Chapter 9 of 12

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Madhya Pradesh (MPBSE)Class 9 Mathematics★ 6–8 marks · CBSE Class 9 board

Circles

The circle is the most symmetric shape — and that symmetry produces a cascade of beautiful theorems. Class 9 chapter on chords, arcs, angles subtended, cyclic quadrilaterals — the foundation of trigonometry, calculus, and physics. 25+ exam problems with full step-by-step proofs.

⏱ 40 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Identify radius, chord, diameter, arc, sector, segment in a circle
2Apply Theorem 1: perpendicular from centre bisects a chord (and converse)
3Apply Theorem 2: equal chords are equidistant from the centre (and converse)
4Apply Theorem 3: equal chords cut off equal arcs (and converse)
5Apply Theorem 4: angle at centre = 2 × angle at circumference (subtended by same arc)
6Use Corollary: angles in the same segment are equal
7Use Corollary: angle in a semicircle is 90° (Thales)
8Apply the Cyclic Quadrilateral theorem (opposite angles sum to 180°)

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Why this chapter matters

Circle theorems are gateway to trigonometry, calculus, physics (orbital motion, wave functions), engineering (gears, wheels) and even modern internet algorithms. Master angle subtention and cyclic-quadrilateral and 90% of geometry questions in Classes 9–12 fall.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Triangles & congruence (Class 9 Ch 7)

SSS, SAS, RHS — used in every circle proof

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Pythagoras' theorem (Class 7-8)

For chord-perpendicular-radius problems

Circles — Class 9 (CBSE)

The circle is mathematics at its purest. A single number — the radius — determines everything: the curve's shape, its symmetry, its area, its circumference. Every theorem about circles flows from one fact: every point on a circle is equidistant from the centre. The richness of what follows from that is one of the wonders of geometry.

1. The story — why circles matter

The wheel revolutionised civilisation around 3500 BCE. The pottery wheel, the cart wheel, the spindle — all circles. Why? Because a circle is the only shape where every point on the boundary is the same distance from the centre — meaning a wheel rolls smoothly, a pot spins evenly, a planet orbits stably.

Modern engineering, physics, astronomy, music theory, GPS, internet routing — all use circles or spheres at their core. Even the Internet's TCP/IP packet routing imagines packets traveling on a globe.

In Class 9 you'll prove the key theorems that link chords, arcs and angles at the centre.

2. The big picture — three foundational theorems

Chords equidistant from the centre are equal (and vice versa).
The angle subtended by a chord at the centre is DOUBLE that at any point on the major arc.
Opposite angles of a cyclic quadrilateral are supplementary.

These three answer 90% of all circle questions in CBSE.

3. Vocabulary — the circle's parts

Circle: the set of all points in a plane at a fixed distance (radius) from a fixed point (centre).
Centre: the fixed point. Notation: $O$ .
Radius: distance from centre to any point on the circle. Notation: $r$ .
Chord: a line segment joining two points on the circle.
Diameter: the longest chord; passes through the centre. Length $= 2 r$ .
Arc: a connected portion of the circle.
Minor arc, major arc: a chord (other than a diameter) divides the circle into a smaller arc (minor) and a larger arc (major).
Semicircle: an arc that's exactly half the circle (cut off by a diameter).
Sector: region bounded by two radii and the arc between them. Like a pizza slice.
Segment: region bounded by a chord and the arc cut off by it.
Subtended angle: the angle formed at a point by the lines/rays drawn to the two endpoints of a chord/arc.

4. Chord and the perpendicular from the centre

Theorem 1. The perpendicular from the centre of a circle to a chord BISECTS the chord.

Proof. Let $O M ⊥ A B$ where $A B$ is a chord and $O$ is the centre.

In $△ O A M$ and $△ O B M$ :

$O A = O B$ (radii).
$∠ O M A = ∠ O M B = 90°$ (given).
$O M = O M$ (common).

By RHS: $△ O A M ≅ △ O B M$ . By CPCT: $A M = M B$ . ∎

Converse. The line from the centre to the midpoint of a chord is PERPENDICULAR to the chord.

