+91 8989006839 edugrownteam@gmail.com
Login

Chapter 5 I’m Up and Down, and Round and Round Quick Revision notes | Class 9th Mathematics (Ganita Manjari) notes

July 29, 2026  •  Admin ETeam  •  Class 9, class 9 Quick revision notes
CLASS 9  •  MATHS  •  CHAPTER 5

I’m Up and Down, and Round and Round

Circles — short, colourful, handwritten-style notes by EduGrown

1Circles All Around Us

Humans have been fascinated by round shapes since the very beginning. In the cave paintings of Gudahandi (Odisha) we already find triangles, squares, circles and ovals — shapes people copied straight from nature.

Fig. 5.1 — Raindrop ripples, a plant stem cross-section and a sunflower head | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.1 — Raindrop ripples, a plant stem cross-section and a sunflower head
🌍 Circles in nature
  • Ripples formed when a raindrop falls on water
  • The cross-section of a plant stem
  • The inflorescence of a sunflower
  • The full moon and the sun
Fig. 5.2 — The Moon, and the Sun during a total solar eclipse | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.2 — The Moon, and the Sun during a total solar eclipse
💡 The one common property Big or small, every circle has a centre, and every point on the circle is at the same distance from that centre. Mathematicians turned this observation around and made it the very definition of a circle.
⇧ back to top

2Basic Definitions & Parts of a Circle

📝 Circle A circle is the set of all points on a plane that are equidistant from a given point on that plane.
Fig. 5.3 — Circle with centre A, chord BC and diameter | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.3 — Circle with centre A, chord BC and diameter
TermMeaning
CentreThe fixed point from which all points of the circle are equidistant (point A)
RadiusDistance from the centre to any point on the circle
ChordA line segment joining any two points on the circle (e.g. BC)
DiameterA chord that passes through the centre — the longest chord
LocusThe set of all points satisfying a given condition
📝 Circle as a locus A circle is the locus of points that are equidistant from a given point.
✔ Two other loci worth remembering:
  • Points equidistant from a fixed point → a circle
  • Points equidistant from two fixed points A, B → the perpendicular bisector of AB
💡 Angle subtended by a chord The angle that chord BC makes at the centre A is ∠BAC. This idea of “subtending an angle” runs through the whole chapter — learn it well!
⇧ back to top

3Symmetries of a Circle

⭐ Two kinds of symmetry
  • Rotational symmetry: rotate a circle about its centre by any angle — it looks exactly the same. This is called complete rotational symmetry.
  • Reflection symmetry: fold a paper circle so the boundaries overlap. The crease is a line of symmetry — and it always passes through the centre. So every diameter is a line of reflection symmetry, and there are infinitely many of them.
✏️ The rotating wheel Look at a moving wheel and note the point touching the ground. Look again later — is it the same point? You cannot tell! A rotating circle looks identical at every instant. That is complete rotational symmetry in action.
💡 Handy fact Longest chord in a circle of radius r = the diameter = 2r. And there is no smallest chord — a chord can shrink towards a single point, getting as tiny as you like.
⇧ back to top

4How Many Circles Through Given Points?

✔ Through TWO points A and B

Fig. 5.4 — Infinitely many circles pass through two points | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.4 — Infinitely many circles pass through two points
⭐ The answer: infinitely many!
  • The centre O must satisfy OA = OB, so O is equidistant from A and B.
  • All such points lie on the perpendicular bisector of AB.
  • Every point on that bisector gives one circle → infinitely many circles.
💡 Two things to notice
  • Smallest circle: centre at the midpoint of AB, radius = 12 AB. Here AB is a diameter.
  • Move further along the bisector → radius grows, and the circle looks less curved near A and B. There is no largest circle.

✔ Through THREE points A, B and C

⚠️ Careful — check collinearity first If A, B and C are collinear (on one straight line), then no circle passes through all three. A straight line can cut a circle in at most 2 points.
THEOREM 1

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

📝 Why it is true (in 3 lines)
  • OA = OB → O lies on the perpendicular bisector of AB.
  • OA = OC → O lies on the perpendicular bisector of AC.
  • The points are not collinear, so these two bisectors meet at exactly one point — that point is O. One centre → one circle.
⇧ back to top

5Circumcircle & Circumcentre

📝 Definitions
  • The circle through the three vertices of ΔABC is its circumcircle — it circumscribes the triangle.
  • Its centre O is the circumcentre, found where the perpendicular bisectors of the sides meet.
  • The triangle is said to be inscribed in the circle.
Fig. 5.5 — Circumcircle of an acute triangle: centre O lies inside | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.5 — Circumcircle of an acute triangle: centre O lies inside

Where the circumcentre sits depends entirely on the type of triangle:

