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# NCERT Solutions Class 9 Maths Chapter 9 Circles

Last updated date: 20th Sep 2024
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## NCERT Solutions for Class 9 Chapter 9 Circles Maths - FREE PDF Download

Circles Class 9 NCERT Solutions, provide comprehensive guidance on understanding the properties and theorems related to circles. This chapter is crucial as it lays the foundation for concepts like tangents, chords, and the different segments of a circle.

Table of Content
1. NCERT Solutions for Class 9 Chapter 9 Circles Maths - FREE PDF Download
2. Glance of NCERT Solutions for Class 9 Maths Chapter 9 Circles | Vedantu
3. Access Exercise wise NCERT Solutions for Chapter 9 Maths Class 9
4. Exercises under NCERT Solutions for Class 9 Maths Chapter 9 Circles
5. Access NCERT Solutions for Class 9 Maths Chapter 9 – Circles
5.1Exercise 9.1
5.2Exercise 9.2
5.3Exercise 9.3
6. Overview of Deleted Syllabus for CBSE Class 9 Maths Circles
7. Class 9 Maths Chapter 9: Exercises Breakdown
8. Other Study Material for CBSE Class 9 Maths Chapter 9
9. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs

These solutions help students to grasp the key points and solve complex problems with ease. Focus areas include practising theorems, understanding their proofs, and applying these concepts to different problems. The Class 9 Maths NCERT Solutions detailed explanations and step-by-step approach make it easier for students to prepare effectively for their exams and build a strong mathematical base.

## Glance of NCERT Solutions for Class 9 Maths Chapter 9 Circles | Vedantu

• In Chapter 9 Class 9 Maths, focused on circles, several fundamental concepts and terminologies are essential for understanding this key geometric shape.

• A circle is defined by all the points in a plane that lie at a fixed distance, known as the radius, from a central point.

• The diameter of a circle is a special type of chord—a line segment whose endpoints are on the circle—that passes through the circle's centre and is twice the length of the radius.

• An arc represents a portion of the circle’s circumference, while a sector is the area enclosed by two radii and the arc between them.

• A segment, on the other hand, is the area bounded by a chord and the arc subtended by that chord.

• This chapter delves into the properties of circles, demonstrating that a line bisects and is perpendicular to a chord, equal chords are equidistant from the centre.

• This chapter explains about circles' theorems, focusing on cyclic quadrilateral angles and the principle that an angle subtended by an arc at the centre is twice that subtended at any other point.

• There are Exercise links provided. It has solutions for each question from Circles.

• There are three exercises (20 fully solved questions) in Chapter 9 Class 9 Maths Circles.

## Access Exercise wise NCERT Solutions for Chapter 9 Maths Class 9

 Current Syllabus Exercises of Class 9 Maths Chapter 9 NCERT Solutions of Class 10 Maths Circles Exercise 9.1 NCERT Solutions of Class 10 Maths Circles Exercise 9.2 NCERT Solutions of Class 10 Maths Circles Exercise 9.3
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## Exercises under NCERT Solutions for Class 9 Maths Chapter 9 Circles

NCERT Solutions for Class 9 Maths Chapter 9, "Circles," contains six exercises that cover different topics related to circles. Here is a brief overview of the types of questions dealt with in each exercise:

• Exercise 9.1: This exercise consists of two questions that are based on the tangents and the properties of tangents of a circle. The questions include finding the length of the tangent, the angle between the tangent and the radius, and the distance of the point from the centre of the circle.

• Exercise 9.2: This exercise consists of three questions that are based on the secants of a circle. The questions include finding the length of the secant, the intersection point of two secants, and the angle between the secant and the tangent.

• Exercise 9.3: This exercise consists of six questions that are based on the chords of a circle. The questions include finding the length of the chord, the angle between the chords, and the perpendicular bisector of the chord.

## Access NCERT Solutions for Class 9 Maths Chapter 9 – Circles

### Exercise 9.1

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Ans:

As we know that a circle is a collection of points therefore, they are equidistant from a fixed point. Now, this fixed point will be the centre of the circle and the equal distance between these points will be the radius of the circle. Hence, the shape of a circle will depend on its radius. Therefore, when we superimpose two circles of equal radius, then both the circles will cover each other. Thus, these two circles will be congruent when they have equal radius. Now, let us assume that two congruent circles have a common centre: $O$ and $O'$, $AB$ and $CD$ are the two chords of same length.

In $\Delta AOB$ and $\Delta CO'D$, we can observe that

$AB=CD$ as they are chords of the same length.

