Chapter 11 — Circles | Class 9 CBSE Maths 2026-27

Section 11.1

Circle — Definitions

A circle is the locus of all points equidistant from a fixed point (centre). Every term here is tested in MCQs.

Key parts of a circle — centre, radius, diameter, chord, arc, sector, segment

📋 All Circle Definitions at a Glance

Term	Definition	Relation
Centre	Fixed point equidistant from all points on circle	O
Radius	Distance from centre to any point on circle	r
Diameter	Longest chord; passes through centre	d = 2r
Chord	Line segment joining two points on circle	Diameter is the longest chord
Arc	Part of the circle (minor: <semicircle, major: >semicircle)	Minor + Major = full circle
Sector	Region bounded by two radii and an arc	Area = (θ/360)πr²
Segment	Region between a chord and the arc	Minor/Major segment
Semicircle	Half circle — diameter creates 2 semicircles	Angle in semicircle = 90°

Section 11.2 — Chord Theorems

Chord Properties

Chords and their relationship to the centre are tested in 2M and 3M questions.

📐 Theorem 11.1

Equal chords of a circle (or congruent circles) are equidistant from the centre.
Converse: Chords equidistant from the centre are equal.

📐 Theorem 11.2

The perpendicular from the centre to a chord bisects the chord.
Converse: The line from the centre bisecting a chord is perpendicular to the chord.

Left: perpendicular from centre bisects chord. Right: equal chords are equidistant from centre

Example 1 · 3M

A chord AB of a circle is 24cm long. The centre is 5cm from the chord. Find the radius.

1

Perp from centre bisects chord: AM = 12cm

2

r² = OM² + AM² = 5² + 12² = 25 + 144 = 169 → r = 13 cm

Radius = 13 cm

Section 11.3 — Most Important

Angle Subtended by an Arc

The central angle / inscribed angle relationship is the core of this chapter — tested in every exam format.

📐 Theorem 11.7 — Central Angle = 2 × Inscribed Angle

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.
If ∠AOB (centre) and ∠ACB (circle) subtend same arc AB:
∠AOB = 2 × ∠ACB

Arc AB subtends angle 2θ at centre O and angle θ at point C on the remaining arc

📐 Theorem 11.8 — Angles in the Same Segment

Angles subtended by the same arc at any points on the same side of the chord are equal.
If C and D are on the major arc side: ∠ACB = ∠ADB

📐 Theorem 11.9 — Angle in a Semicircle

The angle subtended by a diameter at any point on the circle is 90°.
If AB is a diameter and C is on the circle: ∠ACB = 90°
(Special case of Thm 11.7: central angle = 180°, so inscribed = 90°)

Example 2 · 2M

In a circle, ∠AOB = 140°. Find ∠ACB where C is on major arc.

1

∠AOB = 2∠ACB (Thm 11.7)
140 = 2∠ACB → ∠ACB = 70°

∠ACB = 70°

Example 3 · 2M

AB is a diameter. C on circle. ∠BAC = 35°. Find ∠ABC.

1

∠ACB = 90° (angle in semicircle)

2

∠ABC = 180°−90°−35° = 55°

∠ABC = 55°

Section 11.4

Cyclic Quadrilaterals

A cyclic quadrilateral has all four vertices on a circle. The opposite angles theorem is the most tested result.

📐 Theorem 11.11 — Cyclic Quadrilateral

The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
In cyclic quad ABCD: ∠A + ∠C = 180° and ∠B + ∠D = 180°
Converse: If opposite angles of a quadrilateral sum to 180°, it is cyclic.

Example 4 · 3M

In cyclic quad PQRS, ∠P = 75°, ∠Q = 95°. Find ∠R and ∠S.

1

∠P + ∠R = 180° → ∠R = 105°

2

∠Q + ∠S = 180° → ∠S = 85°

3

Check: 75+95+105+85 = 360° ✓

∠R = 105°, ∠S = 85°

Example 5 — Proof · 3M

Prove: Opposite angles of a cyclic quadrilateral sum to 180°.

G

ABCD is a cyclic quad. To prove: ∠A + ∠C = 180°

1

Arc BCD subtends ∠BAD at A and reflex ∠BOD at centre.
By Thm 11.7: reflex ∠BOD = 2∠A

2

Arc BAD subtends ∠BCD at C: ∠BOD = 2∠C (non-reflex)

3

Reflex ∠BOD + ∠BOD = 360° → 2∠A + 2∠C = 360° → ∠A + ∠C = 180° □

More Examples

NCERT Style Problems

Example 6 · 3M

O is the centre. ∠ACD = 40°. Find ∠ABC where ABCD is a cyclic quad on a circle.

1

∠ACD and ∠ABD subtend same arc AD → ∠ABD = ∠ACD = 40° (same segment)

2

∠ABC = ∠ABD + ∠DBC... need more info. If ∠DBC given separately, add.

Example 7 · 2M

Two chords AB and CD of a circle intersect inside at P. If AP=4, PB=3, CP=2, find PD.

1

Intersecting chords theorem: AP × PB = CP × PD

2

4 × 3 = 2 × PD → PD = 6

PD = 6 (Intersecting Chords Theorem: PA×PB = PC×PD)

NCERT Exercise · Solved

Exercise 11.1 & 11.2

Q1 · 2M

In the figure, O is the centre, ∠AOB = 80°. Find ∠ACB.

1

∠ACB = ½∠AOB = ½×80° = 40° (Thm 11.7)

Q2 · 3M

A circle passes through A, B, C, D. ∠ABC = 70°. Find ∠ADC.

1

ABCD is cyclic → ∠ABC + ∠ADC = 180° → ∠ADC = 110°

∠ADC = 110°

Q3 · 3M

Two circles of radii 5cm and 3cm intersect at A and B. If the distance between centres is 4cm, find the length of common chord AB.

1

Let O₁O₂=4, r₁=5, r₂=3. Common chord ⊥ O₁O₂ at M.
O₁M² = r₁²−AM² and O₂M² = r₂²−AM². Let O₁M=x, then O₂M=4−x.

2

25−AM²=x² and 9−AM²=(4−x)²
Subtract: 16 = x²−(4−x)² = 8x−16 → 8x=32 → x=4, AM²=25−16=9 → AM=3
AB = 2×3 = 6 cm

Common chord AB = 6 cm

Smart Study

8 Key Tips — Circles

1

Central ∠ = 2 × Inscribed ∠

The most important theorem. Central angle is always double the inscribed angle on the same arc. Use this to find any unknown angle.

2

Angle in semicircle = 90°

If the chord is a diameter, the angle inscribed at ANY point on the circle is exactly 90°. Check this immediately when you see a diameter.

3

Perpendicular from centre bisects chord

OM⊥AB → AM=MB. Very useful for finding chord lengths given centre distance.

4

Cyclic quad: opposite angles = 180°

∠A+∠C=180° and ∠B+∠D=180°. Memorise — this appears in EVERY exam.

5

Same segment → equal angles

Any two inscribed angles subtending the same arc from the same side are equal.

6

Intersecting chords: PA×PB = PC×PD

When chords intersect inside a circle, the products of their segments are equal.

7

Draw O and all radii in proofs

Most circle proofs involve drawing radii to key points and using isosceles triangles (two radii = equal sides).

8

Four concyclic points = cyclic quad

If opposite angles of a quad sum to 180°, all four points lie on a circle. Used to prove cyclicity.

CBSE Practice

50 Practice Questions

MCQ · 1M · 2M · 3M · 5M · Case-Based