Chapter 9 of 12

75%

CBSE 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

Ask AI about this

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°)

💡

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.

🔗

Triangles & congruence (Class 9 Ch 7)

SSS, SAS, RHS — used in every circle proof

🔗

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.

⚠️

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 (with solutions)

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

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Chord-Pythagoras

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

▶ Show solution

Step 1 — Drop the perpendicular from centre O to chord AB at midpoint M.
  This forms a right triangle OMA with OA = r (radius), OM = 6 (distance), AM = 8 (half-chord).

Step 2 — Apply Pythagoras.
  OA² = OM² + AM²
  r² = 6² + 8² = 36 + 64 = 100.
  r = 10.

✦ Answer: Radius = 10 cm.

Q2EASY· Angle at centre

An arc subtends an angle of 100° at the centre. What angle does it subtend at a point on the major arc?

▶ Show solution

Step 1 — Apply Theorem 4.
  Angle at centre = 2 × angle at any point on the major arc.

Step 2 — Solve.
  Inscribed angle = central angle / 2 = 100° / 2 = 50°.

✦ Answer: 50°.

Q3EASY· Thales

In a circle, AB is a diameter and P is a point on the circle. What is ∠APB?

▶ Show solution

Step 1 — Apply Thales' Theorem (Corollary 2).
  The angle in a semicircle (subtended by a diameter) is 90°.

✦ Answer: ∠APB = 90°.

This is one of the oldest theorems in geometry — attributed to Thales of Miletus circa 600 BCE.

Q4EASY· Cyclic quad

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

▶ Show solution

Step 1 — Apply Cyclic Quadrilateral Theorem.
  Opposite angles of a cyclic quadrilateral sum to 180°.
  ∠A + ∠C = 180°.

Step 2 — Solve.
  80° + ∠C = 180° → ∠C = 100°.

✦ Answer: ∠C = 100°.

Q5EASY· Same segment

In a circle, points A, B, C, D lie on the same arc. If ∠ACB = 40°, find ∠ADB.

▶ Show solution

Step 1 — Recall: angles in the same segment are equal.
  ∠ACB and ∠ADB both subtend chord AB and both vertices C, D are on the same arc.

Step 2 — So ∠ADB = ∠ACB = 40°.

✦ Answer: ∠ADB = 40°.

Q6MEDIUM· Chord-perpendicular

In a circle of radius 13 cm, a chord is at distance 5 cm from the centre. Find the length of the chord.

▶ Show solution

Step 1 — Setup the right triangle OMA.
  OA = 13 (radius), OM = 5 (distance from centre to chord).

Step 2 — Find half-chord AM by Pythagoras.
  AM² = OA² − OM² = 169 − 25 = 144.
  AM = 12.

Step 3 — Chord length AB = 2 × AM = 24.

✦ Answer: Chord length = 24 cm.

Q7MEDIUM· Equal chords

Two chords of a circle, AB = 16 cm and CD = 12 cm. The perpendicular from the centre to AB is 6 cm. Find the perpendicular distance from the centre to CD. (Radius is the same circle.)

▶ Show solution

Step 1 — Use AB to find the radius.
  AB = 16, half-chord = 8. Perpendicular distance from centre = 6.
  r² = 6² + 8² = 36 + 64 = 100 → r = 10.

Step 2 — For chord CD = 12, half-chord = 6.
  Let d = perpendicular distance from centre to CD.
  r² = d² + 6²
  100 = d² + 36 → d² = 64 → d = 8.

✦ Answer: Perpendicular distance from centre to CD = 8 cm.

Observation: longer chord is CLOSER to the centre (AB closer than CD, since shorter chord is farther).

Q8MEDIUM· Subtended angles

In a circle, an arc subtends an angle of 130° at the centre. Find the angles subtended at any point on (a) the major arc, (b) the minor arc.

▶ Show solution

Step 1 — Major arc.
  By Theorem 4, angle at any point on the MAJOR arc = (1/2) × central angle = 130° / 2 = 65°.

Step 2 — Minor arc.
  A point on the MINOR arc subtends an angle related to the REFLEX of the central angle.
  Reflex angle = 360° − 130° = 230°.
  Inscribed angle on minor arc = 230° / 2 = 115°.

Step 3 — Verify by cyclic-quadrilateral logic.
  The angle on major arc (65°) + angle on minor arc (115°) = 180°. ✓ (They subtend supplementary inscribed angles — a special case of the cyclic theorem.)

✦ Answer: 65° on the major arc, 115° on the minor arc.

Q9MEDIUM· Cyclic quad

In cyclic quadrilateral ABCD, ∠A : ∠B = 2 : 3. If ∠C = 70°, find ∠A, ∠B, ∠D.

▶ Show solution

Step 1 — Use Cyclic Theorem for opposite angles.
  ∠A + ∠C = 180° → ∠A = 180° − 70° = 110°.

Step 2 — Find ∠B from the ratio.
  ∠A : ∠B = 2 : 3 → 110 : ∠B = 2 : 3.
  ∠B = (3/2) × 110 = 165°.

Wait — let me check. If ∠A = 110° from the cyclic property, but ∠A : ∠B = 2 : 3 gives ∠B = 165°. Then ∠B + ∠D = 180° → ∠D = 180 − 165 = 15°.

Step 3 — Verify all four angles sum to 360°.
  110 + 165 + 70 + 15 = 360. ✓

✦ Answer: ∠A = 110°, ∠B = 165°, ∠C = 70° (given), ∠D = 15°.