$OA=O'C$ as they are radii of congruent circles,

$OB=O'D$ as they are radii of congruent circles.

Therefore, $\Delta AOB\cong \Delta CO'D$ by the SSS congruence rule. This implies $\angle AOB\cong \angle CO'D$ By CPCT. Hence, equal chords of congruent circles subtend equal angles at their centres.

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Ans:

Let us assume that there are two congruent circles with the same radii that have centres as $O$ and $O'$.

In $\Delta AOB$ and $\Delta CO'D$,

$\angle AOB=\angle CO'D$ (Given)

$OA=O'C$ as they are radii of congruent circles

$OB=O'D$ as they are radii of congruent circles

Therefore,

$\Delta AOB\cong \Delta CO'D$ by the SSS congruence rule.

$\Rightarrow AB=CD$ (By CPCT)

Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

### Exercise 9.2

1. Two circles of radii $5cm$ and $3cm$ intersect at two points and the distance between their centres is $4cm$. Find the length of the common chord.

Ans: Let us assume that the radius of the circle which is centred at $O$ and $O'$ be $5cm$ and $3cm$.

Therefore,

$OA=OB$

$\Rightarrow 5cm$

Similarly,

$O'A=O'B$

$\Rightarrow 3cm$

Now, the line segment $OO'$ will be the perpendicular bisector of the chord $AB$.True.

We know that the points on the circle are always on equal distances from the centre of the circle and hence, this equal distance is defined as the radius of the circle. This is why a line segment joining the centre to any point on the circle is a radius of the circle.

2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Ans:

Let us assume that $PQ$ and $RS$ are the two chords of equal length of a circle and they are intersecting at a common point $T$.

So, let us draw two perpendicular bisectors $OV$ and $OU$ on these chords.

In $\Delta OVT$ and $\Delta OUT$,

We have $OV=OU$ as they are equal chords of a circle and are equidistant from the centre.

Also, $\angle OVT=\angle OUT$.

Therefore,

$\Delta OVT\cong \Delta OUT$, by the RHS congruent rule.

$\Rightarrow VT=UT$ by CPCT.

Now, we have given that –

$PQ=RS$

$\Rightarrow \frac{1}{2}PQ=\frac{1}{2}RS$

$\Rightarrow PV=RU$.

Now, let us add both the conditions as –

$PV+VT=RU+UT$

$\Rightarrow PT=RT$.

On subtracting we get –

$PQ-PT=RS-RT$

This equation indicates that a corresponding segment of the chords are congruent to each other. Hence, proved.

3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Ans:

Let us assume that $PQ$ and $RS$ are the two chords of the same length of a circle which are intersecting at a common point $T$.

So, let us draw two perpendicular bisectors $OV$ and $OU$ on these chords.

In $\Delta OVT$ and $\Delta OUT$,

We have $OV=OU$ as they are equal chords of a circle and are equidistant from the centre.

Also, $\angle OVT=\angle OUT$.

Therefore,

$\Delta OVT\cong \Delta OUT$, by the RHS congruence rule.

Therefore, we can conclude that $\angle OVT=\angle OUT$ by CPCT. Hence, if two equal chords of a circle intersect within the circle, then the line joining the point of intersection to the centre makes equal angles with the chords. Hence, proved.

4. If a line intersects two concentric circles (circles with the same centre) with centre $O$ at $A,B,C$ and $D$, prove that $AB=CD$.

Ans:

In the figure, let us draw a perpendicular $OM$ bisecting the chord $BC$ and $AD$.

We can observe from the figure that $BC<AD$.

Hence, we have –

$BM=MC$ and

$AM=MD$.

On subtracting both equations, we get –

$AM-BM=MD-MC$

$\Rightarrow AB=CD$.

Hence, proved.

5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius $5m$ drawn in a park. Reshma throws a ball to Salma, Salma and Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is $6m$ each, what is the distance between Reshma and Mandip?

Ans:

Let us assume that $OA$ and $OB$ are the two perpendiculars of $RS$ and $SM$ as shown in the figure below.

Hence, we have –

$AR=AS$

$\Rightarrow 3m$.

Also, $OR=OS=OM=5m$.

In $\Delta OAR$,

$O{{A}^{2}}+A{{R}^{2}}=O{{R}^{2}}$

$\Rightarrow O{{A}^{2}}=25-9$

$\Rightarrow OA=4m$.

As, from the figure we can observe that $ORSM$ is a kite. Now, we know that the diagonals of a kite are perpendicular.

Therefore,

$\angle RCS=90{}^\circ$ and $RC=CM$.

