Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Circles Class 9 Notes CBSE Maths Chapter 9 (Free PDF Download)

ffImage
banner

Class 9 Maths Revision Notes for Circles of Chapter 9 - Free PDF Download

Maths has always been challenging as remembering the formulae is even tougher for a lot. But, with Vedantu’s class 9 Maths Chapter circle notes, we help you remember all of it at a glance. Our notes are designed as per the latest CBSE curriculum helping you understand the concepts quickly and easily. Moreover, the download option for CBSE Class 9 Maths Notes Chapter 9 Circles provides ease of offline study. So, if you want to grasp the concepts quickly or require hands-on formulae while studying, Vedantu got you covered. You can also download Class 9 Science Solutions and CBSE Solutions from Vedantu.com.

Watch videos on

Circles Class 9 Notes CBSE Maths Chapter 9 (Free PDF Download)
Previous
Next
Vedantu 9&10
Subscribe
iconShare
Circles L-2 [ Theorems on Circles-2 ] | CBSE Class 9 Maths Chap 10 | Term 2 Preparation | Umang 2021
3.2K likes
86.3K Views
3 years ago
Vedantu 9&10
Subscribe
iconShare
Circles L-1 [ Theorems on Circles ] CBSE Class 9 Maths Chapter 10 | Term 2 Preparation | Umang 2021
4.2K likes
131.5K Views
3 years ago

Access Class 9 Mathematics Chapter 9 - Circles

Introduction

Circle:

  • The locus of the points at a certain distance from a fixed point is defined as a circle.


The locus of the points at a certain distance from a fixed point


Chord:

  • A chord is a straight line that connects any two points on a circle.


A chord


  • A chord is represented by the letters \[\text{AB}\] .

  • If the longest chord passes  through the centre of the circle, it is termed as the diameter.

  • The radius is twice as long as the diameter. 

  • A diameter is referred to as a \[\text{CD}\].

  • A secant is a line that divides a circle in half. 

  • \[\text{PQR}\] is a secant of a circle.


Circumference:

  • Circumference refers to the length of a full circle. 

  • The circumference of a circle is defined as the border curve (or perimeter) of the circle.


The circumference of a circle


Arc:

  • An arc is any section or a part of the circumference.

  • A diameter divides a circle into two equal pieces. 

  • A minor arc is one that is smaller than a semicircle. 

  • A major arc is one that is larger than a semicircle.


An arc


  • \[\overset\frown{\text{ADC}}\] is a minor arc, whereas \[\overset\frown{\text{ABC}}\] is a major arc.


Sector:

  • A sector is the area between an arc and the two radii that connect the arc's centre and end points.

  • A segment is a section of a circle that has been cut off by a chord.


A sector


Concentric Circles:

Concentric circles are circles with the same centre.


Theorem \[\text{1}\]:

A straight line drawn from the centre of a circle to bisect a chord which is not a diameter, is at right angles to the chord.


A straight line drawn from the centre of a circle to bisect a chord which is not a diameter, is at right angles to the chord


  • Given Data:

  • Here, \[\text{AB}\] is a chord of a circle with the centre \[\text{O}\].

  • The midpoint of \[\text{AB}\] is \[\text{M}\]. 

  • \[\text{OM}\] is joined.

  • To Prove:

\[\angle \text{AMO = }\angle \text{BMO = 9}{{\text{0}}^{\text{0}}}\]

  • Construction:

Join \[\text{AO}\] and \[\text{BO}\].

  • Proof:

In \[\Delta \text{AOM }\]and \[\Delta \text{BOM }\]

Statement

Reason

\[\text{AO = BO}\]

radii

\[\text{AM = BM}\]

Data

\[\text{OM = OM}\]

Common

\[\Delta \text{AOM }\cong \Delta \text{BOM }\]

\[\left( \text{S}\text{.S}\text{.S} \right)\]

\[\therefore \angle \text{AMO = }\angle \text{BMO}\]

Statement \[\left( \text{4} \right)\]

But \[\angle \text{AMO + }\angle \text{BMO = 18}{{\text{0}}^{\text{0}}}\]

Linear pair

\[\therefore \angle \text{AMO = }\angle \text{BMO = 9}{{\text{0}}^{\text{0}}}\]

Statements \[\left( 5 \right)\] and  \[\left( 6 \right)\]


Theorem \[2\] (Converse of theorem \[1\]):

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


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


  • Given Data:

  1. Here, \[\text{AB}\] is a chord of a circle with the centre \[\text{O}\].

  2. \[\text{OM}\bot \text{AB}\]

  • To Prove:

\[\text{AM = BM}\]

  • Construction:

Join \[\text{AO}\] and \[\text{BO}\].