Note: the problem's ratio combined with the given ∠C forces ∠B to be quite large. Worth verifying the angles are valid (each < 360° — yes).

Q10MEDIUM· Multi-chord

Two equal chords AB and CD of a circle intersect at point P inside the circle. Prove that PA = PD (or equivalently, the segments from P to corresponding endpoints are equal).

▶ Show solution

Step 1 — Equal chords are equidistant from the centre (Theorem 2).
  Let OM ⊥ AB at M, ON ⊥ CD at N. AB = CD ⇒ OM = ON.

Step 2 — In △OMP and △ONP.
  OM = ON (Step 1).
  ∠OMP = ∠ONP = 90°.
  OP = OP (common).
  By RHS, △OMP ≅ △ONP.

Step 3 — By CPCT, MP = NP.

Step 4 — Since M is midpoint of AB, AM = MB. Since N is midpoint of CD, CN = ND.
  So PA = AM − MP and PD = ND − NP (depending on which side of M and N point P lies; specific arrangement may vary).
  Given the symmetry MP = NP and AM = ND, we get PA = PD.

✦ Answer: PA = PD. ∎

This is a classic 'use perpendicular distance to relate chords' style problem.

Q11HARD· HOTS — Theorem 4

Prove Theorem 4 for the case where the centre lies INSIDE △APB (where ∠APB is the inscribed angle).

▶ Show solution

Step 1 — Setup. Centre O inside △APB. Draw radii OA, OB, OP. Extend PO to meet the circle at Q (on the major arc, opposite side from P).

Step 2 — In △OPA, OA = OP = r (radii).
  So △OPA is isosceles → ∠OAP = ∠OPA. (Isosceles theorem.)

Step 3 — The exterior angle of △OPA at O is ∠AOQ.
  By the Exterior Angle Theorem of a triangle:
  ∠AOQ = ∠OAP + ∠OPA = 2 ∠OPA.

Step 4 — Similarly in △OPB.
  OB = OP → △OPB isosceles → ∠OBP = ∠OPB.
  Exterior angle: ∠BOQ = ∠OBP + ∠OPB = 2 ∠OPB.

Step 5 — Add the two exterior angles at O.
  ∠AOQ + ∠BOQ = 2 ∠OPA + 2 ∠OPB = 2 (∠OPA + ∠OPB).

Step 6 — Recognise: ∠AOQ + ∠BOQ = ∠AOB (since O is inside ∠APB, the two angles at O sum to the central angle).
  And ∠OPA + ∠OPB = ∠APB.

Step 7 — Therefore ∠AOB = 2 ∠APB. ∎

✦ Answer: ∠AOB = 2 ∠APB.

Note: similar proofs handle the other cases (O outside △APB or on a side).

Q12HARD· HOTS — cyclic

Prove: if a parallelogram is cyclic, it must be a rectangle.

▶ Show solution

Step 1 — Setup. Let ABCD be a parallelogram that's also cyclic.

Step 2 — Apply parallelogram property.
  Adjacent angles are supplementary: ∠A + ∠B = 180°.

Step 3 — Apply cyclic property.
  Opposite angles sum to 180°: ∠A + ∠C = 180°.

Step 4 — Also from parallelogram, ∠A = ∠C (opposite angles equal).

Step 5 — Substitute ∠C = ∠A into Step 3:
  ∠A + ∠A = 180° → 2∠A = 180° → ∠A = 90°.

Step 6 — So all four angles are 90° (∠A = ∠C = 90°; ∠B = ∠D = 90° by similar argument).
  A parallelogram with all 90° angles is a rectangle.

✦ Answer: A cyclic parallelogram is necessarily a rectangle. ∎

Elegant interaction between two chapter theorems.

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.

AI helpers for this chapter

Generate more practice on the fly, or have the chapter explained at a different level.

⚡

Generate 5 more questions

Fresh MCQs + short-answer items based on this chapter, plus answer keys.

🧠

Explain at a different level

Re-explain the chapter at the level you actually want.

📝 My notes

Saved on this device — sign in to sync · how this works

0/600

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.

Practice more, ask doubts, find a tutor

CBSE Class 9 mock tests

CBSE Class 9 PYQs

Ask in Q&A

Mathematics flashcards

Related chapters

Students who read Circles usually read these next.

Number Systems

From counting cows to imaginary numbers — how mathematicians extended the number line over 4000 years. Class 9 Number Systems with rationals, irrationals, real numbers, decimal expansions, the number line, surds, rationalisation, and 30+ exam-style problems.

Polynomials

From counting apples to encoding all of algebra — polynomials are the words of mathematics. Class 9 Polynomials with one-variable polynomials, degree, zeros, the Remainder & Factor Theorems, identities for factorisation, and 25+ exam-style problems with full step-by-step solutions.

Coordinate Geometry

From Descartes' fly on the ceiling to plotting your own GPS — Class 9 Coordinate Geometry, taught the way a great teacher would. Cartesian plane, quadrants, plotting, reflections, distance along axes, real-world uses and 30+ worked problems.

Linear Equations in Two Variables

Every train timetable, every demand-and-supply curve, every flight path — they're all linear equations in two variables. Class 9 chapter on solutions, plotting, slope-intercept ideas, geometric meaning, axis-equations, and real-world applications with 25+ worked exam problems.