Why Tangents Matter in Circle Problems
Ever wondered why a straight line can just kiss a circle at one point? That's a tangent, and it pops up in lots of CBSE questions.
💡 In Simple Words: A tangent is a line that touches a circle at exactly one point and never crosses it. Think of a pencil just grazing a round ball – that tiny touch is the tangent.
What Is a Tangent? (Definition)
A tangent to a circle is a line that meets the circle at one point only. That single point is called the point of tangency.
Key Theorems About Tangents
- Perpendicular Radius Theorem: The radius drawn to the point of tangency is perpendicular (forms a 90° angle) to the tangent line.
- Tangent–Chord Angle Theorem: The angle formed between a tangent and a chord through the point of contact equals the angle in the alternate segment of the circle.
- Tangent from an External Point Theorem: From a point outside the circle, the two tangents drawn to the circle are equal in length.
Perpendicular Radius Theorem – Quick Proof
Take a circle with centre O and a tangent line at point P. Draw the radius OP. If OP weren’t perpendicular, you could slide the line a little and still touch the circle, which would give two points of contact – contradicting the definition of a tangent. Hence OP ⟂ tangent.
Tangent–Chord Angle Theorem – Sketch of Proof
Let the tangent at point T meet chord AB at T. Draw radii OA and OB. Because OA = OB (radii of the same circle), triangle OAT is congruent to OBT by side‑angle‑side. This shows that ∠(T A B) equals the angle in the opposite arc, proving the theorem.
Tangent from an External Point – Proof Idea
Let P be an outside point, and PT₁, PT₂ be two tangents touching the circle at T₁ and T₂. Join O to P, O to T₁, and O to T₂. Triangles OPT₁ and OPT₂ share OP and have OT₁ = OT₂ (radii) and right angles at T₁, T₂. By hypotenuse‑leg (HL) congruence, PT₁ = PT₂.
Worked Example
Problem: In a circle with centre O, a tangent at point A meets the extension of radius OB at point C. If OA = 5 cm and ∠AOB = 30°, find the length of AC.
Solution: Since OA is a radius, OA ⟂ tangent at A. So ∠OAC = 90°. Triangle OAC is right‑angled at A. Using the sine rule in ΔOAB: sin 30° = opposite/hypotenuse = AB/5, so AB = 2.5 cm. Because C lies on the line OB, OC = OA + AC = 5 + AC. Also, in ΔOAC, sin 30° = OA/OC = 5/(5+AC). Solve 0.5 = 5/(5+AC) → 5+AC = 10 → AC = 5 cm.
Quick Comparison Table
| Theorem | What It Says | Typical Use in Exams |
|---|---|---|
| Perpendicular Radius | Radius ⟂ tangent at point of contact | Identify right angles, find missing lengths |
| Tangent–Chord Angle | Angle between tangent and chord = angle in opposite arc | Relate angles, solve for unknown arcs |
| Equal Tangents | Two tangents from same external point are equal | Compute lengths of two tangent segments |
Common Mistakes to Avoid
- Assuming a tangent can intersect the circle at two points – it can’t.
- Mixing up the angle formed by tangent‑chord with the angle subtended by the chord at the centre – they’re different.
- Forgetting that the radius to the point of tangency is always perpendicular.
📝 Likely Exam Questions
- State and prove the theorem that the radius drawn to the point of tangency is perpendicular to the tangent.
Answer: (Provide definition, draw figure, argue by contradiction, conclude perpendicular.) - Two tangents are drawn from a point P to a circle, touching at A and B. Show that PA = PB.
Answer: Connect centre O to P, A, B; note right angles; apply HL congruence. - In a circle, a tangent at point T meets a secant through T at point S. If the radius is 7 cm and ∠OTS = 45°, find the length of ST.
Answer: Use right‑triangle OTS, sin 45° = 7/ST, so ST = 7/ sin45° = 7√2 ≈ 9.9 cm. - Given a circle with centre O, a chord AB and a tangent at B. Prove that ∠ABT equals the angle in the alternate segment.
Answer: Use the tangent‑chord theorem, reference the intercepted arc. - Find the length of the tangent from an external point P that is 13 cm from the centre of a circle of radius 5 cm.
Answer: Use PT² = PO² – R² → PT = √(13² – 5²) = √(169‑25) = √144 = 12 cm.