Prove that the tangent at any point on a circle is perpendicular to radius

Question

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

(NCERT Class 10, Circles — Theorem 10.1)

Solution — Step by Step

Let a circle have centre $O$ and radius $r$ . Let $PQ$ be a tangent to the circle at point $P$ . We need to prove that $OP \perp PQ$ (i.e., $\angle OPQ = 90°$ ).

We’ll use proof by contradiction (indirect proof).

Assume that $OP$ is NOT perpendicular to $PQ$ .

If $OP$ is not perpendicular to the tangent $PQ$ , then there exists some other point, say $R$ on $PQ$ , such that $OR \perp PQ$ .

Since $OR \perp PQ$ , the segment $OR$ is the shortest distance from $O$ to the line $PQ$ . This means:

OR < OP

But $OP = r$ (radius of the circle). So $OR < r$ .

This means $R$ is a point on line $PQ$ that is closer to the centre than the radius. But if $OR < r$ , then $R$ lies inside the circle.

If $R$ (a point on line $PQ$ ) lies inside the circle, then the line $PQ$ must enter the circle at $R$ . A line passing through an interior point of a circle must intersect the circle at two points.

But $PQ$ is a tangent — by definition, it touches the circle at exactly one point ( $P$ ). Having a second intersection point contradicts this.

Therefore, our assumption that $OP$ is not perpendicular to $PQ$ must be wrong.

Hence, $OP \perp PQ$ . Proved. $\blacksquare$

Why This Works

The proof rests on two key facts:

The perpendicular from a point to a line is the shortest distance from that point to the line
A tangent touches a circle at exactly one point (any line that enters the circle must cross it at two points)

If the radius weren’t perpendicular to the tangent, we could find a shorter segment from the centre to the tangent line — but that shorter distance would place a point of the tangent inside the circle, forcing the tangent to actually be a secant (intersecting at two points). This contradicts the tangent property.

The converse is also true: a line perpendicular to the radius at its endpoint on the circle is a tangent. This converse is used frequently in construction problems.

Alternative Method

A coordinate geometry approach: place the circle at origin with equation $x^2 + y^2 = r^2$ . At point $P(a, b)$ on the circle, the tangent has equation $ax + by = r^2$ . The slope of $OP$ is $b/a$ . The slope of the tangent is $-a/b$ . Product of slopes $= (b/a)(-a/b) = -1$ . Since the product is $-1$ , they are perpendicular.

This theorem is one of the most important in CBSE Class 10 circles chapter. It’s used as a stepping stone in many proofs: tangent-tangent angle problems, tangent-chord angles, and proving that tangents from an external point are equal. Make sure you can write this proof from memory — it appears as a 3-mark question almost every year.

Common Mistake

Students sometimes try to “prove” this by drawing the figure and saying “it looks perpendicular.” That’s not a proof — it’s an observation. The proof must use logical reasoning (contradiction, in this case) to show WHY it must be perpendicular. Another error: using the result to prove itself (circular reasoning). You cannot assume the tangent is perpendicular to derive that it’s perpendicular.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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