1. Chapter at a Glance
A bicycle wheel touches the road at just one point at any moment. A stone whirled on a string flies off along a straight line that just touches its circular path. Such a line, touching a circle at exactly one point, is a tangent. In Class 9 you studied chords and angles in a circle; now we study tangents. Two theorems carry this chapter — a tangent is perpendicular to the radius at the point of contact, and the two tangents from an external point are equal — and both proofs are often asked in board exams.
- A line and a circle: non-intersecting, secant, tangent.
- Tangent at a point; point of contact.
- Theorem: tangent ⊥ radius at the point of contact.
- Number of tangents from a point inside, on and outside a circle.
- Theorem: lengths of tangents from an external point are equal.
- Applications: quadrilaterals around a circle, concentric circles, angle problems.
2. A Line and a Circle
| Case | Common points | Name |
|---|---|---|
| Line misses the circle | 0 | Non-intersecting line |
| Line cuts the circle | 2 | Secant |
| Line touches the circle | 1 | Tangent |
A tangent can be thought of as a special secant whose two intersection points have moved together into one point.
3. Tangent and Radius
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
- Let XY be the tangent at P to a circle with centre O. Take any other point Q on XY.
- Q lies outside the circle (otherwise XY would be a secant), so OQ > OP (OP is a radius).
- This holds for every point Q ≠ P on XY, so OP is the shortest distance from O to XY.
- The shortest segment from a point to a line is the perpendicular. Hence OP ⊥ XY. ∎
Consequences: exactly one tangent can be drawn at a point on a circle; the line through the point of contact perpendicular to the tangent passes through the centre.
4. Number of Tangents from a Point
| Position of point | Number of tangents |
|---|---|
| Inside the circle | 0 |
| On the circle | Exactly 1 |
| Outside the circle | Exactly 2 |
The length of a tangent from external point P is the length PT from P to the point of contact T.
5. Tangents from an External Point
The lengths of tangents drawn from an external point to a circle are equal.
- Join OQ, OR and OP. ∠OQP = ∠ORP = 90° (Theorem 10.1).
- In right triangles OQP and ORP: OQ = OR (radii), OP is common (hypotenuse).
- By RHS, △OQP ≅ △ORP, so PQ = PR (CPCT). ∎
Also from the congruence: ∠OPQ = ∠OPR — the centre lies on the bisector of the angle between the two tangents; and ∠QOP = ∠ROP.
Angle relation: in quadrilateral OQPR, ∠Q = ∠R = 90°, so ∠QPR + ∠QOR = 180°.
6. Applications
From a point P 25 cm from the centre of a circle, the tangent length is 24 cm. Find the radius.
Solution: r = √(25² − 24²) = √49 = 7 cm.
Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that touches the smaller one.
Solution: The chord is tangent to the small circle, so the radius 3 cm is perpendicular to it and bisects it. Half-chord = √(25 − 9) = 4 → chord = 8 cm.
A circle touches all four sides of quadrilateral ABCD. Prove AB + CD = AD + BC.
Proof: Let the sides touch at P (AB), Q (BC), R (CD), S (DA). Equal tangents: AP = AS, BP = BQ, CR = CQ, DR = DS. Add: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), i.e. AB + CD = AD + BC. ∎
Consequence: a parallelogram circumscribing a circle must be a rhombus (AB + CD = AD + BC with AB = CD and AD = BC gives AB = AD).
Two tangents PA and PB are drawn to a circle with centre O such that ∠APB = 80°. Find ∠POA.
Solution: ∠AOB = 180° − 80° = 100°. OP bisects it, so ∠POA = 50°.
7. Solved Problems (Board Pattern)
Prove that the tangents drawn at the ends of a diameter are parallel.
Solution: Let AB be a diameter with tangents l at A and m at B. OA ⊥ l and OB ⊥ m, and OA, OB lie on the same line AB. Two lines perpendicular to the same line are parallel, so l ∥ m. ∎
Prove that the angle between two tangents from an external point is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution: In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Angle sum 360° → ∠APB + ∠AOB = 180°. ∎
A circle is inscribed in △ABC touching AB, BC, CA at P, Q, R. If AB = 10, BC = 8 and CA = 12 (cm), find AP, BQ and CR.
Solution: Let AP = AR = x, BP = BQ = y, CQ = CR = z. x + y = 10, y + z = 8, z + x = 12. Adding: x + y + z = 15. So x = 7, y = 3, z = 5. AP = 7, BQ = 3, CR = 5 cm.