  • Proof:

In \[\Delta \text{AOM }\]and \[\Delta \text{BOM }\]

Statement

Reason

\[\text{AMO = }\angle \text{BMO }\]

Each \[\text{9}{{\text{0}}^{0}}\] (data)

\[\text{AO = BO}\]

Radii

\[\text{OM = OM}\]

Common

\[\Delta \text{AOM }\cong \Delta \text{BOM }\]

\[\left( \text{R}\text{.H}\text{.S} \right)\]

\[\text{AM = BM}\]

Statement \[\left( \text{4} \right)\]


The transposition of a statement consisting of 'data' and 'to prove' is the converse of a theorem.

We can see how it works by looking at the previous two theorems:

Theorem

Converse of Theorem

\[\text{1}\] Data: \[\text{M}\] is the midpoint of \[\text{AB}\]

To Prove: \[\text{M}\] is the midpoint of \[\text{AB}\]

\[\text{2}\] To Prove: \[\text{OM}\bot \text{AB}\]

Data: \[\text{OM}\bot \text{AB}\]


Theorem \[3\]:

Equal chords of a circle are equidistant from the centre.


Equal chords of a circle are equidistant from the centre


  • Given Data:

  • Here, \[\text{AB}\]and \[\text{CD}\]are equal chords of a circle with centre \[\text{O}\].

  • \[\text{OK}\bot \text{AB}\] and \[\text{OL}\bot \text{CD}\]

  • To Prove:

\[\text{OK = OL}\]

  • Construction:

Join \[\text{AO}\] and \[\text{CO}\].

  • Proof:

Statement

Reason

\[\text{AK = }\dfrac{1}{2}\text{ AB}\]

\[\bot \] from the centre bisects the chord.

\[\text{CL = }\dfrac{1}{2}\text{ CD}\]

\[\bot \] from the centre bisects the chord.

But \[\text{AB = CD}\]

data

\[\therefore \text{AK = CL }\]

Statements \[\left( 1 \right),\left( 2 \right)\] and  \[\left( 3 \right)\]

In \[\Delta \text{AOK}\]and \[\Delta \text{COL }\]

\[\angle \text{AKO = }\angle \text{CLO}\]

Each \[\text{9}{{\text{0}}^{0}}\] (data)

\[\text{AO = CO}\]

radii

\[\text{AK = CL }\]

Statements \[\left( 4 \right)\]

\[\therefore \Delta \text{AOK }\cong \Delta \text{COL}\]

\[\left( \text{R}\text{.H}\text{.S} \right)\]

\[\therefore \text{OK = OL }\]

Statements \[\left( 8 \right)\]


Theorem \[4\](Converse of theorem \[3\]):

Chords which are equidistant from the centre of a circle are equal.


Chords which are equidistant from the centre of a circle are equal


  • Given Data:

  • Here, \[\text{AB}\]and \[\text{CD}\]are equal chords of a circle with centre \[\text{O}\].

  • \[\text{OK}\bot \text{AB}\] and \[\text{OL}\bot \text{CD}\]

  • \[\text{OK = OL}\]

  • To Prove:

\[\text{AB = CD}\]

  • Construction:

Join \[\text{AO}\] and \[\text{CO}\].

  • Proof:

In \[\Delta \text{AOK}\]and \[\Delta \text{COL }\]

Statement

Reason

\[\angle \text{AKO = }\angle \text{CLO}\]

Each \[\text{9}{{\text{0}}^{0}}\] (data)

\[\text{AO = CO}\]

radii

\[\text{OK = OL }\]

data

\[\Delta \text{AOK }\cong \Delta \text{COL}\]

\[\left( \text{R}\text{.H}\text{.S} \right)\]

\[\therefore \text{AK = CL }\]

Statements \[\left( 4 \right)\]

But \[\text{AK = }\dfrac{1}{2}\text{ AB}\]

\[\bot \] from the centre bisects the chord.

\[\text{CL = }\dfrac{1}{2}\text{ CD}\]

\[\bot \] from the centre bisects the chord.

\[\therefore \text{AB = CD}\]

Statements \[\left( 5 \right),\left( 6 \right)\] and  \[\left( 7 \right)\]


Theorem \[5\]:

There is one circle, and only one, which passes through three given points not in a straight line.


One circle passes through three given points not in a straight line


  • Given Data:

Here, \[\text{X, Y}\]and \[\text{Z}\] are three points not in a straight line.

  • To Prove:

A unique circle passes through \[\text{X, Y}\]and \[\text{Z}\].

