Class 10 Mathematics Notes Chapter 10 (Circles) – Mathematics Book

Alright class, let's focus on Chapter 10: Circles. This is a fundamental chapter in geometry, and its concepts are frequently tested in various government exams. Pay close attention to the definitions and theorems.

Chapter 10: Circles - Detailed Notes

1. Introduction & Basic Terminology (Recap)

• Circle: A collection of all points in a plane that are at a fixed distance (radius) from a fixed point (centre).
• Centre (O): The fixed point.
• Radius (r): The fixed distance from the centre to any point on the circle.
• Diameter (d): A chord passing through the centre. It's the longest chord (d = 2r).
• Chord: A line segment joining any two points on the circle.
• Arc: A continuous piece of a circle between two points.
• Secant: A line that intersects a circle at two distinct points.
• Segment: The region between a chord and its corresponding arc.
• Sector: The region between two radii and the corresponding arc.

2. Tangent to a Circle

• Definition: A line that intersects (touches) the circle at exactly one point.
• Point of Contact: The single point where the tangent touches the circle.
• Key Idea: For every point on the circle, there is one and only one tangent passing through it.
• Non-intersecting Line: A line that does not touch or intersect the circle at all.
• Comparison:
  • A secant cuts through the circle (2 points).
  • A tangent just touches the circle (1 point).
  • A tangent can be seen as a special case of a secant where the two endpoints of its corresponding chord coincide.

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.
• Explanation: If 'P' is the point of contact, 'O' is the centre, and line 'L' is the tangent at P, then the radius OP is perpendicular to the tangent L (OP ⊥ L). This means the angle between the radius and the tangent at the point of contact is always 90°.
• Significance: This is a fundamental property used in many problems involving tangents, especially for finding angles and using the Pythagoras theorem. The shortest distance from the centre to a tangent line is the radius at the point of contact.

4. Number of Tangents from a Point to a Circle

This depends on the position of the point relative to the circle:

• Case 1: Point Inside the Circle: No tangent can be drawn to the circle from a point lying inside it. Any line through this point will be a secant.
• Case 2: Point On the Circle: There is exactly one tangent to the circle passing through a point lying on the circle.
• Case 3: Point Outside the Circle: There are exactly two tangents to the circle that can be drawn from a point lying outside the circle.

5. Length of a Tangent

• Definition: The length of the segment of the tangent from the external point to the point of contact is called the length of the tangent from that external point.
• Example: If P is an external point and A is the point of contact of a tangent from P to the circle, then the length of the tangent is PA.

6. 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.
• Explanation: If P is an external point, and PA and PB are the two tangents from P to the circle with centre O (where A and B are points of contact), then PA = PB.
• Proof Outline (using RHS congruence):
  • Consider triangles ΔOAP and ΔOBP.
  • OA = OB (Radii of the same circle).
  • OP = OP (Common side).
  • ∠OAP = ∠OBP = 90° (Theorem 10.1).
  • Therefore, ΔOAP ≅ ΔOBP (By RHS congruence rule).
  • Hence, PA = PB (By CPCT - Corresponding Parts of Congruent Triangles).
• Important Corollaries (Consequences) from the proof:
  • The tangents are equally inclined to the line segment joining the centre to that point (i.e., ∠OPA = ∠OPB). The line OP bisects the angle between the two tangents (∠APB).
  • The line joining the centre to the external point bisects the angle between the two radii drawn to the points of contact (i.e., ∠AOP = ∠BOP).
• Significance: This theorem is extremely important for solving problems involving lengths related to tangents from an external point, perimeters of triangles/quadrilaterals involving tangent segments, etc.

7. Additional Important Properties & Results

• Parallel Tangents: The tangents drawn at the ends of a diameter of a circle are parallel.
• Quadrilateral Circumscribing a Circle: If a quadrilateral ABCD is drawn to circumscribe a circle (touches sides AB, BC, CD, DA at P, Q, R, S respectively), then the sum of opposite sides are equal: AB + CD = AD + BC. (This is derived using Theorem 10.2: AP=AS, BP=BQ, CR=CQ, DR=DS).
• Angle between Tangents: If PA and PB are tangents from P to a circle with centre O, then the angle between the tangents (∠APB) and the angle subtended by the line segment joining the points of contact at the centre (∠AOB) are supplementary. That is, ∠APB + ∠AOB = 180°. (This follows because OAPB is a quadrilateral, and ∠OAP = ∠OBP = 90°).

Key Takeaways for Exams:

• Know the difference between a secant and a tangent.
• Memorize and understand Theorem 10.1 (Tangent ⊥ Radius) and Theorem 10.2 (Equal lengths of tangents from an external point).
• Be able to apply these theorems to find lengths and angles.
• Remember the properties related to the number of tangents from a point.
• Understand the property of quadrilaterals circumscribing a circle.
• Remember the supplementary angle relationship between the angle between tangents and the angle at the centre.

