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

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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.

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Circles Class 9 Notes CBSE Maths Chapter 9 (Free PDF Download)
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Circles L-2 [ Theorems on Circles-2 ] | CBSE Class 9 Maths Chap 10 | Term 2 Preparation | Umang 2021
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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 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 CD.

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

  • 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


  • ADC is a minor arc, whereas 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 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, AB is a chord of a circle with the centre O.

  • The midpoint of AB is M

  • OM is joined.

  • To Prove:

AMO = BMO = 900

  • Construction:

Join AO and BO.

  • Proof:

In ΔAOM and ΔBOM 

Statement

Reason

AO = BO

radii

AM = BM

Data

OM = OM

Common

ΔAOM ΔBOM 

(S.S.S)

AMO = BMO

Statement (4)

But AMO + BMO = 1800

Linear pair

AMO = BMO = 900

Statements (5) and  (6)


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, AB is a chord of a circle with the centre O.

  2. OMAB

  • To Prove:

AM = BM

  • Construction:

Join AO and BO.

  • Proof:

In ΔAOM and ΔBOM 

Statement

Reason

AMO = BMO 

Each 900 (data)

AO = BO

Radii

OM = OM

Common

ΔAOM ΔBOM 

(R.H.S)

AM = BM

Statement (4)


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

1 Data: M is the midpoint of AB

To Prove: M is the midpoint of AB

2 To Prove: OMAB

Data: OMAB


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, ABand CDare equal chords of a circle with centre O.

  • OKAB and OLCD

  • To Prove:

OK = OL

  • Construction:

Join AO and CO.

  • Proof:

Statement

Reason

AK = 12 AB

from the centre bisects the chord.

CL = 12 CD

from the centre bisects the chord.

But AB = CD

data

AK = CL 

Statements (1),(2) and  (3)

In ΔAOKand ΔCOL 

AKO = CLO

Each 900 (data)

AO = CO

radii

AK = CL 

Statements (4)

ΔAOK ΔCOL

(R.H.S)

OK = OL 

Statements (8)


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, ABand CDare equal chords of a circle with centre O.

  • OKAB and OLCD

  • OK = OL

  • To Prove:

AB = CD

  • Construction:

Join AO and CO.

  • Proof:

In ΔAOKand ΔCOL 

Statement

Reason

AKO = CLO

Each 900 (data)

AO = CO

radii

OK = OL 

data

ΔAOK ΔCOL

(R.H.S)

AK = CL 

Statements (4)

But AK = 12 AB

from the centre bisects the chord.

CL = 12 CD

from the centre bisects the chord.

AB = CD

Statements (5),(6) and  (7)


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, X, Yand Z are three points not in a straight line.

  • To Prove:

A unique circle passes through X, Yand Z.

  • Construction:

  • Join XY and YZ.

  • Draw perpendicular bisectors of XY and YZ to meet at O.

  • Proof:

Statement

Reason

OX = OY

O lies on the bisector of XY

OY = OZ

O lies on the bisector of YZ

OX = OY = OZ

Statements (1) and (2)

O is the only point equidistant from X, Yand Z.

Statements (3)

With O as centre and radius OX, a circle can be drawn to pass through X, Yand Z.  

Statements (4)

Therefore, the circle with centre O is a unique circle passing through X, Yand Z.  

Statements (5)


Angle Properties (Angle, Cyclic Quadrilaterals and Arcs):

  • In figure(i) , the straight line ABstudents APB on the circumference.


Angle, Cyclic Quadrilaterals and Arcs


APBcan be said to be subtended by arc AMB, on the remaining part of the circumference.

  • In fig. (ii) , arc AMB subtends APB on the circumference, and it subtends AOB at the centre.

  • In fig. (iii), APB and 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 AMBsubtends AOB at the center O of the circle and APB on the remaining part of circumference.

  • To Prove: 

AOB = 2APB

  • Construction:

Join PO and produce it to Q.

Let, APQ = x and BPQ = y

  • Proof:

Statement

Reason

AOQ = x + A

Ext=sum of the int. opps

x = A

 OA = OP (Radii)

AOQ = 2x

Statements (1) and (2)

BOQ = 2y

Same way as Statements (3)

From figure (i) and (ii)

AOQ + BOQ = 2x + 2y

Statements (3) and (4)

 AOB = 2(x + y)

Statements (5)

From figure (ii)

BOQAOQ = 2y2x

Statements (3) and (4)

AOB = 2(y - x)

Statements (8)

AOB = 2APB

Statements (9)


Theorem 7:

Angles in the same segment of a circle are equal.



Angles in the same segment of a circle are equal


  • Given Data:

APB and AQB are in the same segment of a circle with center O.

  • To Prove: 

APB = AQB

  • Construction:

Join AO and BO.

Let arc AMBsubtend angle x at the center O.

