Chapter 9 — Circles | Class 9 CBSE Maths

Basics

Circle — Key Definitions & Terms

Before studying theorems, master the vocabulary of circles. Every proof uses these terms precisely.

Definition — Circle

A circle is the collection of all points in a plane which are equidistant from a fixed point. The fixed point is called the centre (O) and the fixed distance is called the radius (r).

Essential Circle Vocabulary

Term	Definition	Key Property
Chord	A line segment joining any two points on the circle	Diameter is the longest chord
Diameter	A chord passing through the centre; = 2r	Longest chord; bisects the circle
Arc	A continuous part of the circle	Minor arc < semicircle; Major arc > semicircle
Segment	Region between a chord and its arc	Minor segment; Major segment
Sector	Region between two radii and an arc	Angle of sector = angle at centre
Circumference	Total length of the circle = 2πr	Perimeter of circle
Concentric Circles	Two or more circles with the same centre	Different radii, same centre O
Congruent Circles	Circles with the same radius	Can be made to coincide exactly

Section 9.1

Angle Subtended by a Chord at a Point

Equal chords in the same circle subtend equal angles at the centre — and vice versa. These two paired theorems are foundational.

Theorem 9.1

Equal chords of a circle subtend equal angles at the centre.
If chord AB = chord CD in a circle with centre O, then ∠AOB = ∠COD.

Proof

Theorem 9.1 — Using SSS congruence

In △AOB and △COD: OA=OC (Radii) | OB=OD (Radii) | AB=CD (Given)

∴ △AOB ≅ △COD SSS rule

∴ ∠AOB = ∠COD (CPCT)

Equal chords subtend equal angles at centre. ✓

Theorem 9.2 — Converse of 9.1

If the angles subtended by two chords of a circle at the centre are equal, then the chords are equal.
If ∠AOB = ∠COD, then chord AB = chord CD.

💡

Key insight: Theorems 9.1 and 9.2 together mean: equal chords ↔ equal central angles. They're converses of each other and both proved by SSS (using the two radii and the chord/angle).

Section 9.2

Perpendicular from the Centre to a Chord

The perpendicular from the centre of a circle to a chord always bisects it. This creates a right triangle that enables many calculations.

Theorem 9.3

The perpendicular from the centre of a circle to a chord bisects the chord.
If OM ⊥ AB, then AM = MB (M is the midpoint of AB).

Theorem 9.4 — Converse of 9.3

The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
If M is the midpoint of chord AB and O is the centre, then OM ⊥ AB.

Proof

Theorem 9.4 — Using SSS

M is midpoint of chord AB (AM=BM). Join OA and OB.

In △OAM and △OBM: OA=OB (Radii) | AM=BM (Given) | OM=OM (Common)

∴ △OAM ≅ △OBM (SSS) → ∠OMA=∠OMB (CPCT)

∠OMA+∠OMB=180° (Linear pair) → 2∠OMA=180° → ∠OMA=90° → OM⊥AB ✓

The line from centre through midpoint of chord is perpendicular to the chord.

Distance of a Chord from the Centre

The distance of a chord from the centre = the length of the perpendicular from the centre to the chord.
Formula: If chord length = L and radius = r, then distance d = √(r² − (L/2)²)
Conversely: Chord length L = 2√(r² − d²)

Section 9.3

Equal Chords and Their Distances from the Centre

Equal chords are equidistant from the centre — and any two chords equidistant from the centre are equal in length.

Theorem 9.5

Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).
If AB = CD, then their perpendicular distances from O are equal: OM = ON.

Theorem 9.6 — Converse of 9.5

Chords equidistant from the centre of a circle are equal in length.
If OM = ON (perpendicular distances), then chord AB = chord CD.

Summary: Chord-Distance Relationships

Relationship	Theorem
Equal chords → Equal central angles	Theorem 9.1
Equal central angles → Equal chords	Theorem 9.2
⊥ from centre → bisects chord	Theorem 9.3
Line from centre bisecting chord → ⊥ to chord	Theorem 9.4
Equal chords → Equidistant from centre	Theorem 9.5
Equidistant chords → Equal chords	Theorem 9.6

📐

Key insight: Longer chords are NEARER to the centre. The diameter (longest chord) passes through the centre — distance = 0.

Section 9.4

Angle Subtended by an Arc

The most important theorem in this chapter! The central angle is DOUBLE the inscribed angle subtending the same arc.

Theorem 9.7 — The Key Theorem

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.
∠POQ = 2 × ∠PAQ
(where O is centre and A is any point on the major arc)

Proof Summary

Theorem 9.7 — All 3 cases

Setup

Join AO and extend to point B. In △OAQ: OA=OQ (radii) → ∠OAQ=∠OQA (isosceles).

Case 1

∠BOQ = ∠OAQ + ∠AQO = 2∠OAQ (exterior angle). Similarly ∠BOP = 2∠OAP.
Adding: ∠POQ = ∠BOP + ∠BOQ = 2(∠OAP + ∠OAQ) = 2∠PAQ ✓ (Minor arc)

Case 2

PQ is diameter → ∠POQ = 180° → ∠PAQ = 90° (angle in semicircle). This is Angle in a Semicircle = 90°

Case 3

PQ is major arc → reflex ∠POQ = 2∠PAQ. Same derivation with subtraction.

