Class 9 Mathematics Notes Chapter 10 (Chapter 10) – Examplar Problem (Englisha) Book
Alright class, let's focus on Chapter 10: Circles, from your NCERT Exemplar. This is a crucial chapter, not just for your Class 9 understanding, but also because concepts related to circles frequently appear in various government competitive exams. We need to be thorough with the definitions, theorems, and their applications.
Chapter 10: Circles - Detailed Notes for Competitive Exam Preparation
(Based on NCERT Class 9 Syllabus & Exemplar Problems)
1. Introduction & Basic Terminology:
- Circle: The collection (locus) of all points in a plane that are at a fixed distance (radius) from a fixed point (centre).
- Centre (O): The fixed point inside the circle.
- Radius (r): The fixed distance from the centre to any point on the circle. Plural: Radii.
- Diameter (d): A chord passing through the centre. It's the longest chord of a circle. (d = 2r).
- Chord (AB): A line segment joining any two points on the circle.
- Arc (⌒AB): A continuous piece of a circle between two points.
- Minor Arc: The shorter arc connecting two points.
- Major Arc: The longer arc connecting two points.
- Semicircle: An arc whose endpoints are the ends of a diameter (half the circle).
- Circumference: The total length of the boundary of the circle. (Formula: 2πr).
- Segment: The region between a chord and either of its arcs.
- Minor Segment: Region between the chord and the minor arc.
- Major Segment: Region between the chord and the major arc.
- Sector: The region between two radii and the arc connecting their endpoints on the circle.
- Minor Sector: The region enclosed by two radii and the minor arc.
- Major Sector: The region enclosed by two radii and the major arc.
- Interior/Exterior: Points inside the circle form the interior; points outside form the exterior. Points on the circle are 'on the circle'.
- Concentric Circles: Circles with the same centre but different radii.
2. Key Theorems and Properties (Crucial for Problem Solving):
-
(Theorem 10.1): Equal chords of a circle subtend equal angles at the centre.
- Converse (Theorem 10.2): If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.
- Application: Used to relate chord lengths and central angles.
-
(Theorem 10.3): The perpendicular from the centre of a circle to a chord bisects the chord.
- Diagram: Draw circle, centre O, chord AB, draw OM ⊥ AB. Then AM = MB.
- Converse (Theorem 10.4): The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
- Application: Fundamental for problems involving chord lengths, distances from the centre, and radii (often forms a right-angled triangle OMA or OMB).
-
(Theorem 10.5): There is one and only one circle passing through three given non-collinear points.
- Application: Justifies unique circumcircles for triangles. The construction involves perpendicular bisectors of the sides formed by the points.
-
(Theorem 10.6): Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).
- Diagram: Draw circle, centre O, equal chords AB = CD. Draw OM ⊥ AB and ON ⊥ CD. Then OM = ON.
- Converse (Theorem 10.7): Chords equidistant from the centre of a circle are equal in length.
- Application: Relates chord length to its distance from the centre.
-
(Theorem 10.8): The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
- Diagram: Arc AB subtends ∠AOB at centre O and ∠ACB at point C on the remaining part. Then ∠AOB = 2 * ∠ACB.
- Application: Very important for finding unknown angles. Works for minor and major arcs (reflex angle at centre).
-
(Theorem 10.9): Angles in the same segment of a circle are equal.
- Diagram: Points C and D are on the same arc (major or minor) defined by chord AB. Then ∠ACB = ∠ADB.
- Application: Finding equal angles within a circle diagram.
-
(Theorem 10.10): If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e., they are concyclic).
- Application: Converse of Theorem 10.9, used to prove points are concyclic.
-
(Theorem 10.11): The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
- Definition: A Cyclic Quadrilateral is a quadrilateral whose all four vertices lie on a circle.
- Property: In cyclic quad ABCD, ∠A + ∠C = 180° and ∠B + ∠D = 180°.
- Application: Essential for angle problems involving cyclic quadrilaterals.
-
(Theorem 10.12 - Converse): If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.
- Application: Used to prove a quadrilateral is cyclic.
3. Important Corollaries & Derived Results:
- Angle in a Semicircle: The angle subtended by a diameter (or semicircle) at any point on the remaining part of the circle is a right angle (90°). (This is a special case of Theorem 10.8 where the central angle is 180°).
- Radius to Chord: A radius perpendicular to a chord bisects the chord and also bisects the arc corresponding to the chord (both minor and major arcs).
4. Problem-Solving Strategy for Exams:
- Draw Neat Diagrams: Always draw a clear diagram based on the problem statement. Mark the centre, radii, chords, and given values.
- Identify Given Information & What to Find: Clearly list what is known and what needs to be calculated or proved.
- Look for Key Geometries: Identify radii (creating isosceles triangles with chords), diameters (implying semicircles and 90° angles), perpendiculars from the centre, cyclic quadrilaterals, angles in the same segment, etc.
- Apply Relevant Theorems: Select the appropriate theorem(s) based on the identified geometry and relationships. Often, multiple theorems are needed in sequence.
- Use Basic Geometry: Don't forget properties of triangles (sum of angles, isosceles triangle properties), parallel lines, etc.
- Check Exemplar Problems: The Exemplar book contains higher-level problems. Practice them thoroughly to understand complex applications of these theorems.
Multiple Choice Questions (MCQs)
Here are 10 MCQs based on the concepts from Chapter 10, suitable for practice:
-
In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is:
(A) 2AB
(B) √2 AO
(C) ½ AB
(D) AO -
AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is:
(A) 17 cm
(B) 15 cm
(C) 4 cm
(D) 8 cm -
If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:
(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 12 cm -
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°. Then ∠BAC is equal to:
(A) 80°
(B) 50°
(C) 40°
(D) 30° -
In a circle with centre O, if ∠OAB = 40°, then ∠ACB is equal to (where C is a point on the circle):
(A) 40°
(B) 50°
(C) 80°
(D) 100° -
Two chords AB and CD of a circle are parallel and a line 'l' is the perpendicular bisector of AB. Then 'l':
(A) Bisects CD
(B) Does not bisect CD
(C) Is perpendicular to CD
(D) Both (A) and (C) -
The angle subtended by a major arc at the centre is:
(A) Double the angle subtended by it at any point on the remaining part of the circle.
(B) Equal to the angle subtended by it at any point on the remaining part of the circle.
(C) 90°
(D) Less than 180° -
In the given figure, O is the centre. If ∠BAC = 30°, then ∠BOC is:
(Assume a figure where A, B, C are points on the circle, O is the centre)
(A) 30°
(B) 60°
(C) 90°
(D) 120° -
A chord of length 16 cm is drawn in a circle of radius 10 cm. The distance of the chord from the centre is:
(A) 8 cm
(B) 6 cm
(C) √36 cm
(D) Both (B) and (C) -
If the angles subtended by two chords of a circle at the centre are equal, then the chords are:
(A) Unequal
(B) Equal
(C) Parallel
(D) Perpendicular
Answer Key for MCQs:
- (B)
- (D)
- (C)
- (B)
- (B)
- (D)
- (A)
- (B)
- (D)
- (B)
Make sure you understand the reasoning behind each answer, applying the theorems we discussed. Keep practicing, especially the Exemplar problems, to build confidence and speed. Good luck!