(Proof exercise — uses SSS.)

Worked example. A chord of length 24 cm is at distance 5 cm from the centre. Find the radius.

Half-chord = 12 cm. Distance = 5 cm. By Pythagoras: $r^{2} = 1 2^{2} + 5^{2} = 144 + 25 = 169 \Rightarrow r = 13$ cm.

5. Equal chords and their distances

Theorem 2. Equal chords of a circle are equidistant from the centre. Conversely, chords equidistant from the centre are equal.

Proof sketch. Use the perpendicular distances $O M_{1}$ and $O M_{2}$ (where $M_{1}, M_{2}$ are midpoints of the two chords). In the right triangles formed by half-chord, radius and distance, equal chords ⇔ equal half-chords ⇔ equal distances (by Pythagoras).

Worked example. Two chords of a circle have lengths 6 cm and 8 cm. Their perpendicular distances from the centre are 4 cm and $d$ cm respectively. Find $d$ given the radius is 5 cm.

For the 6 cm chord: $r^{2} = 3^{2} + 4^{2} = 25$ . ✓ (Confirms $r = 5$ .) For the 8 cm chord: $5^{2} = 4^{2} + d^{2} \Rightarrow 25 = 16 + d^{2} \Rightarrow d = 3$ cm.

6. Equal chords subtend equal arcs (and vice versa)

Theorem 3. Equal chords of a circle (or congruent circles) cut off equal arcs.

This is intuitive — the same-length chord cuts off the same-sized "bite" from the circle.

Converse. Equal arcs are cut off by equal chords.

7. Angle subtended at the centre vs at the circumference

This is the central theorem of the chapter.

Theorem 4. The angle subtended by an arc at the centre is DOUBLE the angle subtended by it at any point on the remaining (major) arc.

In symbols: if arc $A B$ subtends $∠ A O B$ at the centre and $∠ A P B$ at a point $P$ on the major arc, then $∠ A O B = 2∠ A P B$ .

Proof. (For the case where $O$ lies inside $△ A P B$ .)

Join $P O$ and extend to meet the circle at $Q$ on the major arc.

In $△ O A P$ : $O A = O P$ (radii). So $△ O A P$ is isosceles. By Isosceles Theorem, $∠ O A P = ∠ O P A$ . Exterior angle $∠ A O Q = ∠ O A P + ∠ O P A = 2∠ O P A$ .

Similarly in $△ O B P$ : $∠ B O Q = 2∠ O P B$ .

Adding: $∠ A O Q + ∠ B O Q = 2 (∠ O P A + ∠ O P B)$ . LHS = $∠ A O B$ . RHS = $2∠ A P B$ .

Hence $∠ A O B = 2∠ A P B$ . ∎

Corollary 1. Angles in the SAME segment of a circle are EQUAL. (Because they all equal half the central angle of the same arc.)

Corollary 2. The angle in a SEMICIRCLE is a RIGHT ANGLE. (Because the angle subtended by a diameter at the centre is $180°$ , so at any point on the semicircle it's $90°$ .)

This second corollary — the angle in a semicircle = $90°$ — is Thales' Theorem, one of the oldest results in geometry (600 BCE).

8. Cyclic Quadrilateral Theorem

A cyclic quadrilateral is a quadrilateral whose four vertices all lie on a circle. Three points always determine a unique circle; four do sometimes.

Theorem 5 (Cyclic Quadrilateral). The sum of either pair of opposite angles of a cyclic quadrilateral is $180°$ .

Proof. Let $A B C D$ be cyclic, centre $O$ . Consider opposite angles $∠ A$ and $∠ C$ .

By Theorem 4:

$∠ A$ subtends arc $B C D$ at point $A$ → $∠ A = (1/2)$ × (central angle of arc $B C D$ ).
$∠ C$ subtends arc $B A D$ at point $C$ → $∠ C = (1/2)$ × (central angle of arc $B A D$ ).