  • Proof:

Statement

Reason

x = 2APB

at center=2×on the circumference

x = 2AQB

at center=2×on the circumference

APB = AQB

Statements (1) and (2)


Theorem 8:

The angle in a semicircle is a right angle.


The angle in a semicircle is a right angle


  • Given Data:

AB is the diameter of a circle with center O.

P is any point on the circle.

  • To Prove: 

APB = 90

  • Proof:

Statement

Reason

APB = 12AOB

at center=2×on the circumference

AOB = 180

AOB is a straight line.

APB = 12×180

Statements (1) and (2)

APB = 90

Statements (3).


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, ABCD is a cyclic quadrilateral. 

The vertices A, B, C and 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:

ABCD is a cyclic quadrilateral.

 O is the center  of a circle.

  • To Prove: 

  1. A+C = 180

  2. B+D = 180

  • Proof:

Statement

Reason

A = 12x

at center=2×on the circumference

C = 12y

at center=2×on the circumference

A + C = 12x + 12y

Statements (1) and (2)

A + C = 12(x + y)

Statements (3)

But x + y=360

at a point

A + C = 12×360

Statements (4) and (5)

A + C = 180

Statements (6)

Also, ABC + ADC = 180

Same way as statements (7)


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:

ABCD is a cyclic quadrilateral.

 BC is produced to E

  • To Prove: 

DCE=A

  • Proof:

Statement

Reason

A +BCD=180

Opp. s of a cyclic quad.

BCD +DCE=180

Linear pair

BCD +DCE=A+BCD

Statements (1) and (2)

DCE=A

Statements (2)


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 SAT touches a given circle with centre O at A. AC is a chord through the point of contact A.

ADC is an angle in the alternate segment to CAT and AECis an angle in the alternate segment to CAS

  • To Prove: 

  1. CAT=ADC

  2. CAS=AEC

  • Construction:

Draw AOB as diameter and join BC and OC.

  • Proof:

Statement

Reason

OAC = OCA=x

Since, OA = OC and supposition

CAT +x=90

Since, tangent-radius property

AOC +x + y=180

Sum of angles of a triangle

AOC=1802x

Statements (3)

Also, AOC=2ADC

at the center=2on the circle

CAT=90x

Statements (2)

2CAT=1802x

Statements (6)

2CAT=2ADC

Statements (4), (5) and (7)

CAT=ADC

Statements (8)

CAS +CAT=180

Linear pair

ADC +AEC=180

Opp. Angles of a cyclic quad

CAS +CAT=ADC +AEC

Statements (10) and (11)

CAS=AEC

Statements (9) and (12)


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:

AXB and CYD are equal circles with centers P and O.

Arcs AMD, CND subtend equal angles APB, CQD.

  • To Prove: 

arc AMD = arc CND

  • Proof:

Statement

Reason

Apply  CYD to  AXB so that center Q falls on center P and QC along PA and D on the same side as B.

Therefore,  CYD overlaps  AXB

Since, circles are equal (data)

C falls on A

Since, PA = QC (data)

APB = CQD

data

QD falls along PB

Statements (1) and (3)

D falls on B

Since, QD = PB (data)

 arc CND coincides with arc AMD

Statements (2) and (5)

arc AMD = arc CND

Statements (6)


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 AXB and CYD, equal arcs AMD and CND subtend APB and CQD at the centers P and Q respectively.

  • To Prove: 

APB = CQD

  • Proof:

Statement

Reason

Apply  CYD to  AXB so that center Q falls on center P and QC along PA and D on the same side as B.

Therefore,  CYD overlaps  AXB

Since, circles are equal (data)

C falls on A

Since, PA = QC (data)

arc AMD = arc CND

data

D falls on B

Statements (1),(2) and (3)

QD coincides with PBand QC coincides with PA

Statements (1),(2) and (4)

APB = CQD

Statements (5)


In case of the same circle:


Equal circles


Figures (ii) and (iii) can be considered to be two equal circles which are obtained from figure (i) 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 AXB, CYD with centers P and Q have

 chord AB = chord CD

  • To Prove: 

arc AMB = arc CND

arc AXB = arc CYD

  • Proof:

In ΔABP and ΔCDQ

Statement

Reason

AP = CQ

Radii of equal circles.

BP = DQ

Radii of equal circles.

AB = CD

Radii of equal circles.

ΔABP  ΔCDQ

(S.S.S)

APB = CQD

Statements (4)

arc AMB = arc CND

Statements (5)

AXB - arc AMB = CYD - arc CND

Equal arcs [Statements 6]

 arc AXB = arc CYD

Statements (7)


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 AXB, CYD with centers P and Q have

 arc AMB = arc CND

  • To Prove: 

chord AB = chord CD

  • Construction:

Join AP, BP, CQ and DQ.

  • Proof:

In ΔABP and ΔCDQ

Statement

Reason

AP = CQ

Radii of equal circles.

BP = DQ

Radii of equal circles.

