Step 1: Let $ABCD$ be a rectangle circumscribing a circle, touching it at points $P$, $Q$, $R$ and $S$ on sides $AB$, $BC$, $CD$ and $DA$ respectively.
Step 2: Use the theorem — the lengths of tangents drawn from an external point to a circle are equal. Therefore
$$AP=AS,\quad BP=BQ,\quad CR=CQ,\quad DR=DS$$
Step 3: Add all four equations.
$$AP+BP+CR+DR=AS+BQ+CQ+DS$$
Step 4: Regroup the terms on both sides.
$$(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)$$
Step 5: This gives
$$AB+CD=AD+BC$$
Step 6: Since $ABCD$ is a rectangle, opposite sides are equal: $AB=CD$ and $AD=BC$. Substituting,
$$2AB=2AD \;\Rightarrow\; AB=AD$$
Step 7: So the rectangle has a pair of adjacent sides equal, and all its angles are $90^{\circ}$.
Step 8: A rectangle with all sides equal is a square.
Hence a rectangle circumscribing a circle is a square. Proved.