∠POQ = 2 × ∠PAQ in all cases.

Theorem 9.8

Angles in the same segment of a circle are equal.
If A and C are both on the major arc, then ∠PAQ = ∠PCQ.
Proof: Both = ½∠POQ (by Theorem 9.7)

Special Case — Angle in a Semicircle

When the chord is a diameter: ∠POQ = 180° (straight angle). By Theorem 9.7: ∠PAQ = 90°.
∴ Angle in a semicircle is always a right angle (90°).
This is used extensively in coordinate geometry and construction problems.

Theorem 9.9

If a line segment joining two points subtends equal angles at two other points lying on the same side of the line, then all four points lie on a circle (they are concyclic).
This is the converse of Theorem 9.8.

Section 9.5

Cyclic Quadrilaterals

A cyclic quadrilateral has all four vertices on a circle. Its opposite angles have a remarkable property — they're supplementary.

Definition — Cyclic Quadrilateral

A quadrilateral is called cyclic if all four of its vertices lie on a circle (also called a circumscribed circle). The quadrilateral is said to be inscribed in the circle.

Theorem 9.10 — Opposite Angles of Cyclic Quad

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

Proof

Theorem 9.10 — Using Theorem 9.7

Let ABCD be a cyclic quadrilateral with centre O. Arc BCD subtends ∠BOD at centre and ∠BAD (= ∠A) at A on the circle.

By Theorem 9.7: ∠BOD = 2∠BAD = 2∠A …(1) (for major arc BCD)

Similarly, reflex ∠BOD = 2∠BCD = 2∠C …(2) (for minor arc BAD)

∠BOD + reflex ∠BOD = 360° (angles at centre) → 2∠A + 2∠C = 360° → ∠A + ∠C = 180° ✓

∠A + ∠C = ∠B + ∠D = 180°

Theorem 9.11 — Converse of 9.10

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

How to Identify a Cyclic Quadrilateral

A quadrilateral is cyclic if ANY of these conditions hold:

1. Opposite angles are supplementary (∠A + ∠C = 180° or ∠B + ∠D = 180°) [Theorem 9.11]
2. An exterior angle equals the interior opposite angle
3. The quadrilateral formed by internal angle bisectors of a quadrilateral is cyclic [Example 5]
4. Four points lying on a circle (by definition)

Textbook Examples

All 5 Examples — Fully Solved

Example 1

Two chords make equal angles with diameter through intersection point. Prove chords are equal.

Chords AB, CD with centre O meet at E. PQ is diameter through E with ∠AEQ=∠DEQ. Draw OL⊥AB, OM⊥CD.

∠LOE = 90°−∠LEO = 90°−∠AEQ = 90°−∠DEQ = 90°−∠MEO = ∠MOE

In △OLE and △OME: ∠LEO=∠MEO | ∠LOE=∠MOE | EO=EO → △OLE≅△OME (AAS) → OL=OM (CPCT)

Chords equidistant from centre are equal Thm 9.6 → AB=CD ✓

If two chords make equal angles with a diameter, the chords are equal.

Example 2

AB is diameter, CD = radius. AC and BD extended meet at E. Prove ∠AEB = 60°.

Join OC, OD, BC. OC=OD=CD (all radii since CD=radius) → △ODC is equilateral → ∠COD=60°.

∠CBD = ½∠COD = 30° Thm 9.7

∠ACB = 90° (angle in semicircle, AB is diameter) → ∠BCE = 180°−90° = 90°

In △BCE: ∠CEB = 90°−∠CBE = 90°−30° = 60° → ∠AEB = 60° ✓

∠AEB = 60°

Example 3

Cyclic quadrilateral ABCD, ∠DBC=55°, ∠BAC=45°. Find ∠BCD.

∠CAD = ∠DBC = 55° Angles in same segment

∠DAB = ∠CAD + ∠BAC = 55° + 45° = 100°

∠DAB + ∠BCD = 180° Opposite angles of cyclic quad (Thm 9.10)

∠BCD = 180° − 100° = 80° ✓

∠BCD = 80°

Example 4

Two circles intersect at A and B. AD and AC are diameters. Prove B lies on DC.

Join AB. ∠ABD = 90° (angle in semicircle, AD is diameter of one circle)

∠ABC = 90° (angle in semicircle, AC is diameter of other circle)

∠ABD + ∠ABC = 90° + 90° = 180° → D, B, C are collinear → B lies on DC ✓

B lies on line segment DC (since ∠DBC = 180°).

Example 5

Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic.

ABCD any quadrilateral. Angle bisectors form quadrilateral EFGH. In △AEB (where E is intersection of bisectors of ∠A and ∠B):
∠EAB = ½∠A, ∠EBA = ½∠B → ∠AEB = 180°−½(∠A+∠B) → ∠FEH = 180°−½(∠A+∠B)

Similarly ∠FGH = 180°−½(∠C+∠D)

∠FEH+∠FGH = 360°−½(∠A+∠B+∠C+∠D) = 360°−½×360° = 360°−180° = 180°

Opposite angles sum to 180° → EFGH is cyclic Thm 9.11 ✓

The quadrilateral formed by angle bisectors is always cyclic.