Fig. 5.6 — Obtuse-angled triangle: circumcentre O lies outside | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.6 — Obtuse-angled triangle: circumcentre O lies outside
Fig. 5.7 — Right-angled triangle: O is the midpoint of the hypotenuse | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.7 — Right-angled triangle: O is the midpoint of the hypotenuse
Type of triangleWhere the circumcentre O lies
Acute-angledInside the triangle
Obtuse-angledOutside the triangle
Right-angledExactly at the midpoint of the hypotenuse
💡 Remember OA = OB = OC always, because all three are radii of the same circle. Every triangle has exactly one circumcircle.
⇧ back to top

6Equal Chords ↔ Equal Angles at the Centre

Tie a thread between two points of a wheel and pull it tight — that thread is a chord. Now rotate the wheel: the chord moves to a new position but its length never changes… and neither does the angle it makes at the centre.

Fig. 5.8 — A thread on a wheel behaves exactly like a chord | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.8 — A thread on a wheel behaves exactly like a chord
THEOREM 2

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

Fig. 5.9 — Equal chords AB and DE subtend equal angles at centre C | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.9 — Equal chords AB and DE subtend equal angles at centre C
📝 Proof idea (SSS) Given AB = DE. In ΔCAB and ΔCDE:
  • CA = CD = r  (radii)
  • CB = CE = r  (radii)
  • AB = DE  (given)
So ΔCAB ≅ ΔCDE by SSS → ∠ACB = ∠DCE ✔
THEOREM 3 — the converse

Chords that subtend equal angles at the centre are equal in length.

Fig. 5.11 — Equal angles at the centre force the chords to be equal | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.11 — Equal angles at the centre force the chords to be equal
📝 Proof idea (SAS) Given ∠ACB = ∠DCE. In ΔACB and ΔDCE:
  • AC = DC = r,   BC = EC = r  (radii)
  • Included angles are equal  (given)
So ΔACB ≅ ΔDCE by SAS → AB = ED ✔
💡 Bonus fact A chord together with the centre always forms an isosceles triangle (two sides are radii). That is why congruence works so smoothly here.
⇧ back to top

7Centre, Midpoint & Perpendicular of a Chord

THEOREM 4

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

Advertisement
Fig. 5.12 — CM joins the centre to midpoint M, so CM ⊥ AB | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.12 — CM joins the centre to midpoint M, so CM ⊥ AB
📝 Proof idea ΔCAB is isosceles (CA = CB = r), and M is the midpoint so AM = BM.
By SAS, ΔCMA ≅ ΔCMB, so ∠CMA = ∠CMB.
But ∠CMA + ∠CMB = 180° (angles on a line), so each is 90° ✔
THEOREM 5 — the converse

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

💡 Two-in-one shortcut From the centre, perpendicular → bisects and bisects → perpendicular. Knowing one of the two automatically gives you the other. Also: the perpendicular bisector of any chord passes through the centre — that is exactly how you find the centre of a circular paper!
⇧ back to top

8Distance of Chords from the Centre

📝 What does “distance” mean here? The distance of a chord from the centre always means the perpendicular distance — i.e. the length from the centre to the midpoint of the chord.

✔ Paper-folding activity

Fig. 5.13 — Paper folding: A the circle, B the chord crease, C the perpendicular | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.13 — Paper folding: A the circle, B the chord crease, C the perpendicular
  • Fold a paper circle inwards from the boundary → the crease is a chord (Fig. 5.13B).
  • Fold again so the two end points of the chord meet → this second crease is the perpendicular from the centre (Fig. 5.13C).
  • The two creases cross exactly at the midpoint of the chord.
THEOREM 6

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

Advertisement
Fig. 5.14 — Equal chords AB and FG are equidistant from centre C | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.14 — Equal chords AB and FG are equidistant from centre C
📝 Two ways to prove it
  1. SSS: ΔCAB ≅ ΔCFG (CA = CF = r, CB = CG = r, AB = FG). Congruent triangles have congruent altitudes → CE = CH.
  2. RHS: In ΔCEA and ΔCHF — AE = FH (halves of equal chords), ∠CEA = ∠CHF = 90°, CA = CF = r → CE = CH.
THEOREM 7 — the converse

Chords that are equidistant from the centre have equal length.

⭐ The must-remember formula If a chord is at perpendicular distance d from the centre of a circle of radius r: Chord length = 2√(r² – d²) It comes straight from the Baudhāyana–Pythagoras Theorem: half the chord, d and r form a right triangle.
✏️ Example r = 5 cm, d = 3 cm → chord = 2√(25 – 9) = 2 × 4 = 8 cm
⇧ back to top

9Longer Chord is Closer to the Centre

THEOREM 8

If AB > DE are two chords of a circle, then AB is nearer to the centre than DE.

