Tangent theorem (radius perpendicular to tangent)

Touch a circle with a ruler at exactly one point, and something exact and unavoidable happens at that point: the line from the centre to the touching point always meets the ruler at a perfect right angle. This is the tangent theorem — the radius drawn to the point of contact is perpendicular to the tangent there, no matter how the circle is sized or how the tangent is tilted. It is one of only two theorems in the CBSE Class 10 Circles chapter that you must be able to prove, not just quote, and it quietly powers almost every tangent-length problem you will meet. The moment you spot a radius meeting a tangent at the point of contact, you have a free $90^\circ$ angle — and a free right triangle to attack with Pythagoras or trigonometry. This section proves the theorem from first principles, applies it to find an unknown distance, and flags the traps students fall into when the right angle in a circle diagram isn't where they assume it is.

The core insight

The radius drawn to the point where a tangent touches a circle is always perpendicular to that tangent, turning every tangent problem into a right-triangle problem.

The core idea

Theorem. The tangent at any point of a circle is perpendicular to the radius through the point of contact. If $O$ is the centre of a circle, $P$ is a point on the circle, and $XY$ is the tangent to the circle at $P$ , this says

OP \perp XY.

where: $O$ is the centre, $P$ is the point of contact (where the tangent touches the circle), $OP$ is the radius to that point, and $XY$ is the tangent line itself. The proof rests on one clean fact: a tangent touches the circle at exactly one point, so every other point on the tangent line lies strictly outside the circle. Combine that with the geometric rule that the shortest distance from a point to a line is always along the perpendicular, and the theorem falls out directly (full proof below). Once you know $OP \perp XY$ , any triangle formed by the centre $O$ , the point of contact $P$ , and a further point $A$ on the tangent is right-angled at $P$ . That hands you the Pythagoras relation for free whenever $PA$ is a tangent segment from an external point $A$ :

OP^2 + PA^2 = OA^2.

where: $OP$ is the radius $r$ , $PA$ is the length of the tangent from the external point, and $OA$ is the straight-line distance from the centre to that external point.

Figure

Centre O joined to point of contact P on a circle with the tangent line XY drawn at P and a right angle mark showing OP perpendicular to XY

Why it works

Derivation AProof: radius is perpendicular to the tangent at the point of contact

The goal is to show rigorously, using only the fact that a tangent touches a circle at a single point, that the segment joining the centre to that single touching point must be the perpendicular from the centre to the tangent line, and nothing else about the circle's size or the tangent's direction is needed to reach that conclusion.

Step 1. Let the circle have centre $O$ and radius $r$ , and let the tangent $XY$ touch the circle at the point $P$ . By the defining property of a tangent, $P$ is the only point that $XY$ shares with the circle.

Step 2. Pick any other point $Q$ on the line $XY$ , with $Q$ different from $P$ . Since $Q$ lies on the tangent but is not the point of contact, $Q$ cannot lie on the circle. Because the interior of the circle is entirely enclosed by the circle and the tangent does not cross into it (a tangent, unlike a secant, never enters the circle), $Q$ must lie outside the circle.

Step 3. Every point outside the circle is farther from the centre than every point on the circle, so $OQ > r$ . But $OP = r$ , because $P$ is on the circle. Hence $OQ > OP$ for this arbitrary point $Q$ .

Step 4. Step 3 holds for every point $Q$ on $XY$ other than $P$ — the choice of $Q$ was arbitrary. So $OP$ is shorter than the segment from $O$ to any other point of the line $XY$ , which means $OP$ is the shortest distance from $O$ to the entire line $XY$ .

Step 5. A basic fact of plane geometry is that the shortest distance from a point to a line is always measured along the perpendicular from that point to the line. Since $OP$ realizes this shortest distance, $OP$ must itself be that perpendicular. Therefore $OP \perp XY$ , which is exactly the tangent theorem. $\blacksquare$

Worked example

Problem. In a circle of radius $6\text{ cm}$ , $PA$ is a tangent at point $P$ on the circle from an external point $A$ , with $\angle OAP = 30^\circ$ , where $O$ is the centre. Find the length of $OA$ .

Step 1. By the tangent theorem, the radius $OP$ is perpendicular to the tangent $PA$ at the point of contact $P$ , so triangle $OPA$ is right-angled at $P$ :

\angle OPA = 90^\circ.

Step 2. In this right triangle, $OP$ is the side opposite $\angle OAP$ , and $OA$ is the hypotenuse. Using the sine ratio for $\angle OAP = 30^\circ$ :

\sin(\angle OAP) = \frac{OP}{OA} \implies \sin 30^\circ = \frac{OP}{OA}.

Step 3. Substitute $OP = 6\text{ cm}$ (the radius) and $\sin 30^\circ = \dfrac{1}{2}$ :

\frac{1}{2} = \frac{6}{OA} \implies OA = 12\text{ cm}.

Answer: $OA = 12\text{ cm}$ .

Competitive stretch

The same figure also lets you find the tangent length $PA$ using the Pythagoras relation, without any new information. Since $\triangle OPA$ is right-angled at $P$ :

OP^2 + PA^2 = OA^2 \implies 6^2 + PA^2 = 12^2 \implies PA^2 = 144 - 36 = 108,

so $PA = \sqrt{108} = 6\sqrt{3}\text{ cm} \approx 10.39\text{ cm}$ . Notice this matches what you'd get directly from $\tan 30^\circ = OP/PA$ , since a $30$ - $60$ - $90$ triangle always has side ratio $1 : \sqrt{3} : 2$ opposite its three angles.

Common traps

Thinking the perpendicularity is only approximate. It is exact and universal at the point of contact — it never depends on the circle's radius or on which direction the tangent runs.
Applying the theorem at the wrong point. The radius is perpendicular to the tangent only at the point of contact; a radius drawn to any other point on the circle has no special angle with a tangent drawn elsewhere.
Skipping the shortest-distance justification in the proof. Jumping straight from "tangent touches at one point" to "so the radius is perpendicular" without the intermediate shortest-distance argument loses marks in a full proof answer.
Misreading which angle is $90^\circ$ in busy figures. When a diagram has a radius, a tangent, and a chord all meeting near one point, identify the radius-tangent pair carefully before assuming any other angle equals $90^\circ$ .

Ask the tutor: If the shortest-distance step of the proof feels shaky, or you want more practice spotting the hidden right angle in tricky circle diagrams, ask the AI tutor for a guided walkthrough.

Why is the radius perpendicular only at the point of contact, and not anywhere else on the circle

A tangent touches the circle at just one point, and the "shortest distance is along the perpendicular" argument in the proof only applies to that specific point. If you draw a radius to a different point on the circle, that radius is not even touching the same tangent line at all, so there is no perpendicularity to claim. The 90 degree angle is a property of the pair "this radius, this tangent, at this shared point" — not a property of circles in general.

Do I need to memorize the full proof for exams, or just the result

CBSE lists this among the theorems you must be able to prove, so yes, memorize the logical chain: pick another point Q on the tangent, show Q lies outside the circle, conclude OP is the shortest distance from O to the line, then invoke the fact that shortest distance to a line is along the perpendicular. Each step earns marks separately, so state them in order rather than jumping straight to the conclusion.

How is this theorem actually used in problems, since it just gives one angle

That single right angle is usually the entire unlock. Once you know a radius and tangent meet at 90 degrees, the triangle formed by the centre, the point of contact, and any external point on the tangent becomes a right triangle, and you can bring in Pythagoras or sine, cosine, and tangent ratios to solve for unknown lengths or angles, exactly as in the worked example above.