Let $O$ be the centre of the circle and $A$ be a point inside it. Let $PQ$ be any chord passing through $A$. Draw a perpendicular from $O$ to $PQ$, meeting it at $M$. As given, a perpendicular from the centre to a chord bisects it, so $M$ is the midpoint of $PQ$.

The distance of the chord $PQ$ from the centre is $OM$. We are given that given two unequal chords, the longer chord is closer to the centre. This implies that the shorter chord is farther from the centre. Therefore, to find the shortest chord passing through $A$, we must maximize its distance from the centre, which is $OM$.

In the right-angled triangle $OMA$, $OA$ is the hypotenuse. Therefore, $OA$ is always greater than $OM$. The maximum possible value of $OM$ occurs when $M$ coincides with $A$, making $OM$ equal to $OA$.

When $M$ coincides with $A$, the perpendicular from $O$ to the chord $PQ$ passes through $A$. This means the chord $PQ$ is perpendicular to $OA$. Since the distance from the centre is maximized (equal to $OA$), the chord is the shortest. Thus, the shortest chord passing through $A$ is the one perpendicular to $OA$.