  • Construction:

  • Join \[\text{XY}\] and \[\text{YZ}\].

  • Draw perpendicular bisectors of \[\text{XY}\] and \[\text{YZ}\] to meet at \[\text{O}\].

  • Proof:

Statement

Reason

\[\text{OX = OY}\]

\[\text{O}\] lies on the \[\bot \] bisector of \[\text{XY}\]

\[\text{OY = OZ}\]

\[\text{O}\] lies on the \[\bot \] bisector of \[\text{YZ}\]

\[\text{OX = OY = OZ}\]

Statements \[\left( 1 \right)\] and \[\left( 2 \right)\]

\[\text{O}\] is the only point equidistant from \[\text{X, Y}\]and \[\text{Z}\].

Statements \[\left( 3 \right)\]

With \[\text{O}\] as centre and radius \[\text{OX}\], a circle can be drawn to pass through \[\text{X, Y}\]and \[\text{Z}\].  

Statements \[\left( 4 \right)\]

Therefore, the circle with centre \[\text{O}\] is a unique circle passing through \[\text{X, Y}\]and \[\text{Z}\].  

Statements \[\left( 5 \right)\]


Angle Properties (Angle, Cyclic Quadrilaterals and Arcs):

  • In figure\[\left( \text{i} \right)\] , the straight line \[\text{AB}\]students \[\angle \text{APB}\] on the circumference.


Angle, Cyclic Quadrilaterals and Arcs


\[\angle \text{APB}\]can be said to be subtended by arc \[\text{AMB}\], on the remaining part of the circumference.

  • In fig. \[\left( \text{ii} \right)\] , arc \[\text{AMB}\] subtends \[\angle \text{APB}\] on the circumference, and it subtends \[\angle \text{AOB}\] at the centre.

  • In fig. \[\left( \text{iii} \right)\], \[\angle \text{APB}\] and \[\angle \text{AQB}\] are in the same segment.

  • Now we will go through the theorems based on the angle properties of the circles.


Theorem \[6\]:

The angle at which an arc of a circle subtends at the centre is double the angle which it subtends at any point on the remaining part of the circumference.


The angle at which an arc of a circle subtends at the centre is double the angle which it subtends at any point on the remaining part of the circumference


  • Given Data:

Arc \[\text{AMB}\]subtends \[\square \text{AOB}\] at the center \[\text{O}\] of the circle and \[\square \text{APB}\] on the remaining part of circumference.

  • To Prove: 

\[\angle \text{AOB = 2}\angle \text{APB}\]

  • Construction:

Join \[\text{PO}\] and produce it to \[\text{Q}\].

Let, \[\angle \text{APQ = x}\] and \[\angle \text{BPQ = y}\]

  • Proof:

Statement

Reason

\[\angle \text{AOQ = }\angle \text{x + }\angle \text{A}\]

\[\text{Ext}\text{. }\angle =\text{sum of the int}\text{. opp}\text{. }\angle \text{s}\]

\[\angle \text{x = }\angle \text{A}\]

\[\because \text{ OA = OP}\] (Radii)

\[\therefore \angle \text{AOQ = 2}\angle \text{x}\]

Statements \[\left( 1 \right)\] and \[\left( 2 \right)\]

\[\therefore \angle \text{BOQ = 2}\angle \text{y}\]

Same way as Statements \[\left( 3 \right)\]

From figure \[\left( \text{i} \right)\] and \[\left( \text{ii} \right)\]

\[\angle \text{AOQ + }\angle \text{BOQ = 2}\angle \text{x + 2}\angle \text{y}\]

Statements \[\left( 3 \right)\] and \[\left( 4 \right)\]

\[\Rightarrow \text{ }\angle \text{AOB = 2}\left( \angle \text{x + }\angle \text{y} \right)\]

Statements \[\left( 5 \right)\]

From figure \[\left( \text{ii} \right)\]

\[\angle \text{BOQ}-\angle \text{AOQ = 2}\angle \text{y}-\text{2}\angle \text{x}\]

Statements \[\left( 3 \right)\] and \[\left( 4 \right)\]

\[\angle \text{AOB = 2}\left( \angle \text{y - }\angle \text{x} \right)\]

Statements \[\left( 8 \right)\]

\[\because \angle \text{AOB = 2}\angle \text{APB}\]

Statements \[\left( 9 \right)\]


Theorem \[7\]:

Angles in the same segment of a circle are equal.



Angles in the same segment of a circle are equal


  • Given Data:

\[\angle \text{APB}\] and \[\angle \text{AQB}\] are in the same segment of a circle with center \[\text{O}\].