Fig. 5.16 — AB > DE, so CF < CG: the longer chord is nearer the centre | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.16 — AB > DE, so CF < CG: the longer chord is nearer the centre
📝 Proof idea Both AC and CD are radii, so AC = CD. By Pythagoras: CF² + AF² = CG² + GD²  (both equal r²) AB > DE means AF > GD (F, G are midpoints). So AF² > GD², which forces CF² < CG², i.e. CF < CG ✔
💡 The two extremes
  • Push the chord towards the centre → it becomes the diameter: longest chord, distance = 0.
  • Push the chord away from the centre → it shrinks to a point: length = 0, distance = r.
So as distance increases, chord length decreases — they move in opposite directions.
⇧ back to top

10Arcs: Major, Minor & the Angles They Subtend

📝 Arc An arc is a connected portion of the circle, fixed by its two end points and the path along the circle joining them.
Fig. 5.17 — Major arc AYB and minor arc AXB | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.17 — Major arc AYB and minor arc AXB
⭐ Major and minor arcs Between two points A and B there are always two paths along the circle:
  • The smaller one is the minor arc — it subtends less than 180° at the centre.
  • The bigger one is the major arc — it subtends more than 180° at the centre.
Fig. 5.18 — The angle an arc sweeps at the centre | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.18 — The angle an arc sweeps at the centre
💡 How to measure the angle of an arc Sweep the radius from OA to OB along that particular arc and measure the angle swept. Go via X → you get the minor arc’s angle. Go via Y → you get the major arc’s (reflex) angle. Together they add to 360°.
⇧ back to top

11Angle at Centre = 2 × Angle on the Circle

THEOREM 9 — the star theorem

The angle subtended by an arc at the centre is double the angle it subtends at any point on the circle outside that arc.

Fig. 5.21 — Proving ∠BCA = 2∠BDA using isosceles triangles | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.21 — Proving ∠BCA = 2∠BDA using isosceles triangles
📝 Proof idea (isosceles + exterior angle) Join D to the centre C and extend it to meet the circle at E.
  • ΔDCB is isosceles (CB = CD = r), so ∠CBD = ∠CDB.
  • By the exterior angle theorem, ∠BCE = ∠CBD + ∠CDB = 2∠BDC.
  • Similarly ∠ACE = 2∠CDA.
  • Adding: ∠BCA = 2(∠BDC + ∠CDA) = 2∠BDA ✔

✔ The beautiful consequence

Fig. 5.23 — ∠AEB = ∠ADB = ∠AFB: angles in the same segment are equal | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.23 — ∠AEB = ∠ADB = ∠AFB: angles in the same segment are equal
⭐ Angles in the same segment are equal Since the centre angle is fixed, every point on the circle outside the arc gives the same angle: ∠AEB = ∠ADB = ∠AFB = 12 ∠ACB It does not matter which point you choose — this is a property no other shape has!
⚠️ Only points ON the circle For points inside or outside the circle the angles are all different. The magic works only for points lying on the circle, outside the arc.
⇧ back to top

12Angle in a Semicircle is 90°

📝 What is a corollary? A corollary is a fact that follows immediately from a result already proved.
COROLLARY of Theorem 9

The angle subtended by a diameter at any point on the circle is 90°.

Advertisement
Fig. 5.24 — AB is a diameter, so ∠ADB = 90° | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.24 — AB is a diameter, so ∠ADB = 90°
⭐ One-line proof AB is a diameter, so the arc from A to B (not containing D) subtends a straight angle of 180° at the centre. ∠ADB = 12 × 180° = 90°
💡 Works both ways If ∠ACB = 90° for a point C on a circle, then AB must be a diameter. Very useful in problems — spot a 90° angle, and you have found a diameter!
⇧ back to top

13Concyclic Points

📝 Definition Points that all lie on the same circle are called concyclic.
THEOREM 10

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

📝 Proof idea (by contradiction)
  • A, B, C are non-collinear, so by Theorem 1 a unique circle passes through them.
  • Suppose D is not on this circle → it is either inside or outside.
  • Either way, the exterior angle theorem forces ∠ACB to be greater than itself — impossible!
  • So D must lie on the circle → all four points are concyclic ✔
⚠️ Two conditions, both needed C and D must be on the same side of AB, and the angles must be equal. Drop either condition and the result fails.
⇧ back to top

14Cyclic Quadrilaterals

📝 Definition When the four vertices of a quadrilateral (a 4-gon) are concyclic, it is called a cyclic quadrilateral.
THEOREM 11

The sum of two opposite angles of a cyclic quadrilateral is 180°.

