Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Step-by-Step Solution
Key Concept: The radius drawn to the point of contact of a tangent is perpendicular to the tangent. Hence the line drawn through the point of contact and perpendicular to the tangent must be the extension of the radius, which passes through the centre of the circle.
1. Let the given circle be $\mathcal{C}$ with centre $O$ and radius $r$.
2. Let $P$ be the point of contact of the tangent $t$ to the circle $\mathcal{C}$.
3. Draw the radius $OP$ joining the centre $O$ to the point of contact $P$.
4. Consider any point $Q$ on the tangent line $t$ different from $P$. Since $P$ is the only common point of the line $t$ and the circle, $PQ$ is a chord of the circle extended beyond $P$.
5. Apply the Pythagoras theorem in triangle $\triangle OQP$:
$$OP^{2}=OQ^{2}+PQ^{2} \quad \text{(if $OP$ were not perpendicular to $t$)}$$
But $OP = r$ is the shortest distance from $O$ to any point on the line $t$. Hence the equality can hold only when $PQ=0$, i.e., when $Q=P$.
6. Therefore the distance from the centre to the line $t$ is minimum at $P$, which implies that $OP$ is perpendicular to the tangent $t$.
7. Hence the line drawn through $P$ perpendicular to the tangent coincides with the radius $OP$ and must pass through the centre $O$.
8. Conclusion: The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
Mathematical statement:
$$\text{If } t \text{ is a tangent to the circle } (O,r) \text{ at } P, \text{ then } OP \perp t \text{ and the line through } P \text{ perpendicular to } t \text{ contains } O.$$
Correct Answer: Since the radius drawn to the point of contact of a tangent is always perpendicular to the tangent, the line through the point of contact and perpendicular to the tangent is the extension of that radius. Hence it passes through the centre of the circle.