  • To Prove: 

\[\angle \text{APB = }\angle \text{AQB}\]

  • Construction:

Join \[\text{AO}\] and \[\text{BO}\].

Let arc \[\text{AMB}\]subtend angle \[\text{x}\] at the center \[\text{O}\].

  • Proof:

Statement

Reason

\[\angle \text{x = 2}\angle \text{APB}\]

\[\angle \text{at center}=2\times \angle \text{on the circumference}\]

\[\angle \text{x = 2}\angle \text{AQB}\]

\[\angle \text{at center}=2\times \angle \text{on the circumference}\]

\[\therefore \angle \text{APB = }\angle \text{AQB}\]

Statements \[\left( 1 \right)\] and \[\left( 2 \right)\]


Theorem \[8\]:

The angle in a semicircle is a right angle.


The angle in a semicircle is a right angle


  • Given Data:

\[\text{AB}\] is the diameter of a circle with center \[\text{O}\].

\[\text{P}\] is any point on the circle.

  • To Prove: 

\[\angle \text{APB = }{{90}^{\circ }}\]

  • Proof:

Statement

Reason

\[\angle \text{APB = }\dfrac{1}{2}\angle \text{AOB}\]

\[\angle \text{at center}=2\times \angle \text{on the circumference}\]

\[\angle \text{AOB = 18}{{\text{0}}^{\circ }}\]

\[\text{AOB}\] is a straight line.

\[\therefore \angle \text{APB = }\dfrac{1}{2}\times {{180}^{\circ }}\]

Statements \[\left( 1 \right)\] and \[\left( 2 \right)\]

\[\therefore \angle \text{APB = }{{90}^{\circ }}\]

Statements \[\left( 3 \right)\].


Cyclic Quadrilaterals:

  • If the vertices of a quadrilateral lie on a circle, the quadrilateral is called a cyclic quadrilateral. 

  • The vertices are known as concyclic points.


Cyclic Quadrilaterals


  • From the above figure, \[\text{ABCD}\] is a cyclic quadrilateral. 

The vertices \[\text{A, B, C}\] and \[\text{D}\] are concyclic points.


Theorem \[9\]:

The opposite angles of a quadrilateral inscribed in a circle (cyclic) are supplementary.


The opposite angles of a quadrilateral inscribed in a circle (cyclic) are supplementary


  • Given Data:

\[\text{ABCD}\] is a cyclic quadrilateral.

 \[\text{O}\] is the center  of a circle.

  • To Prove: 

  1. \[\angle \text{A+}\angle \text{C = 18}{{0}^{\circ }}\]

  2. \[\angle \text{B+}\angle \text{D = 18}{{0}^{\circ }}\]

  • Proof:

Statement

Reason

\[\angle \text{A = }\dfrac{1}{2}\angle \text{x}\]

\[\angle \text{at center}=2\times \angle \text{on the circumference}\]

\[\angle \text{C = }\dfrac{1}{2}\angle \text{y}\]

\[\angle \text{at center}=2\times \angle \text{on the circumference}\]

\[\angle \text{A + }\angle \text{C = }\dfrac{1}{2}\angle \text{x + }\dfrac{1}{2}\angle \text{y}\]

Statements \[\left( 1 \right)\] and \[\left( 2 \right)\]

\[\angle \text{A + }\angle \text{C = }\dfrac{1}{2}\left( \angle \text{x + }\angle \text{y} \right)\]

Statements \[\left( 3 \right)\]

But \[\angle \text{x + }\angle \text{y}={{360}^{\circ }}\]

$\angle $ at a point

\[\therefore \angle \text{A + }\angle \text{C = }\dfrac{1}{2}\times {{360}^{\circ }}\]

Statements \[\left( 4 \right)\] and \[\left( 5 \right)\]

\[\therefore \angle \text{A + }\angle \text{C = }{{180}^{\circ }}\]

Statements \[\left( 6 \right)\]

Also, \[\angle \text{ABC + }\angle \text{ADC = }{{180}^{\circ }}\]

Same way as statements \[\left( 7 \right)\]


Corollary: 

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.


The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle


  • Given Data:

\[\text{ABCD}\] is a cyclic quadrilateral.

 \[\text{BC}\] is produced to \[\text{E}\]

  • To Prove: 

\[\angle \text{DCE}=\angle \text{A}\]

  • Proof:

Statement

Reason

\[\angle \text{A +}\angle \text{BCD}={{180}^{\circ }}\]

Opp. \[\angle \text{s}\] of a cyclic quad.

\[\angle \text{BCD +}\angle \text{DCE}={{180}^{\circ }}\]

Linear pair

\[\therefore \angle \text{BCD +}\angle \text{DCE}=\angle \text{A}+\angle \text{BCD}\]

Statements \[\left( 1 \right)\] and \[\left( 2 \right)\]

\[\therefore \angle \text{DCE}=\angle \text{A}\]

Statements \[\left( 2 \right)\]


Alternate Segment Property 

Theorem \[10\]:  

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.


