**CBSE Class 9 Maths Circles Notes:-**Download PDF Here

## 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. - AÂ
**Fixed point**Â from which the set of points are at fixed distance is called theÂ**centre**Â 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 it’s **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.** – TheÂ **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 aÂ **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 Ï€ times the diameter which is given by the formula 2Ï€r

### 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**. –

-AÂ **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**.

–Â For** 2 equa**l arcs or for semicircles – both the segment and sector is called the** semicircular region.**

## Circles and Their Chords

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

– EqualÂ **ch****ords** subtend equal **angles at the centre**.

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

In Î” AOB and Î” COD

OB = OC [Radii]

OA = OD [Radii]

AB = CD [Given]

Î”AOB â‰… Î”COD (SSS rule)

Hence, âˆ AOB = âˆ 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 Î”AOB and Î”COD

OB = OC [Radii] âˆ AOB=âˆ COD [Given]

OA = OD [Radii]

Î”AOB â‰… Î”COD (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 Î”OMB and Î”OMA,

âˆ OMA=âˆ OMB=90^{0} OA = OB (radii)

OM = OM (common)

Hence, Î”OMB â‰… Î”OMA (RHS rule)

Therefore AM = MB [CPCT]

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

– AÂ **line drawn**Â 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 Î”OMA and Î”OMB,

OA = OB (Radii)

OM = OM (common)

AM = BM (Given)

Therefore, Î”OMA â‰… Î”OMB (SSS rule)

â‡’âˆ OMA=âˆ OMB (C.P.C.T)

But, âˆ OMA+âˆ OMB=180^{0}

Hence, âˆ OMA=âˆ OMB=90^{0} â‡’OMâŠ¥AB

### Circle through 3 points

– There is** one **and **only** one **circle** passing through** three given noncollinear points.** –Â 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:Â **Given, AB = CD,Â OÂ is the centre. Join OA and OC.

Draw, OPâŠ¥AB, OQâŠ¥CD

In Î”OAPÂ and Î”OCQ,

OA=OC (Radii)

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

Î”OAP â‰… Î”OCQ (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:Â **Given OX = OY (The chords AB and CD are at equidistant) OXâŠ¥AB, OYâŠ¥CD

In Î”AOX and Î”DOY

âˆ OXA =âˆ OYD (Both 90^{0})

OA = OD (Radii)

OX = OY (Given)

Î”AOX â‰… Î”DOY (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 âˆ POQ at the centre.

Join AO and extend it to B.

In Î”OAQ OA = OQ….. [Radii]

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

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

Similarly, âˆ BOP = 2âˆ OAP

â‡’âˆ BOQ + âˆ BOP = 2âˆ OAQ + 2âˆ OAP

â‡’âˆ POQ = 2âˆ 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.

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

Join OP and OQ

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

â‡’âˆ PCQ = âˆ 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.

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

âˆ POQ is theÂ angle subtended by diameter PQ at the centre.

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

âˆ PAQ = (1/2) Ã— 180^{0}Â = 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 âˆ ACB=âˆ 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.