NCERT Exercises

Exercises 9.1, 9.2, 9.3 — Key Questions Solved

Exercise 9.2

Two circles of radii 5 cm and 3 cm intersect at two points. Distance between centres = 4 cm. Find the common chord length.

Let O and O' be centres (r=5, r'=3), common chord AB, with perpendicular from O to AB at M. OO'=4.

In △OO'A: OA²=OO'²+O'A²? No — OA=5, O'A=3, OO'=4. Check: 3²+4²=9+16=25=5². So O'A⊥OO' (∠O'AO=90°? Actually ∠at O').
In △OAO': OA=5, O'A=3, OO'=4 → 3²+4²=5² → ∠AO'O=90°.

So O'M = O'A = 3 (M is foot of perp from O' to... wait). The perpendicular from O' to AB, let OM be from O to AB.
Using coordinate approach: Let O at origin, O' at (4,0). A at (x,y): x²+y²=25, (x-4)²+y²=9.
Subtracting: x²−(x-4)²=25−9=16 → 8x−16=16 → x=4. y²=25−16=9 → y=3. AB=2y=6 cm.

Length of common chord = 6 cm

Three girls Reshma (R), Salma (S), Mandip (M) on circle of radius 5m. RS=SM=6m each. Find RM.

RS=SM=6m → equal chords → equidistant from centre O. Let perp from O to RS be OL, to SM be OM.
OL=OM → O lies on bisector of ∠RSM.

For chord RS=6m, radius=5m: distance = √(5²−3²) = √(25−9) = √16 = 4m

By symmetry, RS=SM → ∠ROS=∠SOM. Also RS is a chord with half-length 3m, distance from O = 4m.
Using coordinates: Let S at top. ∠ROS: cos(∠ROS/2) = 4/5 = 0.8 → ∠ROS/2 = 36.87° → ∠ROS ≈ 73.74°. Similarly ∠SOM ≈ 73.74°. ∠ROM = 360°−2×73.74° = 212.52°... wait.
Actually ∠ROM = 360°−∠ROS−∠SOM = 360°−73.74°−73.74°= 212.52°, but that's reflex. Use: RM = 2×5×sin(½∠ROM) where ∠ROM=360°−147.48°=212.52°? Better: ∠RMS = inscribed angle. Let's use direct: in △ORS, cos∠ROS = (5²+5²−6²)/(2×5×5) = (50−36)/50 = 14/50 = 7/25. ∠ROM = 2π−2∠ROS. RM=2×5×sin(∠ROM/2). ∠ROS: cos∠ROS=7/25. RM²=r²+r²−2r²cos∠ROM = 50−50cos(∠ROM). ∠ROM=2π−2∠ROS → cos∠ROM=cos(2∠ROS)=2cos²∠ROS−1=2(7/25)²−1=98/625−1=−527/625. RM²=50−50(−527/625)=50+527×50/625=50+42.16=92.16 → RM=√92.16 ≈ 9.6m.

RM = 2√(5²−0) in direct approach. Using formula: RM ≈ 9.6 m (through coordinate/cosine rule).

Exercise 9.3

A, B, C on circle with centre O. ∠BOC=30°, ∠AOB=60°. D is on arc (not arc ABC). Find ∠ADC.

∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°

D is on the major arc (arc other than ABC). The arc ABC subtends ∠AOC = 90° at centre. At D on the remaining arc: ∠ADC = ½ × ∠AOC Thm 9.7

∠ADC = ½ × 90° = 45°

∠ADC = 45°

Chord of a circle equals radius. Find angle subtended at (a) minor arc point (b) major arc point.

Chord AB = radius r. OA=OB=AB=r → △OAB is equilateral → ∠AOB = 60°.

At major arc point P: ∠APB = ½∠AOB = ½×60° = 30° Thm 9.7

At minor arc point Q: reflex ∠AOB = 360°−60° = 300°. ∠AQB = ½×300° = 150°

Major arc point: 30°; Minor arc point: 150°

Cyclic quad ABCD. ∠DBC=70°, ∠BAC=30°. Find ∠BCD. Also if AB=BC, find ∠ECD.

∠DAC=∠DBC=70° (same segment). ∠DAB=∠DAC+∠BAC=70°+30°=100°

∠BCD = 180°−∠DAB = 180°−100° = 80° (opposite angles of cyclic quad)

AB=BC → ∠BCA=∠BAC=30° (angles opp. equal sides). ∠ECD = ∠BCD−∠BCE = 80°−∠BCA... wait. E is intersection of diagonals. ∠DBC=70°, in △BCE: ∠CBE=70°, ∠BEC=∠AEB (vert.opp.)... Actually ∠ECD=∠ACD−∠ACE=∠ABD−∠BAC... Let me use: ∠BAC=30°, AB=BC → ∠BCA=30°. ∠ACD = ∠ABD (same segment). ∠ABD=∠ABC−∠DBC. ∠ABC=180°−∠ADC. Need ∠ADC. ∠ADC=∠ADB+∠BDC. ∠ADB=∠ACB=30° (same segment). ∠BDC=∠BAC=30°. ∠ADC=60°. ∠ABC=120°. ∠ABD=120°−70°=50°. ∠ACD=50°. ∠ECD=∠ACD−∠ACE... Hmm. ∠ACE=∠ACB=30°. ∠ECD=50°−30°=20°.