Area of the $\Delta ORS=\frac{1}{2}\times OA\times RS$

$\Rightarrow \frac{1}{2}\times RC\times OS=\frac{1}{2}\times 4\times 6$

$\Rightarrow RC=4.8$

Hence,

$RM=2RC$

$\Rightarrow RM=9.6m$.

Therefore, the distance between Reshma and Mandip will be $9.6m$.

6. A circular park of radius $20m$ is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.

Ans:

Let us draw a figure as –

From the figure, we can observe that $AS=SD=DA$.

Hence, $\Delta ASD$ will be an equilateral triangle and $OA=20m$.

Now, we know that the medians of an equilateral triangle will pass through the centre. Also, the medians will intersect each other at the ratio $2:1$.

Therefore, the median $AB$ is –

$\frac{OA}{OB}=\frac{2}{1}$

$\Rightarrow \frac{20}{OB}=\frac{2}{1}$

$\Rightarrow OB=10m$

Hence, $AB=OA+OB$

$\Rightarrow AB=30m$.

In $\Delta ABD$, we have –

$A{{D}^{2}}=A{{B}^{2}}+B{{D}^{2}}$

$\Rightarrow A{{D}^{2}}=900+{{\left( \frac{AD}{2} \right)}^{2}}$

$\Rightarrow 3A{{D}^{2}}=3600$

$\Rightarrow AD=20\sqrt{3}$

Hence, the length of the string of each phone will be $20\sqrt{3}m$.

### Exercise 9.3

1. In the given figure, $A,B,$ and $C$ are three points on a circle with centre $O$ such that $\angle BOC=30{}^\circ$ and $\angle AOB=60{}^\circ$. If $D$ is a point on the circle other than the arc $ABC$, find $\angle ADC$.

Ans:

From the figure, we can observe that –

$\angle AOC=\angle AOB+\angle BOC$

$\Rightarrow \angle AOC=60{}^\circ +30{}^\circ$

$\Rightarrow \angle AOC=90{}^\circ$.

As, the angle subtended by the arc at the centre will be twice the angle on the remaining part. Therefore,

$\Rightarrow \angle ADC=\frac{1}{2}(90{}^\circ )$

$\Rightarrow \angle ADC=45{}^\circ$.

2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Ans:

In $\Delta OAB$,

We have –

$AB=OA=OB$ as radius.

Hence, $\Delta OAB$ will be an equilateral triangle.

This implies that each interior angle of the equilateral triangle will be $60{}^\circ$.

$\Rightarrow \angle AOB=60{}^\circ$

$\Rightarrow \angle ACB=\frac{1}{2}\angle AOB$

$\Rightarrow \frac{1}{2}(60{}^\circ )=30{}^\circ$.

In quadrilateral $ACBD$,

We have –

$\angle ACB+\angle ADB=180{}^\circ$

$\Rightarrow \angle ADB=150{}^\circ$.

Therefore, the angle subtended by the chord on the major and minor arc will be $30{}^\circ$ and $150{}^\circ$.

3. In the given figure, $\angle PQR=100{}^\circ$, where $P,Q,$ and $R$ are points on a circle with centre $O$. Find $\angle OPR$.

Ans:

Let us assume that $PR$ is a chord of the circle and $S$ is any point on the major arc.

$PQRS$ is a cyclic quadrilateral.

Hence, we have –

$\angle PQR+\angle PSR=180{}^\circ$

$\Rightarrow \angle PSR=80{}^\circ$

Now, we know that the angle subtended by the arc at centre will be double the angle subtended by it.

Therefore,

$\angle PQR=2\angle PSR$

$\Rightarrow \angle POR=160{}^\circ$

In $\Delta POR$,

We can observe that –

$OP=PR$.

$\Rightarrow \angle OPR=\angle ORP$ as they are opposite angles of equal sides of a triangle.

$\Rightarrow \angle OPR+\angle ORP+\angle POR=180{}^\circ$ which is the angle sum property of a triangle.

$\Rightarrow 2\angle OPR+160{}^\circ =180{}^\circ$

$\Rightarrow \angle OPR=10{}^\circ$

Therefore, $\angle OPR=10{}^\circ$.

4. In figure, $\angle ABC=69{}^\circ$, $\angle ACB=31{}^\circ$, find $\angle BDC$?

Ans:

From the given figure, we have –

$\angle BAC=\angle BDC$.

In $\Delta ABC,$

$\angle BAC+\angle ABC+\angle ACB=180{}^\circ$

$\Rightarrow \angle BAC=180{}^\circ -69{}^\circ -31{}^\circ$

$\Rightarrow \angle BAC=80{}^\circ$.