Fig. 5.28 — Cyclic quadrilateral ABCD with centre O | Class 9 Maths Chapter 5 Circles notes | EduGrown
Fig. 5.28 — Cyclic quadrilateral ABCD with centre O
⭐ Proof idea (the two arcs together)
  • ∠BAD = 12 (reflex ∠BOD)  — A is outside arc BCD
  • ∠BCD = 12 (∠BOD)  — C is outside arc BAD
  • Together the two angles at O make a complete rotation of 360°.
∠BAD + ∠BCD = 12 × 360° = 180°
THEOREM 12 — the converse

If two opposite angles of a quadrilateral add up to 180°, its vertices are concyclic.

💡 Three super-useful spin-offs
  • Since ∠A + ∠C = 180° and ∠B + ∠D = 180°, all four angles add to 360° — as in any quadrilateral.
  • The exterior angle at any vertex equals the interior opposite angle.
  • A rectangle is the only parallelogram that can be cyclic (its opposite angles are 90° + 90° = 180°), and its diagonals meet at the centre.
✏️ Quick check Can a cyclic quadrilateral have ∠A = 80°, ∠B = 110°, ∠C = 100°, ∠D = 70°?
∠A + ∠C = 180° ✔ but ∠B + ∠D = 180° ✔ — so yes, both opposite pairs work.
But if the pairs did not add to 180°, no such cyclic quadrilateral could exist.
⇧ back to top

15All 12 Theorems at a Glance

#Theorem
1There is a unique circle through three non-collinear points.
2Equal chords subtend equal angles at the centre.
3Chords subtending equal angles at the centre are equal.
4Centre to midpoint of a chord → perpendicular to the chord.
5Perpendicular from the centre to a chord bisects it.
6Equal chords are equidistant from the centre.
7Chords equidistant from the centre are equal in length.
8The longer chord lies closer to the centre.
9Angle at the centre = 2 × angle at any point on the circle outside the arc.
★Corollary: angle subtended by a diameter on the circle = 90°.
10Equal angles on the same side of a segment → the four points are concyclic.
11Opposite angles of a cyclic quadrilateral add to 180°.
12Opposite angles adding to 180° → the quadrilateral is cyclic.
💡 Notice the pattern Theorems come in pairs — 2 & 3, 4 & 5, 6 & 7, 11 & 12. Each pair is a statement and its converse. Learn one, and the other is just the arrow reversed!
⇧ back to top

16Common Mistakes to Avoid

❌ Don’t do this
  • Forgetting that “distance of a chord from the centre” always means the perpendicular distance.
  • Thinking a circle can pass through three collinear points. It cannot — a line meets a circle in at most 2 points.
  • Assuming the circumcentre is always inside the triangle. It is outside for obtuse triangles, and on the hypotenuse for right triangles.
  • Applying “angles in the same segment are equal” to points inside or outside the circle. It works only for points on the circle.
  • Using the minor arc angle when the question needs the reflex (major) angle, especially in cyclic-quadrilateral proofs.
  • Writing chord = √(r² – d²) and forgetting the 2 in front — that formula gives only half the chord.
  • Mixing up radius and diameter in the Pythagoras step. Always draw the right triangle first.
  • Assuming any quadrilateral in a circle is cyclic — all four vertices must lie on the circle.
⇧ back to top

17Chapter Summary — Quick Revision

🏆 All of Chapter 5 in 12 lines

  • A circle is the set of all points in a plane at a fixed distance (the radius) from a fixed point (the centre).
  • A circle has reflection symmetry across every diameter, and complete rotational symmetry about its centre.
  • Infinitely many circles pass through two given points; their centres all lie on the perpendicular bisector of the segment joining them.
  • Through three non-collinear points there is exactly one circle — the circumcircle, centred at the circumcentre where the perpendicular bisectors of the sides meet.
  • The circumcentre is inside an acute triangle, outside an obtuse one, and at the midpoint of the hypotenuse of a right triangle.
  • Equal chords ↔ equal angles at the centre (Theorems 2 and 3).
  • From the centre, perpendicular to a chord ↔ bisects the chord (Theorems 4 and 5).
  • Equal chords ↔ equidistant from the centre (Theorems 6 and 7); and chord length = 2√(r² – d²).
  • The longer the chord, the closer it is to the centre. The diameter is the longest chord of all.
  • Angle subtended by an arc at the centre is twice the angle it subtends anywhere on the remaining circle — so angles in the same segment are equal.
  • The angle in a semicircle is always 90°.
  • In a cyclic quadrilateral opposite angles add to 180°, and the converse is also true.

✨ Happy Learning — @edugrown ✨

Leave a Reply

Your email address will not be published. Required fields are marked *

error: Content is protected !!