Classic NCERT theorem. The radius to the point of contact is perpendicular to the tangent. Proof by contradiction.

Prove: Tangent to a Circle is Perpendicular to Radius — Geometry Proof

Question

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

This is Theorem 10.1 from NCERT Class 10, Chapter 10. It’s a proof-by-contradiction question — the kind that appears directly in board exams as a 3-mark or 5-mark proof.

Solution — Step by Step

Let $O$ be the centre of the circle with radius $r$ , and let $P$ be any point on the circle. Draw tangent $XY$ at $P$ . We need to prove $OP \perp XY$ .

Assume $OP$ is NOT perpendicular to $XY$ . If that’s true, we can drop a perpendicular from $O$ to line $XY$ — call that foot $Q$ . So $OQ \perp XY$ , and $OQ \neq OP$ .

Why do we do this? Because contradiction proofs work by assuming the opposite of what we want to prove, then showing something impossible follows.

In right triangle $OQP$ , the hypotenuse is $OP$ (since $\angle OQP = 90°$ ). In any right triangle, the hypotenuse is always the longest side. So:

OQ < OP

We know $OP = r$ (radius). Since $OQ < OP = r$ , point $Q$ lies strictly inside the circle.

But $Q$ is on line $XY$ , the tangent. A tangent touches the circle at exactly one point and lies completely outside (except at the point of contact). So $Q$ can’t be inside the circle — contradiction!

Our assumption that $OP$ is not perpendicular to $XY$ leads to an impossibility. Therefore, $OP \perp XY$ . Hence proved.

Why This Works

The key property we’re using is the definition of a tangent: it meets the circle at exactly one point, meaning every other point on the tangent line lies outside the circle.

When we assumed $OP$ wasn’t the perpendicular, we were forced to create a shorter line $OQ$ from the centre to the tangent. But “shorter than the radius” means “inside the circle” — which the tangent line can never be (except at $P$ ).

The proof is really about a distance argument. The minimum distance from a point to a line equals the perpendicular distance. If $OP$ were not perpendicular, then $OQ < OP$ would mean $O$ is closer to the tangent at $Q$ than at $P$ — but $P$ is supposed to be the only contact point.

Alternative Method

You can also prove this using the property that $P$ is the closest point on $XY$ to centre $O$ .

For any other point $Q$ on $XY$ , we argue: $OQ^2 = OP^2 + PQ^2$ (if $OP \perp XY$ ) or just directly that since every point other than $P$ on $XY$ lies outside the circle, $OQ > r = OP$ for all $Q \neq P$ .

The foot of perpendicular from $O$ to $XY$ must be the point at minimum distance — and that minimum distance point is $P$ itself. So $OP \perp XY$ .

This approach is slightly more intuitive for students who think geometrically.

Common Mistake

Many students write “since $OQ$ is perpendicular to $XY$ , $OQ$ is the shortest distance, therefore $OQ < OP$ ” — and then forget to connect why that’s a contradiction. The proof breaks if you don’t explicitly state: $Q$ lies inside the circle (since $OQ < r$ ), but a tangent cannot pass through an interior point. Without that sentence, the contradiction is incomplete and you’ll lose marks.

In board exams, always draw and label the diagram in this proof — examiners explicitly award 1 mark for a correct figure. Label $O$ , $P$ , $Q$ , $XY$ , the right angle at $Q$ , and mark $OP = r$ . Takes 30 seconds, gets you a free mark.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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