Therefore, we have $\angle BDC=80{}^\circ$.

5. In the given figure, $A,B,C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC=130{}^\circ$ and $\angle ECD=20{}^\circ$. Find $\angle BAC$.

Ans:

From the given figure, we have –

In $\Delta CDE,$

$\angle CDE+\angle DCE=\angle CEB$

$\Rightarrow \angle CDE=130{}^\circ -20{}^\circ$

$\Rightarrow \angle CDE=110{}^\circ$.

But we know that $\angle CDE=\angle BAC$

Therefore,

$\angle BAC=110{}^\circ$.

6. $ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle DBC=70{}^\circ$, $\angle BAC=30{}^\circ$, find $\angle BCD$. Further, if $AB=BC$, find $\angle ECD$.

Ans:

The figure will be as –

From figure, we can observe that –

$\angle CBD=\angle CAD$

$\Rightarrow \angle CAD=70{}^\circ$.

$\Rightarrow \angle BAD=\angle BAC+\angle CAD$

$\Rightarrow \angle BAD=100{}^\circ$

Therefore, we have –

$\angle BCD+\angle BAD=180{}^\circ$

$\Rightarrow \angle BCD=80{}^\circ$.

Now, in $\Delta ABC,$ we have –

$AB=BC$

$\Rightarrow \angle BCA=\angle CAB$

$\Rightarrow \angle BCA=30{}^\circ$.

Also, we have –

$\angle BCD=80{}^\circ$

$\Rightarrow \angle BCA+\angle ACD=80{}^\circ$

$\Rightarrow \angle ACD=80{}^\circ -30{}^\circ$

$\Rightarrow \angle ACD=50{}^\circ$

$\Rightarrow \angle ECD=50{}^\circ$.

7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans:

Let us assume a cyclic quadrilateral $ABCD$ having diagonals $BD$ and $AC$, intersecting at a common point $O$.

$\angle BAD=\frac{1}{2}\angle BOD$

$\Rightarrow \angle BAD=90{}^\circ$

Now, $\angle BCD+\angle BAD=180{}^\circ$

$\Rightarrow \angle BCD=90{}^\circ$.

$\angle ADC=\frac{1}{2}\angle AOC$

$\Rightarrow \angle ADC=90{}^\circ$

$\Rightarrow \angle ADC+\angle ABC=180{}^\circ$

$\Rightarrow \angle ABC=90{}^\circ$.

Therefore, each interior angle of the quadrilateral is $90{}^\circ$ which implies that $ABCD$ is a rectangle.

8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Ans:

Let us assume a trapezium $ABCD$ with $AB\parallel CD$ and $BC=AD$ as shown in the figure below.

From the figure, we can observe that $AM\bot CD$ and $BN\bot CD$.

Therefore, in $\Delta AMD$ and $\Delta BNC$,

$AD=BC$.

$\Rightarrow \angle AMD=\angle BNC$

$AM=BN$

$\Rightarrow \Delta AMD\cong \Delta BNC$ by the RHS congruence rule.

Therefore, $\angle ADC=\angle BCD$.

$\angle BAD$ and $\angle ADC$ are on the same side.

Therefore, $\angle BAD+\angle ADC=180{}^\circ$

$\angle BAD+\angle BCD=180{}^\circ$

Hence, the angles are supplementary. Therefore, $ABCD$ is a cyclic quadrilateral.

9. Two circles intersect at two points $B$ and $C$. Through $B$, two-line segments $ABD$ and $PBQ$ are drawn to intersect the circles at $A$, $D$ and $P$, $Q$ respectively. Prove that $\angle ACP=\angle QCD$.

Ans:

Let us join the chords $AP$ and $DQ$.

Therefore,

$\angle PBA=\angle ACP$,

Also, $\angle DBQ=\angle QCD$.

Now, we know that $ABD$ and $PBQ$ are the line segments intersecting at common point $B$.

Therefore,

$\angle PBA=\angle DBQ$

Hence, we have $\angle ACP=\angle QCD$.

10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans:

Let us consider a triangle $\Delta ABC$ in the figure given below –

We can observe that two circles are drawn by taking the diameters $AB$ and $AC$. We will let the points $B$ and $C$ intersect each other at a common point $D$ which does not lie on the line segment $BC$.

Therefore, after joining $AD$ we have –

$\angle ADB=90{}^\circ$

$\Rightarrow \angle ADC=90{}^\circ$

$\Rightarrow \angle BDC=\angle ADB+\angle ADC$

$\Rightarrow \angle BDC=180{}^\circ$.