∠BCD = 80°; ∠ECD = 20°

Study Strategy

10 Tips for Class 9 Students

Master Theorem 9.7 First

Central angle = 2 × inscribed angle. This is the KING theorem of circles. Everything else builds on it. Memorize: ∠POQ = 2∠PAQ. Draw it, understand all 3 cases (minor arc, semicircle, major arc).

Angle in Semicircle = 90°

When the chord is a diameter, the angle at circumference = 90°. CBSE uses this constantly — "AB is diameter → ∠ACB = 90°." This is the semicircle case of Theorem 9.7.

Same Segment → Equal Angles

Any two angles inscribed in the same segment (same side of a chord) are equal. ∠APB = ∠AQB if P and Q are on the same arc. Use this when you see multiple points on a circle.

Cyclic Quad: Opposite ∠s Sum to 180°

In cyclic quad ABCD: ∠A+∠C=180° and ∠B+∠D=180°. This is Theorem 9.10. Also: exterior angle = opposite interior angle. CBSE always tests this.

⊥ from Centre Bisects Chord

The perpendicular from the centre to ANY chord bisects that chord (Theorem 9.3). Use this for calculation problems: if OM⊥AB and r=5, OA²=OM²+AM². Then AB=2AM.

Equal Chords Equidistant from Centre

Equal chords → equal distances from centre (Theorem 9.5). This lets you prove two chords are equal by showing their perpendicular distances from O are equal.

Radii Make Isosceles Triangles

Whenever you draw two radii (OA and OB), you get an isosceles triangle. So ∠OAB=∠OBA. This is used in the proof of Theorem 9.7 and many other circle proofs.

Draw All Radii in Diagrams

When attempting a circle proof, always draw: (1) the centre O, (2) radii to key points, (3) perpendiculars from O to chords. These create the triangles you need for SSS/SAS/RHS proofs.

Cyclic Parallelogram is Rectangle

In a cyclic parallelogram: opposite angles are equal (parallelogram) AND opposite angles sum to 180° (cyclic). So each angle = 90° → it's a rectangle. CBSE loves asking this.

Congruent Circles = Same Radius

Two circles are congruent if and only if they have the same radius. Equal chords in congruent circles subtend equal central angles. State "radii of same/congruent circle" when using OA=OB in proofs.

Quick Reference

Chapter 9 — Formula & Fact Sheet

Theorem	Statement	Key Formula
9.1	Equal chords → Equal central angles	AB=CD → ∠AOB=∠COD
9.2	Equal central angles → Equal chords	∠AOB=∠COD → AB=CD
9.3	⊥ from centre bisects chord	OM⊥AB → AM=MB
9.4	Line from centre bisecting chord → ⊥	AM=MB, OA=OB → OM⊥AB
9.5	Equal chords → Equidistant	AB=CD → OM=ON
9.6	Equidistant chords → Equal	OM=ON → AB=CD
9.7	Central angle = 2 × inscribed angle	∠POQ = 2∠PAQ
9.8	Angles in same segment are equal	∠PAQ = ∠PCQ
—	Angle in semicircle	∠ACB = 90° (if AB = diameter)
9.9	Equal angles on same side → Concyclic	∠ACB=∠ADB → A,B,C,D concyclic
9.10	Cyclic quad opposite angles	∠A+∠C = ∠B+∠D = 180°
9.11	Opposite angles 180° → Cyclic	∠A+∠C=180° → ABCD cyclic
Chord formula	Distance from centre to chord	d = √(r² - (L/2)²)

CBSE Pattern Practice

50 Practice Questions

Section A — MCQ

The angle subtended by an arc at the centre is related to the angle subtended at any point on the remaining arc by:

MCQ

(a) Equal

(b) Half

(d) Supplementary

Answer: (c) Double
Theorem 9.7: Central angle = 2 × inscribed angle. ∠POQ = 2 × ∠PAQ.

Angle inscribed in a semicircle equals:

MCQ

(a) 45°

(b) 60°

(d) 180°

Answer: (c) 90°
Angle in a semicircle = 90° (special case of Theorem 9.7, when chord is diameter: ∠AOB=180° → ∠APB = 90°).

In cyclic quadrilateral ABCD, ∠A=80°. Find ∠C.

MCQ

(a) 80°

(b) 100°

(d) 110°

Answer: (b) 100°
Theorem 9.10: ∠A + ∠C = 180° → ∠C = 180° − 80° = 100°.

The perpendicular from the centre to a chord:

MCQ

(a) Bisects the chord

(b) Passes through an endpoint

(d) None of these

Answer: (a) Bisects the chord
Theorem 9.3: The perpendicular from the centre of a circle to a chord bisects the chord. OM⊥AB → AM = MB.

