Notes Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down, and Round and Round

📖 Table of Contents

Definitions & Key Terms
Symmetry of a Circle
How Many Circles?
Circumcircle & Circumcentre
Chords & Angles at Centre
Midpoints & Perpendicular Bisectors
Distance of Chords from Centre
Angles Subtended by Arcs
Angle in a Semicircle
Concyclicity of Points
Cyclic Quadrilateral
All Theorems — Quick Reference

Section 5.1

Definitions & Key Terms

Before studying the properties of circles, let’s understand the important words used in this chapter.

Term	Definition
Circle	Set of all points in a plane that are at equal distance from a fixed point (the centre). It is the locus of points equidistant from the centre.
Centre	The fixed point inside the circle from which all boundary points are equidistant.
Radius (r)	Distance from the centre to any point on the circle. All radii of a circle are equal.
Chord	A line segment joining any two points on the circle.
Diameter	A chord that passes through the centre. It is the longest chord. Diameter = 2 × Radius.
Arc	A connected portion of the circle between two points. The smaller part is minor arc; the larger is major arc.
Locus	The set of all points satisfying a given condition.
Concyclic	Points that all lie on the same circle.
Circumcircle	Circle passing through all three vertices of a triangle.
Circumcentre	Centre of the circumcircle; the point where perpendicular bisectors of all sides of a triangle meet.

Parts of a Circle — Centre O, Radius OA, Chord BC, Diameter DE

Remember: Diameter is the longest possible chord of a circle. D = 2r. The diameter always passes through the centre.

Section 5.2

Symmetries of a Circle

Circles are the most symmetric shapes in geometry. They have two types of symmetry:

Rotational Symmetry

Rotate a circle by any angle about its centre — it looks exactly the same! A circle has infinite rotational symmetry.

Reflection Symmetry

Every diameter is a line of reflection symmetry. Fold any diameter and the two halves overlap perfectly. A circle has infinitely many lines of symmetry (one for each diameter).

A square has 4 rotational symmetries and 4 lines of reflection symmetry.
The longest chord in a circle of radius 5 units is the diameter = 10 units.
There is no smallest chord — as a chord moves to the edge, it can approach zero length.
The locus of points equidistant from two points A and B is the perpendicular bisector of AB.

Section 5.3

How Many Circles Pass Through Given Points?

Through 1 point: Infinitely many circles can be drawn.

Through 2 points: Infinitely many circles can be drawn. Their centres all lie on the perpendicular bisector of the line segment joining the two points. The smallest such circle has those two points as the endpoints of its diameter.

Through 3 non-collinear points: Exactly one unique circle can be drawn (Theorem 1).

Through 3 collinear points: No circle can be drawn (perpendicular bisectors of two segments formed will be parallel and never meet).

Theorem 1

There is a unique circle passing through three non-collinear points.

Why is this true?

For three non-collinear points A, B, C: the centre O must be equidistant from all three. Since OA = OB, O lies on the perpendicular bisector of AB. Since OA = OC, O also lies on the perpendicular bisector of AC. These two perpendicular bisectors meet at exactly one point O (since AB and AC are not parallel). Using O as centre with radius OA, we get the unique circle through A, B, C.

Circumcircle of △ABC — perpendicular bisectors of sides meet at circumcentre O

Section 5.3 continued

Circumcircle & Circumcentre

The circle passing through all three vertices of a triangle is called the circumcircle. Its centre is the circumcentre.

Type of Triangle	Position of Circumcentre
Acute-angled	Inside the triangle
Right-angled	At the midpoint of the hypotenuse
Obtuse-angled	Outside the triangle

To draw a circumcircle: draw perpendicular bisectors of any two sides — they meet at the circumcentre.
For a right-angled triangle, the hypotenuse is the diameter of the circumcircle.
Circumradius = OA = OB = OC (all radii are equal).

Section 5.4

Chords and the Angles They Subtend

Theorem 2

Equal chords of a circle subtend equal angles at the centre.

Proof Idea (SSS Congruence)

GivenAB = DE (equal chords)

To Show∠ACB = ∠DCE

In △CAB and △CDE: CA = CD = CB = CE = r (radii), AB = DE (given). By SSS congruence, △CAB ≅ △CDE ∴ ∠ACB = ∠DCE ✓

Theorem 3 (Converse of Theorem 2)

Chords of a circle that subtend equal angles at the centre are equal.