APB = CQD

 arc AMB = arc CND

ΔABP  ΔCDQ

(S.A.S)

AB = CD

Statements (4)


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:

AB and CD are chords of a circle intersecting externally at P.

  • To Prove: 

AP × BP = CP×DP

  • Construction:

Join AC and BD.

  • Proof:

In ΔAPC and ΔDPB

Statement

Reason

A = D

Angles in the same segment.

C = B

Angles in the same segment.

 ΔAPC ΔDPB

AA similarity

APDP=CPBP

Statements (3)

AP×BP = CP × DP

Statements (4)


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:

AB and CD are chords of a circle intersecting externally at P.

  • To Prove: 

AP × BP = CP×DP

  • Construction:

Join AC and BD.

  • Proof:

In ΔACP and ΔDBP

Statement

Reason

A = BDP

Extof a cyclic quad. = Int. opp

C = DBP

Extof a cyclic quad. = Int. opp

 ΔACP ΔDBP

AA similarity

APDP=CPBP

Statements (3)

AP×BP = CP × DP

Statements (4)


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 AB and a tangent TP at a point T on the circle intersect at P.

  • To Prove: 

AP × BP = PT2

  • Construction:

Join AT and BT.

  • Proof:

Statement

Reason

In ΔAPT and ΔTPB

Angles in alternate segment

A = BTP


P = P

Common

 ΔAPT ΔTPB

AA similarity

APPT=PTBP

Statements (3)

AP×BP = PT2

Statements (4)


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 P=Qand the points P,Q are on the same side of AB, then the points A,B,P and Qare 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 A+C=180 then A, B, C and 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.


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

1. What is a Cyclic Quadrilateral and What is the Theorem Associated with it?

When all the quadrilateral vertices lie on a circle, they are defined as a cyclic quadrilateral.

According to the theorem, any pair of the opposite angles in a cyclic quadrilateral will have the sum of 180°. 

The reverse of this theorem states that if the opposite angles in a quadrilateral have a sum of 180°, then the quadrilateral is cyclic.

2. What Happens When the Angles are Subtended From the Common Chord?

The angles subtended from the common chord and which lie on the same segment are always equal.

3. What Happens When an Angle is Subtended at the Centre and Another Angle is Subtended by the Same Arc on Another Point?

According to a theorem, an angle subtended by the arc at the centre is twice the angle that is subtended by the same arc on any other point of the circle.

4. How are NCERT Solutions for Class 9 Maths Chapter 9 helpful for students?

One of the best Solutions out there is provided by Vedantu and their Solutions for chapter 9 of Class 9 Maths helps students in a thorough understanding of the chapter through an in-depth explanation of concepts in short, precise and simple expert-created answers. The circle is a chapter that requires proper comprehension and forms a base during Class 9 for future classes. Therefore, the student must go through all the solutions available free of cost to ensure they do well.

5. Why should we follow NCERT Solutions for Class 9 Maths Chapter 9?

NCERT Solutions offer many benefits to students. These Solutions help students use their time in a meticulous manner by offering them factual analysis and thorough explanation on the concepts mentioned in Chapter 9 Of Class 9 Mathematics in an accurate and short display of information. The structure of the language used is easy to follow, thereby allowing students to easily grasp the concepts. Students can download these Solutions from the vedantu website (vedantu.com) or the vedantu app, and be rest assured with their exams.

6. What does chapter 9 of class 9 Maths cover?

In class 9, students learn Circles in chapter 9 of the NCERT textbook. Revision Notes are good references to learn and revise about the different concepts taught in the chapter like

  • Circles

  • Angle Subtended by a chord

  • Equal Chords with distances from the centre

  • Angle Subtended by an arc

  • Cyclic quadrilaterals

The chapter also covers some major and important theorems that students need to memorize and then apply for solving problems of circles, quadrilaterals, triangles, etc. from the chapter.

7. What are the important theorems in chapter 9 of class 9 Maths?

Chapter 9 of Circles for class 9 Maths has some major theorems that students need to thoroughly learn and memorize to solve all the questions thoroughly. There are 5 major theorems and their converses and the following points are an attempt at stating the theorems in brief:

  • Equal chords subtending angles at the centre

  • Perpendicular to a chord

  • Equal chords are equidistant from the centre

  • The measure of angles subtended by the same arc

  • Opposite angles are supplementary in quadrilaterals

8. What are the benefits of using revision notes for learning the chapter of circles in Class 9?

Revision notes are very beneficial once students learn how these notes help them with their exam preparations. For chapter 9 of class 9 Maths, which deals with circles, their concepts, and various theorems, things can get very complicated and confusing for the students. The notes offered by Vedantu For Class 9 Maths are one of the best because they provide detailed, but short notes which simplify all these complicated concepts, thereby providing students with an easy learning experience and last-minute help. Also these revision notes are available free of cost to download.