A chord of length 16 cm is at distance 6 cm from the centre. Find the radius.

MCQ

(a) 8 cm

(b) 10 cm

(d) 14 cm

Answer: (b) 10 cm
Half chord = 8 cm. r² = d² + (L/2)² = 6² + 8² = 36 + 64 = 100 → r = 10 cm.

P and Q are points on major arc AB of a circle. Then ∠APB and ∠AQB are:

MCQ

(a) Supplementary

(b) Complementary

(d) Different

Answer: (c) Equal
Theorem 9.8: Angles in the same segment are equal. Since P and Q are both on the major arc (same side), ∠APB = ∠AQB.

Two equal chords of a circle are equidistant from the centre — this is:

MCQ

(a) Theorem 9.3

(b) Theorem 9.4

(d) Theorem 9.7

Answer: (c) Theorem 9.5
Theorem 9.5: Equal chords of a circle (or congruent circles) are equidistant from the centre.

If ∠AOB = 100° where O is the centre, then the angle subtended by arc AB at any point on major arc =

MCQ

(a) 100°

(b) 50°

(d) 80°

Answer: (b) 50°
∠POQ = 2 × ∠PAQ → ∠PAQ = ∠AOB/2 = 100°/2 = 50°.

A cyclic parallelogram is always a:

MCQ

(a) Rhombus

(b) Square

(d) Trapezium

Answer: (c) Rectangle
Cyclic: ∠A+∠C=180°. Parallelogram: ∠A=∠C. So 2∠A=180° → ∠A=90°. All angles are 90° → Rectangle.

O is centre, chord AB = chord CD and OM⊥AB, ON⊥CD. Then:

MCQ

(a) OM > ON

(b) OM < ON

(d) Cannot determine

Answer: (c) OM = ON
Theorem 9.5: Equal chords of a circle are equidistant from the centre. AB=CD → OM=ON.

Section B — 1 Mark

State Theorem 9.7 (Arc angle theorem).

1 Mark

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.
i.e., ∠POQ = 2∠PAQ.

O is centre, ∠AOB = 140°. Find angle subtended at any point on major arc.

1 Mark

70°
∠AOB = 2 × ∠APB → ∠APB = 140°/2 = 70°. (P on major arc)

AB is diameter. C is on circle. ∠BAC = 30°. Find ∠ABC.

1 Mark

∠ABC = 60°
∠ACB = 90° (angle in semicircle). In △ABC: ∠BAC+∠ABC+∠ACB=180° → 30°+∠ABC+90°=180° → ∠ABC=60°.

Chord AB = 24 cm, radius = 13 cm. Find distance from centre to chord.

1 Mark

5 cm
d = √(r²−(L/2)²) = √(13²−12²) = √(169−144) = √25 = 5 cm.

Define a cyclic quadrilateral.

1 Mark

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

In cyclic quad ABCD, ∠B=95°. Find ∠D.

1 Mark

∠D = 85°
∠B + ∠D = 180° → ∠D = 180°−95° = 85°.

What is the distance of the diameter from the centre?

1 Mark

Zero (0)
The diameter passes through the centre, so the perpendicular distance from the centre to the diameter is 0.

∠PAQ = 40°. Find ∠POQ where O is centre and P, Q are on circle.

1 Mark

∠POQ = 80°
Central angle = 2 × inscribed angle → ∠POQ = 2 × 40° = 80°.

Two chords AB and CD of a circle are equidistant from the centre. What can you conclude?

1 Mark

AB = CD (chords are equal in length)
Theorem 9.6: Chords equidistant from the centre of a circle are equal in length.

Can a cyclic quadrilateral be a rhombus? Under what condition?

1 Mark

Yes — only if it is a square.
Cyclic requires ∠A+∠C=180°. Rhombus has ∠A=∠C. So 2∠A=180° → ∠A=90°. All sides equal + all angles 90° = square.

Section C — 2 Mark

O is centre, AB is chord with OM⊥AB and OM=4cm, AB=6cm. Find radius OA.

2 Mark

OM⊥AB → M is midpoint → AM = AB/2 = 3 cm. In △OAM: OA² = OM² + AM² = 4² + 3² = 16+9 = 25 → OA = 5 cm.

AB is chord of circle with centre O. OM⊥AB, OM = 3 cm and r = 5 cm. Find length of chord AB.

2 Mark

AM = √(OA²−OM²) = √(25−9) = √16 = 4 cm. AB = 2×AM = 8 cm.

In a circle with centre O, chord AB subtends ∠AOB = 80° at centre. Find the angle subtended at a point C on the major arc.

2 Mark

By Theorem 9.7: ∠ACB = ½∠AOB = ½×80° = 40°.

ABCD is a cyclic quadrilateral. ∠ABD = 35° and ∠BDA = 25°. Find ∠BCA and ∠DCA.

2 Mark

∠BCA = ∠BDA = 25° (angles in same segment for chord AB).
∠DCA = ∠DBA = 35° (angles in same segment for chord CD).

Chord AB is at 5 cm from centre. Chord CD is at 5 cm from centre. r = 13 cm. Find AB and CD.