Proof Idea (SAS Congruence)

Given∠ACB = ∠DCE

To ShowAB = DE

CA = CD = CB = CE = r. Since ∠ACB = ∠DCE, by SAS congruence, △ACB ≅ △DCE ∴ AB = DE ✓

Equal chords AB and DE subtend equal angles at centre C

Theorems 2 and 3 are converses of each other — both are equally important.
Key tool: SSS congruence for Thm 2; SAS congruence for Thm 3.
Triangle formed by a chord and the centre is always isosceles (two sides are radii).

Section 5.5

Midpoints and Perpendicular Bisectors of Chords

Theorem 4

The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Proof (SAS Congruence)

GivenM is midpoint of chord AB; C is centre

To ShowCM ⊥ AB

△CAB is isosceles (CA = CB = r). M is midpoint so AM = BM. By SAS, △CMA ≅ △CMB. So ∠CMA = ∠CMB. But ∠CMA + ∠CMB = 180°. ∴ Both = 90°. So CM ⊥ AB. ✓

Theorem 5 (Converse of Theorem 4)

The perpendicular from the centre of a circle to a chord bisects the chord.

CM ⊥ AB, where M is the midpoint of chord AB and C is the centre

Worked Example

Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre. Radius = 5 cm. Find the distance between the midpoints of the chords.

Perpendicular from centre to chord of length 6 cm: d₁ = √(5² − 3²) = √(25−9) = √16 = 4 cm

Perpendicular from centre to chord of length 8 cm: d₂ = √(5² − 4²) = √(25−16) = √9 = 3 cm

Chords are on opposite sides ∴ Distance between midpoints = d₁ + d₂ = 4 + 3 = 7 cm

Section 5.6

Distance of Chords from the Centre

Theorem 6

Chords of a circle having the same length are all at the same distance from the centre.

Proof (RHS Congruence)

For equal chords AB = FG: Draw perpendiculars CE and CH. Since AE = FH (half of equal chords) and CA = CF (radii), by RHS congruence △CEA ≅ △CHF. ∴ CE = CH. ✓

Theorem 7 (Converse of Theorem 6)

Chords of a circle that are equidistant from the centre have equal length.

Theorem 8

If AB > DE (two chords), then the distance from the centre to AB is less than the distance from the centre to DE. (Longer chord is closer to the centre.)

Proof using Pythagoras

Let CF ⊥ AB and CG ⊥ DE. Then CF² + AF² = CA² = r² and CG² + GD² = CD² = r². Since AB > DE, AF > GD (half-lengths). ∴ AF² > GD² ⟹ CF² < CG² ⟹ CF < CG. ✓

Chord Length = 2√(r² − d²)

where r = radius, d = perpendicular distance from centre to chord

Key Observations:
• The diameter (d=0) is the longest chord.
• As a chord moves away from centre, it gets shorter.
• When d = r, the chord reduces to a point (length = 0).

Worked Example · Exercise 5.5 Q1

Find the length of chord when radius = 7 cm and perpendicular distance = 6 cm.

Use: Chord = 2√(r² − d²) = 2√(49 − 36) = 2√13 ≈ 7.21 cm

Section 5.7

Angles Subtended by an Arc

Minor & Major Arc

Two points A and B on a circle divide it into two arcs. The smaller is the minor arc; the larger is the major arc. The angle subtended by the minor arc at the centre is ∠AOB (less than 180°). The major arc subtends the reflex angle (greater than 180°).

Theorem 9 ⭐ (Most Important!)

The angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining part of the circle.

Proof Idea (Exterior Angle Theorem)

GivenArc AFB; ∠ACB at centre C; D on remaining circle

To Show∠ACB = 2∠ADB

Join DC, extend to E on arc. △DCB is isosceles (CB=CD). Exterior angle ∠BCE = 2∠BDC. Similarly ∠ACE = 2∠ADC. Adding: ∠BCA = 2∠BDA. ✓

∠AOB (centre) = 2 × ∠ADB (on circle) — Theorem 9

Corollary: Angles subtended by the same arc at different points on the remaining circle are all equal. If D, E, F are all on the same arc and ∠ADB, ∠AEB, ∠AFB are on the other arc — they are all equal!

Theorem 9 is the most frequently tested theorem in board exams.
Central angle = 2 × inscribed angle (on same arc).
If central angle = 80°, inscribed angle = 40°.
All angles in the same segment are equal.

Important Corollary

Angle in a Semicircle = 90°

Thales’ Theorem (Corollary of Theorem 9)

The angle subtended by a diameter at any point on the circle is 90°.
If AB is a diameter and C is any point on the circle, then ∠ACB = 90°.