The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment


  • Given Data:

A straight line \[\text{SAT}\] touches a given circle with centre \[\text{O}\] at \[\text{A}\]. \[\text{AC}\] is a chord through the point of contact \[\text{A}\].

\[\angle \text{ADC}\] is an angle in the alternate segment to \[\angle \text{CAT}\] and \[\angle \text{AEC}\]is an angle in the alternate segment to \[\angle \text{CAS}\]

  • To Prove: 

  1. \[\angle \text{CAT}=\angle \text{ADC}\]

  2. \[\angle \text{CAS}=\angle \text{AEC}\]

  • Construction:

Draw \[\text{AOB}\] as diameter and join \[\text{BC}\] and \[\text{OC}\].

  • Proof:

Statement

Reason

\[\angle \text{OAC = }\angle \text{OCA}=\text{x}\]

Since, \[\text{OA = OC}\] and supposition

\[\angle \text{CAT +}\angle \text{x}={{90}^{\circ }}\]

Since, tangent-radius property

\[\angle \text{AOC +}\angle \text{x + }\angle \text{y}={{180}^{\circ }}\]

Sum of angles of a triangle

\[\angle \text{AOC}={{180}^{\circ }}-2\angle \text{x}\]

Statements \[\left( 3 \right)\]

Also, \[\angle \text{AOC}=2\angle \text{ADC}\]

\[\angle \text{at the center}=2\angle \text{on the circle}\]

\[\angle \text{CAT}={{90}^{\circ }}-\text{x}\]

Statements \[\left( 2 \right)\]

\[2\angle \text{CAT}={{180}^{\circ }}-2\text{x}\]

Statements \[\left( 6 \right)\]

\[\therefore 2\angle \text{CAT}=2\angle \text{ADC}\]

Statements \[\left( 4 \right)\], \[\left( 5 \right)\] and \[\left( 7 \right)\]

\[\angle \text{CAT}=\angle \text{ADC}\]

Statements \[\left( 8 \right)\]

\[\angle \text{CAS +}\angle \text{CAT}={{180}^{\circ }}\]

Linear pair

\[\angle \text{ADC +}\angle \text{AEC}={{180}^{\circ }}\]

Opp. Angles of a cyclic quad

\[\angle \text{CAS +}\angle \text{CAT}=\angle \text{ADC +}\angle \text{AEC}\]

Statements \[\left( 10 \right)\] and \[\left( 11 \right)\]

\[\therefore \angle \text{CAS}=\angle \text{AEC}\]

Statements \[\left( 9 \right)\] and \[\left( 12 \right)\]


Theorem \[11\]:  

In equal circles (or in the same circle), if two arcs subtend equal angles at the centres, they are equal.


Two arcs subtend equal angles at the centres


  • Given Data:

\[\text{AXB}\] and \[\text{CYD}\] are equal circles with centers \[\text{P}\] and \[\text{O}\].

Arcs \[\text{AMD, CND}\] subtend equal angles \[\text{APB, CQD}\].

  • To Prove: 

\[\text{arc AMD = arc CND}\]

  • Proof:

Statement

Reason

Apply \[\bigcirc \text{ CYD}\] to \[\bigcirc \text{ AXB}\] so that center \[\text{Q}\] falls on center \[\text{P}\] and \[\text{QC}\] along \[\text{PA}\] and \[\text{D}\] on the same side as \[\text{B}\].

Therefore, \[\bigcirc \text{ CYD}\] overlaps \[\bigcirc \text{ AXB}\]

Since, circles are equal (data)

\[\therefore \text{C}\] falls on \[\text{A}\]

Since, \[\text{PA = QC}\] (data)

\[\angle \text{APB = }\angle \text{CQD}\]

data

\[\therefore \text{QD}\] falls along \[\text{PB}\]

Statements \[\left( 1 \right)\] and \[\left( 3 \right)\]

\[\therefore \text{D}\] falls on \[\text{B}\]

Since, \[\text{QD = PB}\] (data)

\[\therefore \text{ arc CND}\] coincides with \[\text{arc AMD}\]

Statements \[\left( 2 \right)\] and \[\left( 5 \right)\]

\[\text{arc AMD = arc CND}\]

Statements \[\left( 6 \right)\]


Theorem \[12\] (Converse of \[11\]):  

In equal circles (or in the same circle) if two arcs are equal, they subtend equal angles at the centres.


