Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Chapter Notes: Circles

Circles Class 9 Notes Maths Chapter 10

Introduction to Circles

A circle is a unique figure; it is everywhere around us. We see the dials of clocks, buttons of shirts, coins, wheels of a vehicle, etc. All these are in the shape of a circle.

Circles Class 9 Notes Maths Chapter 10

Terms related to circles

1. Chord: The chord of a circle is a straight line segment whose endpoints lie on the circle.

Circles Class 9 Notes Maths Chapter 10

2. Diameter: The chord, which passes through the center of the circle is called a diameter of the circle. Diameter is the longest chord and all diameters have the same length, which is equal to two times the radius of the circle.
Circles Class 9 Notes Maths Chapter 10The length of the complete circle is called its circumference.

Question for Chapter Notes: Circles
Try yourself:What is the theorem that states the sum of either pair of opposite angles of a cyclic quadrilateral?
View Solution

3. Arc: The arc of a circle is a portion of the circumference of a circle.

Or

A piece of a circle between two points is also called an arc.
Circles Class 9 Notes Maths Chapter 10Two points lying on the circle define two arcs: The shorter one is called a minor arc and the longer one is called a major arc.
Circles Class 9 Notes Maths Chapter 10The minor arc AB is also denoted by Circles Class 9 Notes Maths Chapter 10 and the major arc AB by Circles Class 9 Notes Maths Chapter 10 where D is some point on the arc between A and B. When A and B are ends of a diameter, then both arcs are equal and each is called a semicircle.
Circles Class 9 Notes Maths Chapter 10

4. Segment: The region between a chord and either of its arc is called a segment of the circle. There are two types of segments also: which are the major segment and the minor segment.
Circles Class 9 Notes Maths Chapter 10

5. Sector: The region between the arc and the two radii, joining the center to the ends points of the arc is called a sector. The minor arc corresponds to the minor sector and the major arc corresponds to the major sector.
Circles Class 9 Notes Maths Chapter 10When the two arcs are equal, then both segments and both sectors become equal and each is known as a semicircle.
Circles Class 9 Notes Maths Chapter 10

Angle Subtended by a Chord at a Point

Theorem 1: Equal chords of circle subtend equal angles at the centre.

Given: 

A circle with centre H.

Two chords KL and JI are equal

To Prove: ∠KHL = ∠JHI
Proof:
We are given two chords KL and JI. We need to prove that ∠KHL = ∠JHI.
Circles Class 9 Notes Maths Chapter 10In triangles KHL and JHI,
HL = HJ
HI = HK
and we are given that KL = JI.
So, both the triangles are congruent. Thus, both of these angles are equal. Hence, proved.

Theorem 2: If the angles subtended by two chords at the centre are equal, then the two chords are equal

Given: 

A circle with centre O.

∠COD= ∠AOB  are equal

To Prove: AB=CD

Proof: 

We are given two chords AB and CD. We need to prove that two chords AB and CD are equal

Circles Class 9 Notes Maths Chapter 10

∠AOB = ∠COD (Equal angle subtended at centre O) ……………(1)

OA = OB = OP = OQ (Radii of the same circle) ……………(2)

From eq. 1 and 2, we get;

∆AOB ≅ ∆POQ (SAS Axiom of congruency)

Hence,

AB = PQ (By CPCT)

Perpendicular from the Centre to a Chord

Theorem 3: The perpendicular from the centre of a circle to a chord bisects the chord

Circles Class 9 Notes Maths Chapter 10Given: 

A circle with centre O.

PQ is a chord such that OM is perpendicular to PQ

To Prove: OM bisects chord PQ i.e. PM=MQ

Proof: Given, in ∆AOD and ∆BOD,
∠ADO = ∠BDO = 90° (OD ⊥ AB) ………(1)
OA = OB (Radii of the circle) ……….(2)
OD = OD (Common side) ………….(3)
From eq. (1), (2) and (3), we get;
∆AOB ≅ ∆POQ (R.H.S Axiom of congruency)
Hence, AD = DB (By CPCT)

Theorem 4: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord

Circles Class 9 Notes Maths Chapter 10

Given: 

A circle with centre O.

 OM bisects chord PQ i.e. PM=MQ

To Prove: PQ is a chord such that OM is perpendicular to PQ

Proof:
Let PQ be the chord of a circle and OM be the line from the centre that bisects the chord. Here, M is the mid point of the chord, and we have to prove that ∠OMQ = 90°.
In triangles ΔOPM and ΔOQM.
PM = MQ (perpendicular bisects the chord)
OP = OQ (radius of the circle)
OM = OM (common side of both the triangles)
So, both these triangles are congruent. This gives, angles ∠OMQ and ∠OMP as 90°

Equal Chords and Distance from the Centre

If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal. Similar to the case of chords, equal arcs also subtend equal angles at the centre.