8. Master Formula Table
| Result | Statement |
|---|---|
| Tangent ⊥ radius | OP ⊥ tangent at P |
| Tangent length | PT = √(OP² − r²) |
| Equal tangents | PQ = PR from external P |
| Angle relation | ∠APB + ∠AOB = 180° |
| Circumscribed quadrilateral | AB + CD = AD + BC |
| Concentric chord | Chord = 2√(R² − r²) |
| Incircle tangent lengths | x = s − a, y = s − b, z = s − c (s = semi-perimeter) |
9. Master Summary Table
| Topic | Key idea | Exam focus |
|---|---|---|
| Tangent definition | One common point | MCQ / 1 mark |
| Theorem 10.1 | Shortest distance is perpendicular | Proof; right triangles |
| Number of tangents | 0, 1, 2 by position | MCQ |
| Theorem 10.2 | RHS congruence | Proof (3 marks) |
| Applications | Equal tangent segments | Quadrilaterals, triangles, angles |
10. Mind Map
- Secant (2)
- Tangent (1)
- Tangent ⊥ radius
- PT² = OP² − r²
- Equal tangents
- OP bisects angle
- AB + CD = AD + BC
- Concentric circles
11. Tips and Tricks
- Whenever a tangent appears, join the centre to the point of contact — you get a right angle.
- Label equal tangent segments with the same letter (x, y, z) at each vertex.
- In angle problems, use ∠APB + ∠AOB = 180° directly.
- Assuming the tangent is perpendicular to a chord — it is perpendicular to the radius.
- Saying infinitely many tangents can be drawn from an external point (only two).
- Forgetting that the radius bisects the chord of the bigger circle in concentric circle problems.
12. Important One-liners
- A tangent touches a circle at exactly one point.
- A circle can have infinitely many tangents, but only one at a given point.
- The tangent at a point is perpendicular to the radius there.
- Exactly two tangents can be drawn from an external point, and they are equal.
- Tangents at the ends of a diameter are parallel.
- A parallelogram circumscribing a circle is a rhombus.
13. Exam-pattern Questions with Detailed Answers
How many tangents can a circle have?
- One
- Two
- Infinitely many
- None
Show answer
One at each point, and a circle has infinitely many points.
Answer: (C)
The length of a tangent from a point 5 cm from the centre of a circle of radius 3 cm is:
Show answer
√(25 − 9) = 4 cm.
Answer: 4 cm
If tangents PA and PB from P make ∠APB = 60°, find ∠AOB.
Show answer
180° − 60° = 120°.
Answer: 120°
Two concentric circles have radii 13 cm and 5 cm. Find the length of the chord of the larger circle that touches the smaller one.
Show answer
2√(169 − 25) = 2 × 12 = 24 cm.
Answer: 24 cm
A quadrilateral ABCD circumscribes a circle with AB = 6, BC = 7 and CD = 4 cm. Find AD.
Show answer
AB + CD = AD + BC → 10 = AD + 7 → AD = 3 cm.
Answer: 3 cm
PQ is a tangent at Q to a circle with centre O. If OP = 12 cm and PQ = 6√3 cm, find the radius and ∠OPQ.
Show answer
r = √(144 − 108) = 6. sin ∠OPQ = 6/12 = ½ → ∠OPQ = 30°.
Answer: 6 cm; 30°
In the figure for Theorem 10.2, if ∠QOR = 110°, find ∠QPR and ∠OPQ.
Show answer
∠QPR = 70°; OP bisects it → ∠OPQ = 35°.
Answer: 70° and 35°
Prove that a parallelogram circumscribing a circle is a rhombus.
Show answer
AB + CD = AD + BC. In a parallelogram AB = CD and AD = BC, so 2AB = 2AD → AB = AD. All sides equal → rhombus. ∎
Answer: Proved
Prove that the tangents at the ends of a chord of a circle make equal angles with the chord.
Show answer
Let the tangents at A and B meet at P. PA = PB (Theorem 10.2), so △PAB is isosceles and ∠PAB = ∠PBA. ∎
Answer: Proved
A circle of radius 4 cm is inscribed in △ABC. The point of contact D divides BC into BD = 8 cm and DC = 6 cm. Find AB and AC.
Show answer
Let the tangent length from A be x. Sides: AB = x + 8, AC = x + 6, BC = 14. s = x + 14. Area by Heron = √[(x + 14)(x)(6)(8)]; also area = r × s = 4(x + 14). Square: 48x(x + 14) = 16(x + 14)² → 3x = x + 14 → x = 7. AB = 15 cm, AC = 13 cm.
Answer: AB = 15 cm, AC = 13 cm
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.
Show answer
Joining O to the vertices and to the points of contact makes 8 angles at O in 4 equal pairs (by RHS congruences): call them a, a, b, b, c, c, d, d with 2(a + b + c + d) = 360°. ∠AOB + ∠COD = (a + b) + (c + d) = 180°. ∎
Answer: Proved
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.
Show answer
Let ∠PTQ = θ. TP = TQ, so ∠TPQ = ∠TQP = (180° − θ)/2 = 90° − θ/2. ∠OPT = 90°, so ∠OPQ = 90° − ∠TPQ = θ/2. Hence ∠PTQ = 2∠OPQ. ∎
Answer: Proved
14. Quick Revision Summary
- Tangent: one common point; secant: two.
- Tangent ⊥ radius at the point of contact.
- 0, 1, 2 tangents from a point inside, on, outside.
- Tangents from an external point are equal; centre lies on the angle bisector.
- Circumscribed quadrilateral: AB + CD = AD + BC.