Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.

Let $r = OB$ be the radius.

Consider any chord PQ passing through A, and let M be the foot of perpendicular from O to PQ.

Let $d = OM$.

By Theorem, $M$ is the midpoint of $PQ$, and $PQ = 2\sqrt{r^2 - d^2}$.

Now in right triangle $OMA$, $OA^2 = OM^2 + MA^2 = d^2 + MA^2$,
so $d^2 = OA^2 - MA^2 \leq OA^2$ (equality iff $MA = 0$, i.e., $M = A$).

Actually we need the other direction.

Let $OA = a$ (fixed).
Then $OM^2 + MA^2 = a^2$, so $OM^2 = a^2 - MA^2 \leq a^2$, with equality iff $MA = 0$, i.e., the foot of perpendicular from $O$ to $PQ$ coincides with $A$.
That happens precisely when $OA \perp PQ$.

We want to minimise $PQ = 2\sqrt{r^2 - d^2}$, which is minimised when $d$ is maximised.
From $d^2 = a^2 - MA^2$, $d$ is maximum when $MA = 0$, i.e., $d = a = OA$.

This maximum is achieved when the foot of perpendicular from $O$ to the chord $PQ$ is $A$ itself, i.e., $OA \perp PQ$.

Hence the shortest chord through $A$ is perpendicular to $OA$, and its length is $2\sqrt{r^2 - OA^2}$.