Theorem 5: Equal chords of a circle are equidistant (equal distance) from the centre of the circle. 

Circles Class 9 Notes Maths Chapter 10

Given: 

A circle with centre O.

AB and CD are two equal chords of a circle i.e. AB=CD and OM and ON are perpendiculars to AB and CD respectively.

To Prove: OM=ON

Proof: 

Given, In ∆OMB and ∆OND

BM = 1/2 AB (Perpendicular to a chord bisects it) ……..(1) 

DN = 1/2 CD (Perpendicular to a chord bisects it) ……..(2) 

AB = CD (Given)

BM = DN (from eq 1 and 2) 

OB = OD (Radii of the same circle)

∠OMB = ∠OND = 90° (OM ⊥ AB and ON ⊥ CD) 

∆OMB ≅ ∆OND ( By R.H.S Axiom of Congruency) 

Hence, OM = ON ( By CPCT)

Theorem 6: Chords of a circle, which are at equal distances from the centre are equal in length

Circles Class 9 Notes Maths Chapter 10Given: 

A circle with centre O.

AB and CD are two equal chords of a circle i.e. AB=CD and OM and ON are perpendiculars to AB and CD respectively.

OM=ON

To Prove: AB=CD

Proof: Given, in ∆OMB and ∆OND, OM = ON ………….(1) 

∠OMB = ∠OND = 90° ………..(2) 

OB = OD (Radii of the same circle) ………..(3) 

Therefore, from eq. 1, 2 and 3, we get; 

∆OMB ≅ ∆OND (By R.H.S Axiom of Congruency) 

BM = DN ( By CPCT) 

1/2 AB = 1/2 CD (Perpendicular from center bisects the chord) 

Hence, AB = CD

Theorem 7: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Given: An arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle. 

To Prove :∠POQ = 2∠PAQ.

Proof:

Let’s consider three cases,

  • Arc PQ is major arc.
  • Arc PQ is minor arc.
  • Arc PQ is semi-circle.

Circles Class 9 Notes Maths Chapter 10
Let’s join AO and extend it to B.
In all three cases, ∠BOQ = ∠OAQ + ∠OQA. (Exterior angle of a triangle is equal to the sum of the two interior opposite angles).
Also in triangle ΔOAQ,
OA = OQ (Radii of Circle)
Therefore, ∠ OAQ = ∠ OQA
this gives, ∠ BOQ = 2∠OAQ
∠ BOP = 2∠OAP
from (1) and (2), ∠ BOP + ∠ BOQ = 2(∠ OAP + ∠ OAQ)
∠POQ = 2 ∠PAQ
For the case (iii), where PQ is the major arc, (3) is replaced by
Reflex angle POQ = 2∠PAQ


Example 2: What is the value of ∠ABC?
Circles Class 9 Notes Maths Chapter 10Solution: According to the theorem, ∠AOC = 2 ∠ABC
Therefore, ∠ABC = 60°/2 = 30° 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).

Some other properties

  • Angles in the same segment of a circle are equal.
  • 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).

Theorem 8: Angles in the same segment of a circle are equal. 

Circles Class 9 Notes Maths Chapter 10

Given: 

A circle with centre O

Points P and Q on this circle subtends ∠PAQ = ∠PBQ at points A and B respectively. 

To Prove : ∠PAQ = ∠PBQ

Proof: Let P and Q be any two points on a circle to form a chord PQ, A and C any other points on the remaining part of the circle and O be the centre of the circle. Then,

∠POQ = 2∠PAQ  ...... (i)

And ∠POQ = 2∠PCQ  ....... (ii) 

From above equations, we get

2∠PAQ = 2∠PCQ

Hence ∠PAQ = ∠PCQ

Theorem 9: If a line segment joining two points subtend equal angles at two other points lying on the same side of the lien containing the line segment the four points lie on a circle.

Circles Class 9 Notes Maths Chapter 10Given: 

AB is a line segment, which subtends equal angles at two points C and D. i.e., ∠ACB = ∠ADB. 

To Prove: 

The points A, B, C and D lie on a circle.

Proof: 

Let us draw a circle through the points A, C and B.

Suppose it does not pass through the point D.

Then it will intersect AD (or extended AD) at a point, say E (or E’).

If points A,C,E and B lie on a circle,

∠ACD = ∠AEB [∴ Angles in the same segment of circle are equal]

But it is given that ∠ACB = ∠ADB

Therefore, ∠AEB = ∠ADB

This is possible only when E coincides with D. [As otherwise ∠AEB >∠ADB]

Similarly, E’ should also coincide with D. So A, B, C and D are concyclic. 