Two arcs are equal, they subtend equal angles at the centres


  • Given Data:

In equal circles \[\text{AXB}\] and \[\text{CYD}\], equal arcs \[\text{AMD}\] and \[\text{CND}\] subtend \[\angle \text{APB}\] and \[\angle \text{CQD}\] at the centers \[\text{P}\] and \[\text{Q}\] respectively.

  • To Prove: 

\[\angle \text{APB = }\angle \text{CQD}\]

  • Proof:

Statement

Reason

Apply \[\bigcirc \text{ CYD}\] to \[\bigcirc \text{ AXB}\] so that center \[\text{Q}\] falls on center \[\text{P}\] and \[\text{QC}\] along \[\text{PA}\] and \[\text{D}\] on the same side as \[\text{B}\].

Therefore, \[\bigcirc \text{ CYD}\] overlaps \[\bigcirc \text{ AXB}\]

Since, circles are equal (data)

\[\therefore \text{C}\] falls on \[\text{A}\]

Since, \[\text{PA = QC}\] (data)

\[\text{arc AMD = arc CND}\]

data

\[\therefore \text{D}\] falls on \[\text{B}\]

Statements \[\left( 1 \right),\left( 2 \right)\] and \[\left( 3 \right)\]

\[\therefore \text{QD}\] coincides with \[\text{PB}\]and \[\text{QC}\] coincides with \[\text{PA}\]

Statements \[\left( 1 \right),\left( 2 \right)\] and \[\left( 4 \right)\]

\[\angle \text{APB = }\angle \text{CQD}\]

Statements \[\left( 5 \right)\]


In case of the same circle:


Equal circles


Figures \[\left( \text{ii} \right)\] and \[\left( \text{iii} \right)\] can be considered to be two equal circles which are obtained from figure \[\left( \text{i} \right)\] and then the above proofs may be applied.


Theorem \[13\]):

In equal circles (or in the same circle), if two chords are equal, they cut off equal arcs.


Two chords are equal, they cut off equal arcs


  • Given Data:

In equal circles \[\text{AXB, CYD}\] with centers \[\text{P}\] and \[\text{Q}\] have

 \[\text{chord AB = chord CD}\]

  • To Prove: 

\[\text{arc AMB = arc CND}\]

\[\text{arc AXB = arc CYD}\]

  • Proof:

In \[\Delta \text{ABP}\] and \[\Delta \text{CDQ}\]

Statement

Reason

\[\text{AP = CQ}\]

Radii of equal circles.

\[\text{BP = DQ}\]

Radii of equal circles.

\[\text{AB = CD}\]

Radii of equal circles.

\[\Delta \text{ABP }\cong \text{ }\Delta \text{CDQ}\]

\[\left( \text{S}\text{.S}\text{.S} \right)\]

\[\therefore \angle \text{APB = CQD}\]

Statements \[\left( 4 \right)\]

\[\text{arc AMB = arc CND}\]

Statements \[\left( 5 \right)\]

\[\bigcirc \text{AXB - arc AMB = }\bigcirc \text{CYD - arc CND}\]

Equal arcs [Statements 6]

\[\therefore \text{ arc AXB = arc CYD}\]

Statements \[\left( 7 \right)\]


Theorem \[14\] (Converse of \[13\]):

In equal circles (or in the same circle) if two arcs are equal, the chords of the arcs are equal.


Two arcs are equal, the chords of the arcs are equal


  • Given Data:

Equal circles \[\text{AXB, CYD}\] with centers \[\text{P}\] and \[\text{Q}\] have

 \[\text{arc AMB = arc CND}\]

  • To Prove: 

\[\text{chord AB = chord CD}\]

  • Construction:

Join \[\text{AP, BP, CQ}\] and \[\text{DQ}\].

  • Proof:

In \[\Delta \text{ABP}\] and \[\Delta \text{CDQ}\]

Statement

Reason

\[\text{AP = CQ}\]

Radii of equal circles.

\[\text{BP = DQ}\]

Radii of equal circles.

\[\angle \text{APB = CQD}\]

\[\because \text{ arc AMB = arc CND}\]

\[\therefore \Delta \text{ABP }\cong \text{ }\Delta \text{CDQ}\]

\[\left( \text{S}\text{.A}\text{.S} \right)\]

\[\therefore \text{AB = CD}\]

Statements \[\left( 4 \right)\]


Theorem \[15\]:

If two chords of a circle intersect internally, then the product of the length of the segments are equal.


