pi d < 2r \arccos \left(\frac{d}{2r} \right)$$ $$\Leftrightarrow \cos \left(\frac{\pi d}{2r} \right) > \frac{d}{2r}$$
where the last line follows because $\cos$ is decreasing on the range $\left(0, \frac{\pi}{2} \right)$.
This last inequality holds whenever $\frac{d}{2r}$ is less than the positive real solution to the equation $\cos \pi x = x$ and so the cut $C_2$ is shorter than $C_1$ whenever
$$\frac{d}{2r} < \alpha \approx 0.376967$$
Cut 3 (Florian F's cut)
In this case, we cut two arbitrarily close line segments parallel and equidistant to the line $OA$ such that the line segment extend from the circle, at points $P$ and $Q$ to a length of approximately $2r-d$, within the circle, as shown in the diagram. In fact, if the angle $POQ$ is $\theta$ then this length is $2r - d - 2r\sin \left(\frac{\theta}{2} \right) \sin\left(\frac{\theta}{4} \right)$. We then round off the cut with a reflection of the arc of the circle $PQ$ extending between the endpoints of the two line segments. We can slide out this piece containing $O$ and $A$ from this disc, rotate it through an angle of $\pi$ and slide it back in so that $A$ becomes the new centre of the disc.
The length of this cut is approximately given by $$C_3 = 4r-2d.$$ In fact, if $\theta$ is as before, then the formula for the length of the cut is $$C_3 = 4r - 2d + 2r\left(\frac{\theta}{2} - 2\sin \left(\frac{\theta}{2} \right) \sin\left(\frac{\theta}{4} \right) \right)$$ which is greater than $4r - 2d$ but we can make it arbitrarily close as $\theta \rightarrow 0$.
Then we have
$$C_3 < C_1$$ $$\Leftrightarrow 4r-2d < 2r \arccos \left(\frac{d}{2r} \right)$$ which holds whenever $\frac{d}{2r}$ is greater than the smallest positive solution to the equation $\cos(2-2x) = x$ and so cut $C_3$ is shorter than cut $C_1$ whenever $$\frac{d}{2r} > \beta \approx 0.445428$$
NB One thing that Florian F seemed to hint at in the comments was whether we could consider this cut to have length $2r-d$ in the limit as the lines get closer together. This is an interesting point but my suspicion would be that all cut pieces should be two-dimensional so this case is not allowed.