Hence Proved.

Cyclic Quadrilaterals

A quadrilateral is called cyclic if all the four vertices of it lie on a circle.
Circles Class 9 Notes Maths Chapter 10They are also called inscribed quadrilaterals.

Theorem 10: The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.

Given: PQRS is a cyclic quadrilateral with centre O.

To Prove: 

∠PSR + ∠PQR = 360°

∠SPQ + ∠QRS = 360°
Proof:

Circles Class 9 Notes Maths Chapter 10Considering arc PQR, ∠POR = 2 ∠PSR,
Similarly, considering PSR, reflex ∠POR = 2∠PQR
We know, ∠POR + relfex ∠POR = 360°.
⇒ 2 ∠PSR + 2∠PQR = 360°
⇒ 2(∠PSR + ∠PQR) = 360°
⇒ ∠PSR + ∠PQR = 180°
Thus, sum of opposite angles of cyclic quadrilateral is 180°.


Example 3: In the figure below, BC is the diameter of the circle, ED is a chord equal to the radius of the circle. BE and CD when extended intersect at a point F. Prove that ∠BFC = 60°.
Circles Class 9 Notes Maths Chapter 10Solution:
In the figure, join AE, AD and EC. Triangle AED is an equilateral triangle.
Therefore, ∠EAD = 60°. Now, ∠ECD becomes 30°.
We know that ∠BEC = 90°.
So, by the property of exterior angles of triangle,
∠BEC = ∠ECD + ∠BFC,
90° = 30° + ∠BFC
⇒ 60° = ∠BFC
Hence, Proved.

Theorem 11: If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

Circles Class 9 Notes Maths Chapter 10

Given: ABPQ is a quadrilateral, such that ∠ ABP + ∠ AQP =180 ∘ and ∠ QAB + ∠ QPB = 180 ∘ 

To prove: The points A, B, P and Q lie on the circumference of a circle. 

Proof: Assume that point P does not lie on a circle drawn through points A, B and Q. 

Let the circle cut QP at point R. Join BR. ∠ QAB + ∠ QRB = 180 ∘ [opposite angles of cyclic quadrilateral.]

∠ QAB + ∠ QPB = 180 ∘ [given]

∴ ∠ QRB = ∠ QPB 

But this cannot be true since ∠ QRB = ∠ QPB + ∠ RBP (exterior angle of the triangle) 

∴ Our assumption that the circle does not pass through P is incorrect and A, B, P and Q lie on the circumference of a circle.

∴ ABPQ is a cyclic quadrilateral.

The document Circles Class 9 Notes Maths Chapter 10 is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Circles Class 9 Notes Maths Chapter 10

1. What is the angle subtended by a chord at a point on the circle?
Ans. The angle subtended by a chord at a point on the circle is the angle formed by two lines drawn from the endpoints of the chord to the point on the circle. This angle depends on the position of the point in relation to the chord. According to the Circle Theorem, the angle subtended at the circumference by a chord is half the angle subtended at the center by the same chord.
2. How can we prove that a perpendicular from the center of a circle to a chord bisects the chord?
Ans. To prove this, consider a circle with center O and a chord AB. If we draw a perpendicular line from O to AB, meeting it at point D, we need to show that AD = DB. By the property of right triangles, triangles OAD and OBD are congruent (OA = OB as radii, OD is common, and angles OAD and OBD are 90 degrees). Therefore, AD = DB, meaning the perpendicular from the center bisects the chord.
3. What is the relationship between equal chords and their distances from the center of the circle?
Ans. The relationship is that equal chords of a circle are equidistant from the center. If two chords are equal in length, the perpendicular distances from the center of the circle to these chords will also be equal. This is a fundamental property of circles and helps in various geometric constructions and proofs.
4. What are cyclic quadrilaterals, and what is their significance in circle geometry?
Ans. A cyclic quadrilateral is a quadrilateral whose vertices all lie on the circumference of a circle. The significance of cyclic quadrilaterals in circle geometry includes that the opposite angles of a cyclic quadrilateral are supplementary (they add up to 180 degrees). This property is useful in solving problems related to angles and can be applied in various geometric proofs.
5. How is the angle subtended by a chord at the center related to the angles subtended at other points on the circle?
Ans. The angle subtended by a chord at the center of the circle is twice the angle subtended by the same chord at any point on the remaining part of the circle. This means if we have a chord AB and we measure the angle ∠ACB at point C on the circle, this angle will be half of the angle ∠AOB at the center O (i.e., ∠AOB = 2 ∠ACB). This relationship is crucial for understanding the properties of angles in circle geometry.
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