2 Mark

Equal distances → AB = CD (Theorem 9.6).
Half chord = √(13²−5²) = √(169−25) = √144 = 12 cm.
AB = CD = 24 cm.

A chord is at distance 4 cm from centre of a circle with radius √41 cm. Find chord length.

2 Mark

Half chord = √(r²−d²) = √(41−16) = √25 = 5 cm. Chord = 10 cm.

AB is a diameter. P is on the circle such that ∠PAB = 50°. Find ∠PBA and ∠APB.

2 Mark

∠APB = 90° (angle in semicircle).
∠PBA = 180°−90°−50° = 40°.

Two equal chords AB and CD of a circle with centre O. OM⊥AB and ON⊥CD. If OM = 4 cm, find ON.

2 Mark

AB=CD → Equal chords → Equal distances (Theorem 9.5).
OM = ON = 4 cm.

PQRS is cyclic. ∠P = (2x+10)° and ∠R = (3x−40)°. Find x and ∠P.

2 Mark

∠P+∠R=180° → 2x+10+3x−40=180 → 5x−30=180 → 5x=210 → x=42.
∠P = 2(42)+10 = 94°.

Prove that equal chords of a circle subtend equal angles at the centre.

2 Mark

Theorem 9.1: Given AB=CD, centre O. In △AOB and △COD: OA=OC (radii), OB=OD (radii), AB=CD (given) → △AOB≅△COD (SSS) → ∠AOB=∠COD (CPCT). ✓

Section D — 3 Mark

O is centre of circle. ∠PQR = 100° where P, Q, R are on circle. Find ∠OPR.

3 Mark

∠PQR = 100° is inscribed angle for arc PR (minor). Reflex ∠POR = 2×100° = 200°. ∠POR = 360°−200° = 160°.
In △OPR: OP=OR (radii) → ∠OPR=∠ORP. ∠OPR+∠ORP+∠POR=180° → 2∠OPR = 180°−160° = 20° → ∠OPR = 10°.

A, B, C, D are concyclic. AC and BD intersect at E. ∠BEC=130°, ∠ECD=20°. Find ∠BAC.

3 Mark

In △BCE: ∠BEC=130°, ∠ECD=20° → ∠EBC=180°−130°−20°=30°... wait ∠EBC is at B. ∠BEC+∠ECB+∠CBE=180°. We need ∠ECB=∠ECD=20°. So ∠CBE=180°−130°−20°=30°. ∠BAC=∠BDC? No. ∠BAC and ∠BDC are in same segment. ∠ABD+... ∠BAC=∠BDC (same segment of BD and AC). In △BEC: ∠BCE=20°, ∠BEC=130° → ∠CBE=30°. ∠DBC=∠CBE=30°. ∠BAC=∠BDC=? In △DECnot helpful. ∠BAC=∠BEC−∠ABE in exterior angle... ∠BAC is the angle in segment for arc BC not containing A. ∠BAC = ∠BEC−∠ACE = 130°−20°=110°? No. Using exterior angle: ∠BAC = ∠BEC − ∠ABE? Let me use: ∠BAC = ∠BDC (same segment). ∠BDC = 180°−∠BEC = 180°−130° = 50°... that's not right either. Actually ∠ECD=∠ACB... in △AEB: ∠AEB=180°−130°=50°(vertical), ∠EAB+∠ABE=130°. ∠BAC = ∠BAE. Using: in △ABE, ext ∠BEC=∠BAC+∠ABC → 130°=∠BAC+∠ABC. ∠ABD=∠ABE=∠ABC−∠DBC. Hmm. Let ∠BAC=x. ∠BDC=x (same segment). In △DEC: ∠DEC=130°(vert.opp to ∠BEC), ∠ECD=20°, ∠EDC=∠BDC=x → 130°+20°+x=180° → x=30°. ∠BAC = 30°.

Prove that the perpendicular from the centre of a circle to a chord bisects the chord (Theorem 9.3).

3 Mark

Given: Circle with centre O, chord AB, OM⊥AB. To prove: AM=BM.
Join OA and OB. In △OAM and △OBM: OA=OB (radii) | ∠OMA=∠OMB=90° (OM⊥AB) | OM=OM (common) → △OAM≅△OBM (RHS) → AM=BM (CPCT). ✓

Two circles of radii r₁ and r₂ intersect at A and B. Prove the line joining centres is perpendicular bisector of common chord AB.

3 Mark

Let centres O₁ and O₂. OA=OB=r₁ (O₁ equidistant from A,B) → O₁ lies on perp bisector of AB. O₂A=O₂B=r₂ → O₂ also on perp bisector of AB. A unique line passes through O₁ and O₂ → line O₁O₂ IS the perpendicular bisector of AB. ✓

ABCD is a cyclic quad. Prove that ∠DAB + ∠BCD = 180° (Theorem 9.10).

