Class 10 Mathematics Notes Chapter 10 (Chapter 10) – Examplar Problems (English) Book

Alright class, let's focus on Chapter 10: Circles, from your NCERT Exemplar book. This chapter is crucial not just for your board exams but also forms the basis for many geometry questions in government competitive exams. We'll cover the key concepts, theorems, and problem-solving techniques emphasized in the Exemplar, which often tests deeper understanding.

Chapter 10: Circles - Detailed Notes for Competitive Exams

1. Basic Definitions (Recap):

• Circle: A collection of all points in a plane that are at a fixed distance (radius) from a fixed point (centre).
• Radius (r): The fixed distance from the centre to any point on the circle.
• Chord: A line segment joining any two points on the circle.
• Diameter (d): A chord passing through the centre. It's the longest chord (d = 2r).
• Secant: A line that intersects a circle at two distinct points.
• Tangent: A line that intersects (touches) the circle at exactly one point. This point is called the Point of Contact.

2. Tangent Properties:

• There is only one tangent possible at any given point on the circle.
• The tangent at a point on a circle is a special case of a secant where the two endpoints of its corresponding chord coincide.
• A circle can have infinitely many tangents.
• There can be two tangents parallel to a given secant.

3. Theorem 10.1: Tangent-Radius Perpendicularity

• Statement: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• Diagram:

  graph TD
      O((Centre O)) -- Radius 'r' --> P(Point of Contact)
      T1---P---T2(Tangent Line)
      style O fill:#fff,stroke:#333,stroke-width:2px
      style P fill:#f9f,stroke:#333,stroke-width:2px
  

• Explanation: If PT is a tangent to the circle with centre O at point P, then the radius OP is perpendicular to the tangent PT.
• Mathematical Representation: OP ⊥ PT (or ∠OPT = 90°).
• Significance: This theorem is fundamental for solving problems involving angles between tangents, radii, and chords connected to the point of contact. It often helps form right-angled triangles.

4. Number of Tangents from a Point to a Circle:

• Case 1: Point Inside the Circle: No tangent can be drawn. Any line through this point will be a secant.
• Case 2: Point On the Circle: Exactly one tangent can be drawn through this point.
• Case 3: Point Outside the Circle: Exactly two tangents can be drawn from this external point.

5. Theorem 10.2: Lengths of Tangents from an External Point

• Statement: The lengths of tangents drawn from an external point to a circle are equal.
• Diagram:

  graph TD
      O((Centre O))
      P(External Point) -- Tangent 1 --> A(Point of Contact)
      P -- Tangent 2 --> B(Point of Contact)
      O -- Radius --> A
      O -- Radius --> B
      O --- P
  
      style O fill:#fff,stroke:#333,stroke-width:2px
      style A fill:#f9f,stroke:#333,stroke-width:2px
      style B fill:#f9f,stroke:#333,stroke-width:2px
  

• Explanation: If PA and PB are two tangents drawn from an external point P to a circle with centre O, then their lengths are equal.
• Mathematical Representation: PA = PB.
• Proof Hint: Join OA, OB, and OP. Prove ΔOAP ≅ ΔOBP using RHS congruence (OA=OB radii, OP common, ∠OAP = ∠OBP = 90° by Theorem 10.1).
• Corollaries/Related Results (Very Important for Exams):
  • The two tangents subtend equal angles at the centre (∠AOP = ∠BOP).
  • The line segment joining the centre to the external point (OP) bisects the angle between the two tangents (∠APO = ∠BPO).
  • The line segment joining the centre to the external point (OP) bisects the angle between the radii to the points of contact (∠AOP = ∠BOP).
  • The angle between the two tangents drawn from an external point P (∠APB) and the angle subtended by the line segment joining the points of contact at the centre (∠AOB) are supplementary.
    ∠APB + ∠AOB = 180°

6. Important Concepts from Exemplar Problems:

• Circumscribed Figures:
  • If a quadrilateral ABCD circumscribes a circle (touches the circle at points P, Q, R, S on sides AB, BC, CD, DA respectively), then the sum of opposite sides are equal:
    AB + CD = AD + BC
    (Proof uses Theorem 10.2: AP=AS, BP=BQ, CR=CQ, DR=DS)
  • A parallelogram circumscribing a circle is always a Rhombus. (Since AB+CD = AD+BC, and for a parallelogram AB=CD, AD=BC, this implies 2AB = 2AD, hence AB=AD. A parallelogram with adjacent sides equal is a rhombus).
• Inscribed Circle (Incircle): A circle inside a polygon that touches all its sides.
• Perimeter Problems: Theorem 10.2 is frequently used to find perimeters of figures (like triangles) formed by tangents. For example, if a circle touches side BC of ΔABC at P, and touches AB and AC extended at Q and R respectively, then AQ = AR = (1/2) * Perimeter of ΔABC.

