**According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.**

**CBSE Class 9 Maths Circles Notes**

## Introduction to Circles

### Circles

- The
**set of all the points**in a plane that is at a**fixed distance**from a**fixed point**makes a circle. - A
**fixed point**from which the set of points are at a fixed distance is called the**centre**of the circle. - A circle divides the plane into 3 parts:
**interior**(inside the circle), the**circle**itself and**exterior**(outside the circle)

### Radius

The **distance** between the **centre** of the circle and any **point on its edge** is called the **radius.**

### Tangent and Secant

A **line** that **touches** the circle at **exactly one point** is called its **tangent**. A **line** that **cuts** a circle at **two points** is called a **secant.**

In the above figure: PQ is the tangent, and AB is the secant.

### Chord

The **line segment** within the circle joining any 2 points on the circle is called the chord.

### Diameter

A **chord** passing through the centre of the circle is called the **diameter.** The **Diameter is 2 times the radius**, and it is the **longest chord**.

### Arc

The** portion** of a circle(curve)** between 2 points **is called an **arc**. Among the two pieces made by an arc, the** longer** one is called a **major arc** and the **shorter** one is called a** minor arc.**

### Circumference

The **perimeter **of a circle is the **distance** covered by going around its** boundary once**. The perimeter of a circle has a special name: **Circumference**, which is π times the diameter which is given by the formula 2πr

### Segment and Sector

A circular **segment** is a region of a circle which is “**cut off**” from the rest of the circle by a secant or a chord. **Smaller region **cut off by a chord is called **minor segment** and the **bigger region** is called **major segment**.

A **sector** is the portion of a circle **enclosed by two radii and an arc**, where the **smaller area** is known as the **minor sector** and the **larger** being the **major sector**.

For** 2 equal** arcs or semicircles, both the segment and sector are called the** semicircular region.**

## Circles and Their Chords

### Theorem of equal chords subtending angles at the centre

Equal **ch****ords** subtend equal **angles at the centre**.

**Proof**: AB and CD are the 2 equal chords.

In Δ AOB and Δ COD

OB = OC [Radii]

OA = OD [Radii]

AB = CD [Given]

ΔAOB ≅ ΔCOD (SSS rule)

Hence, ∠AOB = ∠COD [CPCT]

### Theorem of equal angles subtended by different chords

If the** angles** subtended by the chords of a circle at the centre are **equal**, then the **chords are equal.**

Proof: In ΔAOB and ΔCOD

OB = OC [Radii] ∠AOB=∠COD [Given]

OA = OD [Radii]

ΔAOB ≅ ΔCOD (SAS rule)

Hence, AB=CD [CPCT]

### Perpendicular from the centre to a chord bisects the chord

**Perpendicular **from the **centre** of a circle to a** chord bisects the chord**.

Proof: AB is a chord, and OM is the perpendicular drawn from the centre.

From ΔOMB and ΔOMA,

∠OMA=∠OMB=90^{0} OA = OB (radii)

OM = OM (common)

Hence, ΔOMB ≅ ΔOMA (RHS rule)

Therefore AM = MB [CPCT]

### A Line through the centre that bisects the chord is perpendicular to the chord

A **line drawn** through the centre of a circle to **bisect** a chord is **perpendicular** to the chord.

**Proof: **OM drawn from the center to bisect chord AB.

From ΔOMA and ΔOMB,

OA = OB (Radii)

OM = OM (common)

AM = BM (Given)

Therefore, ΔOMA ≅ ΔOMB (SSS rule)

⇒∠OMA=∠OMB (C.P.C.T)

But, ∠OMA+∠OMB=180^{0}

Hence, ∠OMA=∠OMB=90^{0} ⇒OM⊥AB

### Circle through 3 points

There is** one **and **only** one **circle** passing through** three given noncollinear points.** A unique circle passes through 3 vertices of a triangle ABC called as the** circumcircle. **The **centre** and **radius** are called the **circumcenter** and **circumradius** of this triangle, respectively.

### Equal chords are at equal distances from the centre

**Equal chords** of a circle(or of congruent circles) are **equidistant from the centre** (or centres).

**Proof: **Given, AB = CD, O is the centre. Join OA and OC.

Draw, OP⊥AB, OQ⊥CD

In ΔOAP and ΔOCQ,

OA=OC (Radii)

AP=CQ ( AB = CD ⇒(1/2)AB = (1/2)CD since OP and OQ bisects the chords AB and CD.)

ΔOAP ≅ ΔOCQ (RHS rule)

Hence, OP=OQ (C.P.C.T.C)

### Chords equidistant from the centre are equal

Chords **equidistant** from the centre of a circle are **equal in length.**

**Proof: **Given OX = OY (The chords AB and CD are at equidistant) OX⊥AB, OY⊥CD

In ΔAOX and ΔDOY

∠OXA =∠OYD (Both 90^{0})

OA = OD (Radii)

OX = OY (Given)

ΔAOX ≅ ΔDOY (RHS rule)

Therefore AX = DY (CPCT)

Similarly XB = YC

So, AB = CD

## Circles and Quadrilaterals

### The angle subtended by an arc of a circle on the circle and at the centre

The** angle** subtended by **an arc** at the **centre is double** the angle subtended by it on any **part of the circle**.

Here PQ is the arc of a circle with centre O, that subtends ∠POQ at the centre.

Join AO and extend it to B.

In ΔOAQ OA = OQ….. [Radii]

Hence, ∠OAQ = ∠OQA ……..[Property of isosceles triangle]

Implies ∠BOQ = 2∠OAQ …..[Exterior angle of triangle = Sum of 2 interior angles]

Similarly, ∠BOP = 2∠OAP

⇒∠BOQ + ∠BOP = 2∠OAQ + 2∠OAP

⇒∠POQ = 2∠PAQ

Hence proved.

### Angles in the same segment of a circle

**Angles** in the **same segment** of a circle are **equal.**

Consider a circle with centre O.

∠PAQ and ∠PCQ are the angles formed in the major segment PACQ with respect to the arc PQ.

Join OP and OQ

∠POQ = 2∠PAQ = 2∠PCQ ……….[ Angle subtended by an arc at the centre is double the angle subtended by it in any part of the circle]

⇒∠PCQ = ∠PAQ

Hence proved

### The angle subtended by diameter on the circle

**Angle **subtended by **diameter** on a circle is a **right angle**. (Angle in a semicircle is a right angle)

Consider a circle with centre O, POQ is the diameter of the circle.

∠PAQ is the angle subtended by diameter PQ at the circumference.

∠POQ is the angle subtended by diameter PQ at the centre.

∠PAQ = (1/2)∠POQ……..[Angle subtended by arc at the centre is double the angle at any other part]

∠PAQ = (1/2) × 180^{0} = 90^{0}

Hence proved

### Line segment that subtends equal angles at two other points

If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e they are concyclic).

Here ∠ACB=∠ADB and all 4 points A, B, C, D are concyclic.

### Cyclic Quadrilateral

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

In a circle, if all **four points** A, B, C and D lie **on the circle**, then quadrilateral ABCD is a **cyclic quadrilateral.**

### Sum of opposite angles of a cyclic quadrilateral

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

### Sum of pair of opposite angles in a quadrilateral

The sum of either pair of opposite angles of a cyclic quadrilateral is 180 degree.