Adding: $∠ A + ∠ C = (1/2)$ × (sum of the two arcs) = $(1/2)$ × $360°$ = $180°$ . ∎

Converse. If a pair of opposite angles of a quadrilateral sums to $180°$ , the quadrilateral is cyclic.

9. Seven worked exam examples

Example 1 — Chord distance (2 marks)

A chord 10 cm long is at distance 12 cm from the centre. Find the radius. Half-chord = 5. $r^{2} = 5^{2} + 1 2^{2} = 25 + 144 = 169 \Rightarrow r = 13$ cm.

Example 2 — Angle at centre (2 marks)

An arc subtends an angle of 70° at the centre. Find the angle it subtends at any point on the major arc. By Theorem 4: angle at circumference = $70°/2 = 35°$ .

Example 3 — Thales (2 marks)

In a circle with diameter $P Q$ , point $R$ is on the circle. Find $∠ P R Q$ . By Corollary 2 (angle in semicircle = 90°): $∠ P R Q = 90°$ .

Example 4 — Cyclic quadrilateral (3 marks)

In cyclic quadrilateral $A B C D$ , $∠ A = 75°$ and $∠ B = 120°$ . Find $∠ C$ and $∠ D$ . $∠ A + ∠ C = 180° \Rightarrow ∠ C = 105°$ . $∠ B + ∠ D = 180° \Rightarrow ∠ D = 60°$ .

Example 5 — Angle in same segment (3 marks)

In a circle, $A, B, C, D$ are on the same major arc. $∠ A D B = 50°$ . Find $∠ A C B$ . Angles in the same segment are equal → $∠ A C B = 50°$ .

Example 6 — Equal chords (3 marks)

Two chords of lengths 16 cm and 30 cm are at distances 15 cm and 8 cm from the centre. Find the radius. For the 16 cm chord: $r^{2} = 8^{2} + 1 5^{2} = 64 + 225 = 289 \Rightarrow r = 17$ . For the 30 cm chord: $r^{2} = 1 5^{2} + 8^{2} = 225 + 64 = 289 \Rightarrow r = 17$ . ✓ Both confirm $r = 17$ cm.

Example 7 — HOTS (4 marks)

Prove that if two equal chords intersect inside a circle, the line through the point of intersection and the centre makes equal angles with the chords.

(Sketch.) Drop perpendiculars from the centre to both chords. Equal chords are equidistant from the centre. Use congruent triangles to show angles are equal. (Full proof is more involved — see textbook.)

10. Common pitfalls

Halving the wrong angle. Theorem 4 says CENTRE angle = $2 \times$ circumference angle. Not the other way around.
Forgetting 'major arc' condition. The angle on the MINOR arc subtends a reflex angle at the centre. Be careful which side of the chord you're on.
Confusing chord with diameter. A diameter is a special chord. Not every chord is a diameter.
Misapplying Cyclic Quadrilateral Theorem to non-cyclic quadrilaterals. Only works if all four vertices are on a circle.
Drawing diagrams without marking the centre. Always mark $O$ and any radii / chord midpoints / perpendiculars in your diagrams.
Wrong perpendicular. It's the perpendicular FROM THE CENTRE to the chord that bisects the chord — not any other perpendicular.

11. Beyond NCERT — stretch problems

Stretch 1 — Tangent-chord angle (preview)

The angle between a tangent to a circle and a chord through the point of tangency equals the angle in the alternate segment. (Class 10 / JEE Foundation.)

Stretch 2 — Power of a point

If a chord through external point $P$ cuts the circle at $A$ and $B$ , then $P A \cdot P B$ is constant for all chords through $P$ . (Used in olympiads.)

Stretch 3 — Ptolemy

For a cyclic quadrilateral $A B C D$ : $A C \cdot B D = A B \cdot C D + A D \cdot B C$ . (Beautiful identity, olympiad classic.)