3 Mark

Join OA and OC. Arc BCD subtends ∠BOD at centre and ∠BAD at A (circumference). By Thm 9.7: ∠BOD = 2∠BAD …(1). Reflex ∠BOD subtends ∠BCD at C: reflex ∠BOD = 2∠BCD …(2). ∠BOD + reflex ∠BOD = 360° → from (1)+(2): 2∠BAD + 2∠BCD = 360° → ∠BAD+∠BCD = 180°. ✓

∠ABC = 69° and ∠ACB = 31°. D is on circle (same as A,B,C). Find ∠BDC.

3 Mark

In △ABC: ∠BAC = 180°−69°−31° = 80°. ∠BDC = ∠BAC = 80° (angles in same segment — both subtend arc BC). ✓

Prove that angles in the same segment of a circle are equal (Theorem 9.8).

3 Mark

Given: Arc PQ, points A and C on the major arc. To prove: ∠PAQ = ∠PCQ.
By Theorem 9.7 applied to point A: ∠POQ = 2∠PAQ …(1). By Theorem 9.7 applied to point C: ∠POQ = 2∠PCQ …(2). From (1) and (2): 2∠PAQ = 2∠PCQ → ∠PAQ = ∠PCQ. ✓

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.

3 Mark

△ABC. Circle on AB as diameter, circle on AC as diameter. These circles intersect at A. Let them intersect again at D. ∠ADB=90° (angle in semicircle on AB) and ∠ADC=90° (angle in semicircle on AC). ∠BDC = ∠ADB+∠ADC=90°+90°=180°. So B,D,C are collinear → D lies on BC. ✓

In the given figure, O is centre, ∠BOC = 30° and ∠AOB = 60°. If D is on the circle other than arc ABC, find ∠ADC.

3 Mark

∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°. D is on the remaining arc. ∠ADC = ½ × ∠AOC = ½ × 90° = 45°.

Show that a cyclic parallelogram is a rectangle.

3 Mark

Let ABCD be cyclic parallelogram. Cyclic: ∠A+∠C=180° (Thm 9.10). Parallelogram: ∠A=∠C (opposite angles equal). From both: ∠A+∠A=180° → 2∠A=180° → ∠A=90°. Since ∠A=90°, all angles = 90° (parallelogram with one right angle is rectangle). ✓

Section E — 5 Mark

Prove Theorem 9.7: The angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle.

5 Mark

Given: Arc PQ, centre O, point A on remaining arc. To prove: ∠POQ = 2∠PAQ.
Construction: Join AO and extend to point B.
Case I (minor arc): In △OAQ: OA=OQ (radii) → ∠OAQ=∠OQA. Ext angle ∠BOQ = ∠OAQ+∠OQA = 2∠OAQ (1). Similarly ∠BOP = 2∠OAP (2). Adding: ∠BOP+∠BOQ = 2(∠OAP+∠OAQ) → ∠POQ = 2∠PAQ ✓
Case II (semicircle): ∠POQ = 180° = 2×90° = 2∠PAQ → Angle in semicircle = 90° ✓
Case III (major arc): Reflex ∠POQ = 2∠PAQ (by similar argument). ✓

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

5 Mark

Given: Equal chords AB and CD intersect at E inside circle. To prove: AE=DE and BE=CE.
Draw OM⊥AB and ON⊥CD. Equal chords → OM=ON (Thm 9.5). ME=½AB−AM... Let M be midpoint of AB, N midpoint of CD. AM=BM=AB/2 and CN=DN=CD/2. Since AB=CD: AM=DN and BM=CN.
In △OME and △ONE: OM=ON | ∠OME=∠ONE=90° | OE=OE → △OME≅△ONE (RHS) → ME=NE.
AE = AM−ME = DM−NE... Hmm. AE=AM+ME or AM−ME depending on position. Since E is intersection: AE=AM−ME and CE=CN+NE. But AM=DN, so AE=AM−ME=DN−NE... Let me use: AE=AM−ME and BE=AM+ME; CE=CN+NE and DE=DN−NE. Since AM=DN and ME=NE: AE=AM−ME=DN−NE=DE ✓, and BE=AM+ME=DN+NE=CE ✓.

Prove that if two chords of a circle intersect within the circle, the line joining the point of intersection to the centre makes equal angles with both chords.

5 Mark

Given: Equal chords AB, CD intersect at E. O is centre. OE is drawn. To prove: ∠OEA=∠OEC (OE makes equal angles with AB and CD).
Draw OM⊥AB and ON⊥CD. Equal chords → OM=ON (Thm 9.5).
In △OME and △ONE: OM=ON (shown) | ∠OME=∠ONE=90° | OE=OE → △OME≅△ONE (RHS) → ∠MOE=∠NOE (CPCT).
But ∠MOE = ∠OEM (= 90°−∠OEM from △OME... actually ∠MOE is the angle at O). The equal angles we need: ∠OEA and ∠OEC. Since OM⊥AB: ∠OEA = 90°−∠MOE. Since ON⊥CD: ∠OEC = 90°−∠NOE. And ∠MOE=∠NOE → ∠OEA=∠OEC ✓

ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

5 Mark

Given: ∠ABC=∠ADC=90°. AC is common hypotenuse. To prove: ∠CAD=∠CBD.
Since ∠ABC=90°, B lies on a circle with AC as diameter (angle in semicircle = 90°). Since ∠ADC=90°, D also lies on the same circle with AC as diameter.
So B, D are both on the circle with diameter AC. Hence A, B, C, D are concyclic (all on the same circle).
∠CAD and ∠CBD both subtend arc CD (on the same side). By Theorem 9.8 (angles in same segment): ∠CAD = ∠CBD. ✓

A line intersects two concentric circles (same centre O) at A, B, C, D (A outermost). Prove AB = CD.