Two chords of a circle intersect internally then the product of the length of the segments are equal


  • Given Data:

\[\text{AB}\] and \[\text{CD}\] are chords of a circle intersecting externally at \[\text{P}\].

  • To Prove: 

\[\text{AP }\!\!\times\!\!\text{ BP = CP}\times \text{DP}\]

  • Construction:

Join \[\text{AC}\] and \[\text{BD}\].

  • Proof:

In \[\Delta \text{APC}\] and \[\Delta \text{DPB}\]

Statement

Reason

\[\angle \text{A = }\angle \text{D}\]

Angles in the same segment.

\[\angle \text{C = }\angle \text{B}\]

Angles in the same segment.

\[\therefore \text{ }\Delta \text{APC  }\!\!\sim\Delta \text{DPB}\]

\[\text{AA}\] similarity

\[\therefore \dfrac{\text{AP}}{\text{DP}}\text{=}\dfrac{\text{CP}}{\text{BP}}\]

Statements \[\left( 3 \right)\]

\[\therefore \text{AP}\times \text{BP = CP }\!\!\times\!\!\text{ DP}\]

Statements \[\left( 4 \right)\]


Theorem \[16\]:

If two chords of a circle intersect externally, then the product of the lengths of the segments are equal.


Two chords of a circle intersect externally and the product of the lengths of the segments are equal


  • Given Data:

\[\text{AB}\] and \[\text{CD}\] are chords of a circle intersecting externally at \[\text{P}\].

  • To Prove: 

\[\text{AP }\!\!\times\!\!\text{ BP = CP}\times \text{DP}\]

  • Construction:

Join \[\text{AC}\] and \[\text{BD}\].

  • Proof:

In \[\Delta \text{ACP}\] and \[\Delta \text{DBP}\]

Statement

Reason

\[\angle \text{A = }\angle \text{BDP}\]

\[\text{Ext}\text{. }\angle \text{of a cyclic quad}\text{. = Int}\text{. opp}\text{. }\angle \]

\[\angle \text{C = }\angle \text{DBP}\]

\[\text{Ext}\text{. }\angle \text{of a cyclic quad}\text{. = Int}\text{. opp}\text{. }\angle \]

\[\therefore \text{ }\Delta \text{ACP  }\!\!\sim\Delta \text{DBP}\]

\[\text{AA}\] similarity

\[\therefore \dfrac{\text{AP}}{\text{DP}}\text{=}\dfrac{\text{CP}}{\text{BP}}\]

Statements \[\left( 3 \right)\]

\[\text{AP}\times \text{BP = CP }\!\!\times\!\!\text{ DP}\]

Statements \[\left( 4 \right)\]


Theorem \[17\]:

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square on the length of the tangent from the point of contact to the point of intersection.


A chord and a tangent intersect externally


  • Given Data:

A chord \[\text{AB}\] and a tangent \[\text{TP}\] at a point \[\text{T}\] on the circle intersect at \[\text{P}\].

  • To Prove: 

\[\text{AP }\!\!\times\!\!\text{ BP = P}{{\text{T}}^{\text{2}}}\]

  • Construction:

Join \[\text{AT}\] and \[\text{BT}\].

  • Proof:

Statement

Reason

In \[\Delta \text{APT}\] and \[\Delta \text{TPB}\]

Angles in alternate segment

\[\angle \text{A = }\angle \text{BTP}\]


\[\angle \text{P = }\angle \text{P}\]

Common

\[\therefore \text{ }\Delta \text{APT  }\!\!\sim\Delta \text{TPB}\]

\[\text{AA}\] similarity

\[\dfrac{\text{AP}}{\text{PT}}\text{=}\dfrac{\text{PT}}{\text{BP}}\]

Statements \[\left( 3 \right)\]

\[\text{AP}\times \text{BP = P}{{\text{T}}^{2}}\]

Statements \[\left( 4 \right)\]


Test for Concyclic Points:

  1. Conversely, the statement, 'Angles in the same segment of a circle are equal', is one test for concyclic points. 

We state:

If two equal angles are on the same side of a line and are subtended by it, then the four points are concyclic. 

In the figure, if $\angle P=\angle Q$and the points $P,Q$ are on the same side of $AB$, then the points $A,B,P$ and $Q$are concyclic.


Concyclic Points


  1. Converse of 'opposite angles of a cyclic quadrilateral are supplementary' is one more test for concyclic points.

We state:

If the opposite angles of a quadrilateral are supplementary, then its vertices are concyclic. 

In the figure, if $\angle \text{A}+\angle C={{180}^{\circ }}$ then $\text{A, B, C}$ and $\text{D}$ are concyclic points.