  graph TD
      subgraph Triangle ABC with extended sides
          A --- B --- Q(Tangent Point)
          A --- C --- R(Tangent Point)
          B --- P(Tangent Point) --- C
      end
      O((Incircle Center))
      style Q fill:#f9f,stroke:#333,stroke-width:2px
      style R fill:#f9f,stroke:#333,stroke-width:2px
      style P fill:#f9f,stroke:#333,stroke-width:2px
  

  Here AQ = AB + BQ = AB + BP, and AR = AC + CR = AC + CP.
  AQ + AR = AB + BP + AC + CP = AB + AC + (BP+CP) = AB + AC + BC = Perimeter(ΔABC).
  Since AQ = AR (Tangents from A), 2AQ = Perimeter(ΔABC).
• Angle Problems: Combining Theorem 10.1 (90° angle) with properties of triangles (sum of angles = 180°), quadrilaterals (sum of angles = 360°), and isosceles triangles (angles opposite equal sides are equal, e.g., in ΔOAB where OA=OB=radius).

Key Takeaways for Exams:

• Memorize the statements and implications of Theorems 10.1 and 10.2 thoroughly.
• Practice problems involving finding lengths and angles using these theorems, especially in combination with properties of triangles and quadrilaterals.
• Pay special attention to the properties of circumscribed figures (quadrilaterals, parallelograms).
• Understand how to apply Pythagoras theorem in right-angled triangles formed by radius, tangent, and the line joining the centre to the external point (like ΔOAP).

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Chapter 10 concepts, similar to what you might encounter:

1. A line intersecting a circle in two distinct points is called a:
   (A) Tangent
   (B) Secant
   (C) Chord
   (D) Diameter

2. From a point P outside a circle, the maximum number of tangents that can be drawn to the circle is:
   (A) 1
   (B) 2
   (C) 0
   (D) Infinite

3. If PT is a tangent to a circle with centre O at point T, then the value of ∠OTP is:
   (A) 45°
   (B) 60°
   (C) 90°
   (D) 180°

4. Two tangents PA and PB are drawn from an external point P to a circle with centre O. If ∠AOB = 110°, then ∠APB is equal to:
   (A) 55°
   (B) 70°
   (C) 90°
   (D) 110°

5. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to:
   (A) 40°
   (B) 50°
   (C) 80°
   (D) 100°

6. A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. The radius of the circle is:
   (A) 1 cm
   (B) 25 cm
   (C) √313 cm
   (D) 5 cm

7. A quadrilateral ABCD is drawn to circumscribe a circle. If AB = 7 cm, BC = 8 cm, CD = 9 cm, then AD is equal to:
   (A) 10 cm
   (B) 8 cm
   (C) 7 cm
   (D) 9 cm

8. A parallelogram circumscribing a circle must be a:
   (A) Rectangle
   (B) Square
   (C) Rhombus
   (D) Trapezium

9. In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 120°, then ∠PTQ is equal to:

   graph TD
       O((O))
       T(T) -- TP --> P
       T -- TQ --> Q
       O -- Radius --> P
       O -- Radius --> Q
       style P fill:#f9f,stroke:#333,stroke-width:2px
       style Q fill:#f9f,stroke:#333,stroke-width:2px
   

   (Assume standard figure for tangents from T to points P, Q on circle with centre O)
   (A) 60°
   (B) 70°
   (C) 80°
   (D) 90°

10. A circle touches the side BC of a ΔABC at P and touches AB and AC produced at Q and R respectively. If the perimeter of ΔABC = 20 cm, then the length of AQ is:
    (A) 10 cm
    (B) 20 cm
    (C) 5 cm
    (D) 15 cm

Answers to MCQs:

1. (B) Secant
2. (B) 2
3. (C) 90° (Theorem 10.1)
4. (B) 70° (Since ∠AOB + ∠APB = 180°)
5. (B) 50° (∠APB = 80°, so ∠APO = 40°. In ΔOAP, ∠OAP = 90°, so ∠POA = 180° - 90° - 40° = 50°)
6. (D) 5 cm (Let radius be r. By Pythagoras in ΔOTP, OP² = OT² + PT². 13² = r² + 12². 169 = r² + 144. r² = 25. r = 5)
7. (B) 8 cm (For circumscribed quadrilateral, AB + CD = BC + AD. 7 + 9 = 8 + AD. 16 = 8 + AD. AD = 8 cm)
8. (C) Rhombus
9. (A) 60° (∠POQ + ∠PTQ = 180°. 120° + ∠PTQ = 180°. ∠PTQ = 60°)
10. (A) 10 cm (AQ = (1/2) * Perimeter of ΔABC = (1/2) * 20 = 10 cm)

Revise these concepts thoroughly and practice problems from the Exemplar book. Good luck!