12. Real-world circles

GPS. Each satellite signal traces a sphere of constant signal-time. Three intersecting spheres = a unique location.
Wheels and gears. Why every wheel is circular — only shape with constant radius.
Astronomy. Planets orbit in ellipses — but circles are the limiting case (eccentricity 0).
Architecture. Domes, arches, rotundas use circular symmetry for strength.
Music. Frequencies of musical notes form 'circles of fifths' — geometric arrangement of pitch classes.
Internet. Server load-balancing uses 'consistent hashing' — placing nodes on a circle.
Sports. Discus, hammer throw use circular motion. Basketball hoop is circular.

13. CBSE exam blueprint

Type	Marks	Typical question	Time
VSA	1	Identify parts; basic theorem application	30 sec
SA-I	2	Chord length, perpendicular distance, Pythagoras	2 min
SA-II	3	Cyclic quadrilateral angles; equal chords problem	4–5 min
LA	4	Multi-step proof using Theorem 4; HOTS	6–8 min

Total marks: 6–8 / 80 in Class 9 finals. The chapter is moderately weighted but conceptually powerful.

Three exam-day strategies:

Always draw the circle with the centre clearly marked. Most theorems involve the centre.
For chord problems, immediately drop perpendiculars from the centre — it sets up Pythagoras instantly.
For cyclic quadrilateral problems, recall the 180° pair. It's almost always the first equation to write down.

14. NCERT exercise walkthrough

Exercise 9.1: 6 questions — chord and perpendicular; equal chords distance.
Exercise 9.2: 12 questions — angles subtended by arcs; cyclic quadrilateral problems.

15. 60-second recap

Circle = points equidistant from centre. Radius $r$ , diameter $2 r$ .
Chord = segment between two points on the circle. Diameter = longest chord.
Perpendicular from centre bisects the chord (and converse).
Equal chords ↔ equidistant from centre (and ↔ equal arcs cut off).
Angle at centre = 2 × angle at circumference (subtended by the same arc, major arc side).
Angles in same segment are equal.
Angle in semicircle = 90° (Thales).
Cyclic quadrilateral: opposite angles sum to 180°.

Take the practice quiz and the flashcard deck. Next: Heron's Formula.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Circle definition

Set of points at distance r from centre O

Diameter

d = 2r

Longest chord.

⊥ from O bisects chord

OM ⊥ AB at midpoint M

Theorem 1.

Equal chords

AB = CD ⇔ d(O, AB) = d(O, CD)

Theorem 2.

Equal chords ↔ equal arcs

AB = CD ⇔ arc(AB) = arc(CD)

Theorem 3.

Centre vs circumference

∠centre = 2 × ∠circumference (same arc)

Theorem 4.

Same segment

Angles in the same segment are equal

Corollary of Theorem 4.

Thales' Theorem

Angle in a semicircle = 90°

Cyclic quadrilateral

∠A + ∠C = 180°, ∠B + ∠D = 180°

Theorem 5 — opposite angles supplementary.

Chord-perpendicular-radius (Pythagoras)

r² = (chord/2)² + d²

Where d = distance from centre to chord.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Halving the wrong angle in Theorem 4

✓ Central angle = 2 × inscribed angle (NOT the other way). Bigger angle at centre.

WATCH OUT

✗ Ignoring 'major arc' qualifier

✓ Theorem 4 says: angle at any point on the MAJOR arc. Inscribed angle on minor arc subtends the REFLEX central angle.

WATCH OUT

✗ Applying cyclic theorem to non-cyclic quadrilateral

✓ All 4 vertices must lie on a circle. Otherwise the 180° rule doesn't apply.

WATCH OUT

✗ Forgetting Pythagoras setup

✓ For chord/perpendicular/radius, ALWAYS form a right triangle: radius is hypotenuse, half-chord and distance are legs.

WATCH OUT

✗ Mixing up chord midpoint

✓ Only the perpendicular FROM THE CENTRE bisects a chord. A perpendicular at the midpoint doesn't necessarily pass through the centre.