Concyclic points


Key Takeaways of NCERT Solutions Class 9 Maths Chapter 9 - Circles Free PDF

The circles class 9 questions with solutions allow the students to take several benefits. They are:


  • Students can score better marks and gain good knowledge.

  • These PDFs are available for free.

  • Students can take either a soft copy or hard copy.


Conclusion 

NCERT Solutions for Class 9 Maths Chapter 9 Circles is a valuable resource for students who want to gain a sound knowledge in logical thinking and problem solving. It provides students with a clear and concise explanation of the concepts, as well as a large number of solved and unsolved problems. Students can download the solutions for free from the official website, which makes it a convenient and affordable resource. In addition to NCERT Solutions for Class 9 Maths Chapter 9 Circles, Vedantu also offers NCERT Solutions for Class 9 Science and CBSE Solutions for all subjects. These solutions are up-to-date and are sure to help students in their academic journey.


Related Study Materials for Class 9 Maths Chapter 9 Circles



Chapter-wise Links for Class 9 Maths Notes



Related Important Links for Maths Class 9

Along with this, students can also download additional study materials provided by Vedantu for Maths Class 9–


WhatsApp Banner

FAQs on Circles Class 9 Notes CBSE Maths Chapter 9 (Free PDF Download)

1. What are the key concepts covered in the Class 9 Circles revision notes?

The revision notes for Class 9 Maths Chapter 9 Circles include core concepts like the definition of a circle, properties of chords, diameters, radii, arcs, sectors, and segments, and important theorems such as angles subtended by chords and arcs, properties of cyclic quadrilaterals, and equality of chords and arcs. The notes focus on the relationships between these properties and provide concise summaries for easy revision.

2. How do revision notes help in quick preparation for the Circles chapter in Class 9 Maths?

Revision notes are designed for efficient last-minute review. They consolidate all the main formulas, definitions, and theorems from the Circles chapter into short, organized summaries. This allows students to recap important points quickly before exams and focus on the most test-relevant areas.

3. What is the relationship between a chord and its distance from the centre of a circle, as highlighted in revision notes?

The revision notes explain that equal chords of a circle are equidistant from the centre. Conversely, chords that are equidistant from the centre are equal in length. This relationship helps students solve problems involving chords and their placement within the circle.

4. How do the theorems about angles in circles support problem-solving in Chapter 9?

The revision notes summarize theorems like “the angle subtended by an arc at the centre is twice the angle at the circumference” and “angles in the same segment are equal”. These theorems are essential for solving questions involving measurements of angles within different parts of the circle.

5. Why is it important to understand the concept of cyclic quadrilaterals in Class 9 Circles?

Understanding cyclic quadrilaterals is important because many exam problems involve identifying such quadrilaterals and applying the property that the sum of opposite angles is 180°. This concept connects circles with quadrilaterals and broadens your problem-solving ability in geometry.

6. How should students use revision notes to build connections between different theorems in circles?

Students should review the sequence in which theorems are derived and understand their logical relationships. For example, the converse of a theorem is often tested, so recognizing how one property leads to another strengthens concept retention and application in varied questions.

7. What misconceptions should students avoid when revising the Circles chapter?

Common misconceptions include

  • Confusing the difference between radius and diameter
  • Assuming every quadrilateral inscribed in a circle is cyclic without checking the angle property
  • Misapplying theorems about arcs and chords without noting the required conditions (like ‘not a diameter’ or ‘equal arcs’)
Carefully reading the conditions given in theorems avoids errors in problem-solving.

8. How can concept maps or summary tables aid revision for Circles Class 9?

Concept maps or tables in revision notes allow students to visually organize relationships between definitions, theorems, and properties. This helps in quick recall and clarifies how topics such as chords, arcs, and cyclic quadrilaterals are interconnected within the chapter.

9. What is the best order to revise Circles Class 9 to maximize understanding before exams?

Start with the basic definitions (circle, radius, chord, diameter), then move to properties of chords and their theorems, followed by theorems about arcs and angles. Finally, revise cyclic quadrilaterals and their properties. This logical order mirrors how concepts build upon each other in the NCERT syllabus, ensuring you establish a strong foundation for solving complex questions.

10. What types of questions from Circles commonly appear in Class 9 exams, as per revision notes focus?

Exams typically include

  • Statements and proofs of key theorems
  • Application-based problems involving angles, chords, and cyclic quadrilaterals
  • Short concept-based questions on definitions and properties
  • Diagram-based problems where students must label or identify properties
Revision notes are structured to target these core question types for CBSE Class 9 2025–26 exams.