Why?

The diameter subtends an angle of 180° (straight angle) at the centre. By Theorem 9, the angle at any point on the circle = ½ × 180° = 90°. ✓

Angle in a semicircle is always 90° — AB is diameter, C is any point on circle

This is used very frequently: If AB is a diameter and C is on the circle ⟹ ∠ACB = 90°.
Converse: If ∠ACB = 90° and A, B, C are on a circle, then AB is the diameter!

Section 5.8

Concyclicity of Points

Concyclic Points

Points that all lie on the same circle are called concyclic.

Theorem 10

If a line segment AB subtends equal angles at two points C and D on the same side of AB, then A, B, C, D are concyclic (lie on the same circle).

Proof by contradiction

Draw circle through A, B, C. Suppose D is not on the circle. If D is outside, then ∠AEB > ∠ADB (E where AD meets circle) — but ∠AEB = ∠ACB = ∠ADB, contradiction! Similarly for D inside. So D must be on the circle. ✓

Section 5.8 continued

Cyclic Quadrilateral

A quadrilateral whose all four vertices lie on a circle is called a cyclic quadrilateral.

Theorem 11 ⭐ (Very Important!)

The sum of opposite angles of a cyclic quadrilateral is 180°.
∠A + ∠C = 180° and ∠B + ∠D = 180°

Proof using Theorem 9

∠BAD = ½ (reflex ∠BOD) and ∠BCD = ½ (∠BOD). Together: ∠BAD + ∠BCD = ½ × 360° = 180°. ✓

Theorem 12 (Converse of Theorem 11)

If the sum of two opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

Cyclic Quadrilateral ABCD — opposite angles are supplementary

In a cyclic quadrilateral: ∠A + ∠C = 180° and ∠B + ∠D = 180°.
Exterior angle of cyclic quad = interior opposite angle.
A rectangle is a cyclic quadrilateral (all 4 vertices on a circle).
A parallelogram is cyclic only if it is a rectangle.
If opposite angles sum to 180° ⟹ it is cyclic (Theorem 12).

Worked Example

In cyclic quad ABCD, ∠A = 75°. Find ∠C. Also if ∠B = 110°, find ∠D.

∠A + ∠C = 180° ⟹ ∠C = 180° − 75° = 105°

∠B + ∠D = 180° ⟹ ∠D = 180° − 110° = 70°

Worked Example · Arc & Inscribed Angle

An arc subtends 70° at the centre. What angle does it subtend at a point on the circle?

By Theorem 9: Inscribed angle = ½ × Central angle = ½ × 70° = 35°

⚡ All 12 Theorems — Quick Reference

There is a unique circle passing through three non-collinear points.

Equal chords subtend equal angles at the centre.

Chords subtending equal angles at the centre are equal. (Converse of Thm 2)

Line from centre to midpoint of chord is perpendicular to the chord.

Perpendicular from centre to a chord bisects the chord. (Converse of Thm 4)

Equal chords are equidistant from the centre.

Chords equidistant from the centre have equal length. (Converse of Thm 6)

Longer chord is closer to the centre. (If AB > DE, dist(C,AB) < dist(C,DE))

Central angle = 2 × Inscribed angle on same arc. (∠AOB = 2∠ACB)

★

Corollary: Angle in a semicircle = 90°. (Angle subtended by diameter at any point on circle is 90°)

If AB subtends equal angles at C and D on same side ⟹ A, B, C, D are concyclic.

In a cyclic quadrilateral, sum of opposite angles = 180°.

If opposite angles of a quadrilateral sum to 180° ⟹ it is a cyclic quadrilateral. (Converse of Thm 11)

📝 Chapter Summary

A circle = locus of all points at distance r from a fixed centre.
Circle has infinite lines of symmetry (all diameters) and complete rotational symmetry.
Through 2 points: infinitely many circles; centres on perpendicular bisector of segment.
Through 3 non-collinear points: exactly one unique circle (circumcircle).
Equal chords ↔ equal angles at centre ↔ equal distances from centre.
Line from centre to midpoint of chord is perpendicular to it (and vice versa).
Longer chord is closer to centre; diameter is closest (distance = 0).
Central angle = 2 × inscribed angle on same arc.
Angle in a semicircle = 90° (Thales’ theorem).
Opposite angles of cyclic quadrilateral are supplementary (sum = 180°).

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