Multiple Choice Questions (MCQs)

1. A line that intersects a circle at exactly one point is called a:
   (A) Secant
   (B) Chord
   (C) Tangent
   (D) Diameter

2. How many tangents can be drawn to a circle from a point lying inside it?
   (A) 0
   (B) 1
   (C) 2
   (D) Infinite

3. The angle between the tangent at a point on a circle and the radius through the point of contact is:
   (A) 0°
   (B) 45°
   (C) 90°
   (D) 180°

4. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
   (A) 7 cm
   (B) 12 cm
   (C) 15 cm
   (D) 24.5 cm

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. How many parallel tangents can a circle have at the most?
   (A) 1
   (B) 2
   (C) 4
   (D) Infinite

7. The lengths of tangents drawn from an external point to a circle are:
   (A) Always different
   (B) Equal
   (C) Perpendicular
   (D) Parallel

8. A quadrilateral ABCD circumscribes a circle. If AB = 6 cm, CD = 5 cm, and AD = 7 cm, then the length of BC is:
   (A) 4 cm
   (B) 6 cm
   (C) 3 cm
   (D) 8 cm

9. PQ is a tangent to a circle with centre O at point P. If ΔOPQ is an isosceles triangle, then ∠OQP is equal to:
   (A) 30°
   (B) 45°
   (C) 60°
   (D) 90°

10. If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is equal to:
    (A) 3 cm
    (B) 6 cm
    (C) 3√3 cm
    (D) 2√3 cm

Answers to MCQs:

1. (C) Tangent
2. (A) 0
3. (C) 90° (Theorem 10.1)
4. (A) 7 cm (Let O be the centre, P be the point of contact. Then ΔOPQ is right-angled at P. OQ² = OP² + PQ². 25² = r² + 24². 625 = r² + 576. r² = 49. r = 7 cm)
5. (B) 50° (∠APB = 80°. Since OP bisects ∠APB, ∠OPA = 40°. In ΔOAP, ∠OAP = 90°. So, ∠POA = 180° - 90° - 40° = 50°. Alternatively, ∠AOB = 180° - 80° = 100°. Since OP bisects ∠AOB, ∠POA = 100°/2 = 50°)
6. (D) Infinite (A circle can have infinitely many pairs of parallel tangents, one pair for each diameter). Correction: The question likely means at the same time or in a given orientation. A circle has infinitely many diameters, and tangents at the ends of each diameter are parallel. However, if the question implies how many tangents can be parallel to each other simultaneously, it's typically considered as pairs. A better phrasing might be "How many pairs of parallel tangents can a circle have?". If it means parallel to a given line, then there are exactly 2. Given the usual context of Class 10 questions, the most common interpretation intended is often related to the ends of a diameter, implying pairs. Let's re-evaluate. If we consider any tangent, we can find one other tangent parallel to it (at the other end of the diameter through the first point of contact). So, there isn't just one set of parallel tangents. There are infinitely many such pairs. If the question means the maximum number of tangents that can all be parallel to each other simultaneously, the answer is 2. Let's stick with 2 as the most likely intended answer in a typical exam context, referring to a single pair. Self-correction 2: Rereading standard interpretations, "How many parallel tangents can a circle have" usually implies "How many tangents can be drawn parallel to a given line", which is 2. Or it refers to tangents at the end of a diameter, which is 2. Let's assume the intended answer is 2 based on typical problem contexts. Final Decision: Sticking with 2 as the most practical interpretation for Class 10 exams. Let's change the answer key. (B) 2
7. (B) Equal (Theorem 10.2)
8. (A) 4 cm (For a circumscribing quadrilateral, AB + CD = AD + BC. 6 + 5 = 7 + BC. 11 = 7 + BC. BC = 4 cm)
9. (B) 45° (PQ is tangent at P, so OP ⊥ PQ, meaning ∠OPQ = 90°. Since ΔOPQ is isosceles, the equal sides must be OP (radius) and PQ (tangent segment), because OQ (hypotenuse) is the longest side. Thus, OP = PQ. This is generally not true. Let's re-read. If ΔOPQ is isosceles, the possibilities are OP=PQ, OQ=OP, or OQ=PQ. We know ∠OPQ = 90°. If OP=PQ, then ∠POQ = ∠OQP = (180-90)/2 = 45°. If OQ=OP, this is impossible as OQ > OP (hypotenuse > side). If OQ=PQ, then ∠POQ = ∠OPQ = 90°, which is impossible in a triangle. So, the only valid isosceles case here is OP=PQ, leading to ∠OQP = 45°.)
10. (C) 3√3 cm (Let P be the external point, A and B be points of contact, O be the centre. ∠APB = 60°. OP bisects this angle, so ∠APO = 30°. In right-angled ΔOAP, OA = 3 cm (radius), ∠OAP = 90°. We need PA. tan(∠APO) = OA/PA. tan(30°) = 3/PA. (1/√3) = 3/PA. PA = 3√3 cm)

Revise these concepts thoroughly. Practice problems based on these theorems, especially those involving finding lengths and proving geometric properties. Good luck!