NCERT exercises

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 9.1

Exercise 9.1

Chords and perpendicular distance from centre

Questions

Ex 9.2

Exercise 9.2

Angles subtended; cyclic quadrilateral problems

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Circles?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min worth ~8 marks in Madhya Pradesh (MPBSE) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Circle: points equidistant from centre. Radius r, diameter 2r.
•Chord = segment between 2 points on circle. Diameter is longest chord.
•Perpendicular from centre BISECTS the chord (and converse).
•Equal chords ↔ equal distances from centre ↔ equal arcs cut.
•Theorem 4: ∠centre = 2 × ∠any point on major arc (same arc).
•Angles in same segment are equal.
•Thales: angle in a semicircle = 90°.
•Cyclic quadrilateral: opposite angles sum to 180°.
•Chord-perpendicular setup ⇒ use Pythagoras with radius/half-chord/distance.

Madhya Pradesh (MPBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6–8 marks

Question type	Marks each	Typical count	What it tests
MCQ	1	1	Definition or formula identification
Short answer (2-mark)	2	1–2	Direct formula application
Long answer (3-mark)	3	1–2	Multi-step problem solving
Long answer (4-mark)	4	1	Proof or complex application

Prep strategy

Write all key formulas on a revision sheet — recall speed matters in board exams
Show ALL working steps clearly — partial marks are awarded for method even if final answer is wrong
Practise past 5 years of CBSE board questions for this chapter
For proof questions: state the theorem/property used at each step by name
Time management: allocate time based on marks — 1 min per mark is a good rule

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

GPS triangulation

Each satellite signal traces a sphere (3D circle). 4+ satellite intersections pin down location.

Wheels and gears

Only shape with constant radius — rolls smoothly without bumping.

Astronomy

Planetary orbits are ellipses; circles are the limiting case (eccentricity 0).

Architecture (domes)

Circular cross-sections distribute load evenly — used in domes from Rome's Pantheon to modern stadiums.

Music — circle of fifths

Pitch classes arranged on a circle; key changes are 'rotations' on this circle.

Internet — consistent hashing

Server load-balancing places nodes on a virtual circle; data points hash to nearest node.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

ALWAYS draw the circle with the centre clearly marked. Most circle theorems need O.

For chord problems, IMMEDIATELY drop perpendicular from centre — it sets up Pythagoras (radius² = half-chord² + distance²).

For cyclic quadrilateral problems, write the 180° pair as the first equation. It often unlocks the whole problem.

When you see TWO equal chords, use 'equidistant from centre' to set up equal perpendiculars and find congruent triangles.

Always cite the theorem by number ('by Theorem 4', 'Cyclic Quadrilateral Theorem') to earn proof marks.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Tangent-chord angle theorem (Class 10): the angle between a tangent and a chord at the point of tangency = the angle in the alternate segment.

STRETCH

Power of a Point: for any line through external point P intersecting a circle at A, B: PA · PB is constant (depends only on P and the circle).

STRETCH

Ptolemy's Theorem: for a cyclic quadrilateral ABCD, AC · BD = AB·CD + AD·BC.

STRETCH

Nine-point circle: in any triangle, the midpoints of the sides, feet of the altitudes, and midpoints of vertex-to-orthocentre segments all lie on a circle of half the triangle's circumradius.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

JEE FoundationHigh — Theorem 4 and cyclic quad problems are standard

NTSEHigh — basic circle MCQs are popular

IMO FoundationVery high — circle geometry features in every olympiad

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

A diameter is a special chord — one that passes through the centre. So every diameter is a chord, but not every chord is a diameter.

The diameter subtends a central angle of 180°. By Theorem 4, the inscribed angle is half — 90°. This is Thales' Theorem.

Yes — if opposite angles of the quadrilateral they form sum to 180°, then yes (converse of Cyclic Theorem).

In the simple case where O is inside ∆APB, both arms PA and PB lie on the same side of O. If P is on the minor arc, the geometry shifts and the relation involves the reflex angle at O.

No. The diameter is THE LONGEST chord, by definition. Any other chord is strictly shorter.

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