Ever wondered why a straight line just kisses a circle at one point and never crosses it? That magical line is called a tangent, and it hides some neat tricks useful for your CBSE exams.
💡 In Simple Words: A tangent touches a circle at exactly one point. At that point the radius (the line from the centre to the edge) stands straight up like a flagpole, making a perfect right angle with the tangent.
What is a Tangent to a Circle?
A tangent (first time: a line that meets a circle at just one point, called the point of contact) is different from a secant, which cuts the circle twice. The point where they meet is often labeled T. The line from the centre O to T is the radius (the distance from the centre to any point on the circle).
Key Tangent Theorems for CBSE Class 10
- Theorem 1: The radius drawn to the point of contact is perpendicular (forms a 90° angle) to the tangent.
- Theorem 2: From an external point, the two tangents drawn to a circle are equal in length.
- Theorem 3: The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment (the angle subtended by the chord in the opposite arc).
Why these theorems matter
They pop up in many NCERT exercises and board questions. Knowing them saves you time and helps you spot shortcuts.
Proof of Theorem 1 – Radius ⟂ Tangent
Let’s prove the right‑angle claim step by step.
**Proof:**
1. Draw the radius OT.
2. Suppose a line through T (the supposed tangent) meets the circle again at P.
3. In triangle OTP, OT is a side opposite the angle at T.
4. By the Triangle Inequality (the sum of any two sides of a triangle is greater than the third), OT is the smallest distance from the centre to any point on the line TP.
5. The only way for a side to be the shortest distance to a line is when it meets the line at a right angle.
6. Hence, OT ⟂ TP.
That’s why the radius stands upright like a flagpole at the touching point.
Proof of Theorem 2 – Equal Tangents from an External Point
Take a point P outside the circle. Draw two tangents PA and PB touching the circle at A and B respectively.
**Proof:**
1. Join the centre O to P, A, B.
2. Triangles OAP and OBP share side OP.
3. By Theorem 1, OA ⟂ PA and OB ⟂ PB, so each triangle has a right angle at A and B.
4. The hypotenuse of both right triangles is OP.
5. The two triangles are therefore congruent (same hypotenuse and one side equal – OA = OB because both are radii).
6. Congruent triangles give PA = PB.
So any external point gives two equal tangents – a handy fact for distance problems.
Proof of Theorem 3 – Tangent‑Chord Angle
Consider tangent PT at point T and chord TC. Let the angle between them be ∠PTC. The theorem says this equals the angle in the opposite segment, i.e., the angle subtended by chord TC at any point on the circle on the other side, say X (∠TXC).
**Proof Sketch:**
1. Draw the centre O and radius OT.
2. Triangle OTC is isosceles (radii equal).
3. The angle at the centre subtended by chord TC is twice any angle on the circumference (the Inscribed Angle Theorem).
4. Because OT is perpendicular to the tangent, ∠PTC = ½∠TOC.
5. But ½∠TOC is exactly the angle ∠TXC on the opposite arc.
6. Hence, ∠PTC = ∠TXC.
Quick Comparison Table
| Theorem | Statement | Typical Use in Exams |
|---|---|---|
| Radius ⟂ Tangent | Radius drawn to point of contact is perpendicular to tangent. | Prove right angles, find distances. |
| Equal Tangents | Tangents from same external point have equal lengths. | Find unknown lengths, solve geometry problems. |
| Tangent‑Chord Angle | Angle between tangent and chord equals angle in alternate segment. | Angle chasing, prove cyclic quadrilaterals. |
Worked Example
Problem: In a circle with centre O, a tangent at point T meets a line through an external point P. If OP = 13 cm and the radius OT = 5 cm, find the length of the tangent PT.
Solution:
1. From Theorem 2, PT is the distance from P to the point of contact.
2. Triangle OPT is right‑angled at T (Theorem 1).
3. Use Pythagoras: OP² = OT² + PT².
4. Plug values: 13² = 5² + PT² → 169 = 25 + PT² → PT² = 144.
5. Hence PT = 12 cm.
Simple, right? This exact setup appears in many CBSE questions.
How to Remember the Three Tangent Theorems
- Think of a flagpole: the pole (radius) stands straight up, making a right angle with the rope (tangent) that just brushes the flag.
- Imagine two equal‑length strings tied to a pole from the same outside point – they’re the equal tangents.
- Visualise a slice of pizza: the crust (tangent) and a slice edge (chord) create an angle that matches the angle you see on the opposite side of the pizza.
📝 Likely Exam Questions
- State and prove the theorem that a radius drawn to the point of contact of a tangent is perpendicular to the tangent.
Model Answer: (Provide proof as in Theorem 1.) - Two tangents drawn from a point P outside a circle touch the circle at A and B. Prove PA = PB.
Model Answer: (Use congruent right‑triangles as in Theorem 2.) - In a circle, a tangent at T meets a chord TC. Show that ∠PTC equals the angle in the alternate segment.
Model Answer: (Refer to Theorem 3 proof.) - Given OP = 15 cm and radius = 9 cm, find the length of the tangent from P to the circle.
Model Answer: Apply Pythagoras in right triangle OPT → PT = √(15²‑9²)=12 cm. - Explain why the lengths of tangents from an external point are useful in solving coordinate geometry problems involving circles.
Model Answer: They provide equal distances, allowing substitution into distance formulas and simplifying equations.