5 Mark

Given: Two concentric circles, centre O. Line intersects outer at A,D and inner at B,C. To prove: AB=CD.
Draw OM⊥AD (perpendicular from centre to line AD).
M is midpoint of chord AD in outer circle: AM=MD (Theorem 9.3) …(1). M is midpoint of chord BC in inner circle: BM=MC …(2).
AB = AM−BM and CD = MD−MC. From (1): AM=MD and from (2): BM=MC → AB = AM−BM = MD−MC = CD. ✓

Two circles intersect at B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q. Prove ∠ACP = ∠QCD.

5 Mark

∠ACP = ∠ABP (angles in same segment of circle passing through A,B,C,P — ∠ACP and ∠ABP both subtend AP).
∠QCD = ∠QBD (angles in same segment of the other circle — ∠QCD and ∠QBD both subtend QD... let me reconsider).
Since ABD is a straight line: ∠ABP = ∠DBQ (vertically opposite... or same angle). Since PBQ is a straight line: ∠ABP = ∠ABP. In fact ∠ACP = ∠ABP (same chord AP, same circle through ABCP) and ∠QCD = ∠QBD (same chord QD, circle through BCDQ). Since ABD is straight: ∠ABP and ∠PBD... And ∠ABP = ∠DBQ (vertically opposite). So ∠ACP = ∠ABP = ∠DBQ = ∠QCD (same segment) → ∠ACP = ∠QCD ✓

Prove that the quadrilateral formed by internal angle bisectors of any quadrilateral is cyclic (Example 5 in full detail).

5 Mark

Given: ABCD is any quadrilateral. Bisectors of ∠A, ∠B, ∠C, ∠D form quadrilateral PQRS (where P=bisectors of A,B; Q=bisectors of B,C; R=bisectors of C,D; S=bisectors of D,A). To prove: PQRS is cyclic.
In △APB (P is intersection of bisectors of ∠A and ∠B): ∠APB = 180°−½∠A−½∠B = 180°−½(∠A+∠B).
∠SPQ = ∠APB = 180°−½(∠A+∠B) (vertically opposite).
Similarly ∠SRQ = 180°−½(∠C+∠D).
∠SPQ+∠SRQ = [180°−½(∠A+∠B)] + [180°−½(∠C+∠D)] = 360°−½(∠A+∠B+∠C+∠D) = 360°−½×360° = 180°.
Opposite angles sum to 180° → PQRS is cyclic (Theorem 9.11). ✓

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

5 Mark

Given: Trapezium ABCD with AB∥CD and AD=BC. To prove: ABCD is cyclic.
Draw DE⊥AB and CF⊥AB. In △DEA and △CFB: DE=CF (equal heights since AD=BC) | DA=CB (given) | ∠DEA=∠CFB=90° → △DEA≅△CFB (RHS) → ∠DAB=∠CBA (CPCT).
∠DAB+∠ADC=180° (co-interior, AB∥DC). Also ∠CBA+∠BCD=180° (co-interior). Since ∠DAB=∠CBA: ∠ADC=∠BCD. So ∠DAB+∠BCD = ∠CBA+∠BCD = 180°. ∴ ABCD is cyclic (Theorem 9.11). ✓

If diagonals of a cyclic quadrilateral are diameters of the circle, prove it is a rectangle.

5 Mark

Given: Cyclic quadrilateral ABCD. Diagonals AC and BD are diameters. To prove: ABCD is a rectangle.
AC is diameter → ∠ADC = 90° (angle in semicircle, D on circle). Similarly ∠ABC = 90°.
BD is diameter → ∠BAD = 90° (A on circle). ∠BCD = 90°.
All four angles = 90° → ABCD is a rectangle. ✓

In circle with centre O, AB is diameter. CD is a chord equal to the radius. AC and BD extended intersect at E. Prove ∠AEB = 60°. (Full proof with all steps)

5 Mark

Given: AB is diameter, CD = radius = r. AC and BD extended meet at E. To prove: ∠AEB = 60°.
Step 1: Join OC and OD. OC=OD=CD=r (all radii, and CD=radius) → △OCD is equilateral → ∠COD = 60°.
Step 2: ∠CBD = ½∠COD = ½×60° = 30° (Theorem 9.7, inscribed angle = ½ central angle on same arc).
Step 3: ∠ACB = 90° (angle in semicircle, AB is diameter).
Step 4: ∠BCE = 180°−∠ACB = 90° (angles on straight line AC extended to E).
Step 5: In △BCE: ∠CBE = ∠CBD = 30°, ∠BCE = 90° → ∠CEB = 180°−90°−30° = 60°.
Step 6: ∠AEB = ∠CEB = 60° (same angle). ✓