Hence, we have a straight line as $BDC$. This implies that the assumption that we considered was wrong.

Therefore, the point of intersection $D$ will lie on the line segment $BC$.

11. $ABC$ and $ADC$ are two right triangles with common hypotenuse $AC$. Prove that $\angle CAD=\angle CBD$.

Ans:

From the figure we know that in $\Delta ABC$,

$\angle ABC+\angle BCA+\angle CAB=180{}^\circ$

$\Rightarrow \angle BCA+\angle CAB=90{}^\circ$.

In $\Delta ADC,$

$\angle CDA+\angle ACD+\angle DAC=180{}^\circ$

$\Rightarrow \angle ACD+\angle DAC=90{}^\circ$

After adding both the conditions, we get –

$\angle BCA+\angle CAB+\angle ACD+\angle DAC=180{}^\circ$

$\Rightarrow \left( \angle BCA+\angle ACD \right)+\left( \angle CAB+\angle DAC \right)=180{}^\circ$

$\Rightarrow \angle BCD+\angle DAB=180{}^\circ$.

Now, we know that $\angle B+\angle D=180{}^\circ$.

Therefore, we can observe from the sum of each interior angle that it is a cyclic quadrilateral.

Hence,

$\angle CAD=\angle CBD$.

12. Prove that a cyclic parallelogram is a rectangle.

Ans:

Let us assume a cyclic parallelogram $ABCD$ as shown in the figure below –

We have –

$\angle A+\angle C=180{}^\circ$.

Now, we know that in a parallelogram opposite angles are always equal.

Therefore,

$\angle A=\angle C$ and

$\angle B=\angle D$.

$\Rightarrow \angle A+\angle C=180{}^\circ$

$\Rightarrow \angle A=90{}^\circ$.

Similarly,

$\Rightarrow \angle B=90{}^\circ$.

Therefore, all the interior angles of the parallelogram are $90{}^\circ$ which implies it is a rectangle. Hence, proved.

## Overview of Deleted Syllabus for CBSE Class 9 Maths Circles

 Chapter - 9 Dropped Topics Circles Exercise 9.1 Introduction Exercise 9.2 Circles and its related terms: Review and Circle through three points.

## Class 9 Maths Chapter 9: Exercises Breakdown

 Chapter 9 -  Circles Exercises in PDF Format Exercise 9.1 2 Questions with Solutions Exercise 9.2 6 Questions with Solutions Exercise 9.3 12 Questions with Solutions

## Conclusion

NCERT Chapter 9 of Class 9 Maths, "Circles," covers essential concepts and theorems related to circles. Key topics include properties of chords, angles subtended by chords, and cyclic quadrilaterals. Important theorems to focus on are those involving the perpendicular from the centre of a chord, equal chords and their distances from the centre, and the unique circle passing through three non-collinear points. Understanding these concepts and practising the related exercises can enhance logical thinking and problem-solving skills. NCERT solutions offer detailed explanations and a variety of solved and unsolved problems, which are crucial for thorough preparation.

## Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths Chapter 9. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 9 Maths Chapter 9 Circles

1. Define Sectors?

The sector is an interior part of a circle which is covered by two radii and one arc of that Circle. These are of two types. The minor sector, which covers less area. The major sector, which covers a large area.

2. Distinguish Between Chord and Diameter?

A chord is a line segment drawn between any two points in the circle. At the same time, the diameter is drawn in the middle of the circle. It cuts the circle into two halves. The radius of a circle is defined as a half part of the diameter.

d = 2r.

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

Students who aim to excel in their CBSE Board exams have NCERT textbooks as the best study material at their disposal. In order to fully comprehend the solutions of NCERT textbook problems, students can refer to Vedantu's NCERT Solutions for Class 9 Maths Chapter 9 and download it for free. These solutions offer on point solutions prepared by experts that can help you ace the material in no time.

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

Here is why you should follow Vedantu's NCERT Solutions:

• If you are attempting CBSE Board exams, then a rigorous study of the NCERT syllabus should be your first most priority.

• It is equally important to practice the NCERT problems and properly understand their solutions.

• Vedantu's NCERT Solutions for Class 9 Maths are curated by Mathematics experts.

• The solutions are offered in a direct, logical and understandable way making the concepts easy for students to grasp.

5. What are the basics of Class 9 Maths Chapter 9 Circles?

Chapter 9 Circles talks about the basic concepts and terms related to a circle. The chapter has some important theorems for students to learn. Basic concepts of the chapter include:

• Angle Subtended by the Chord at a Point

• Perpendicular From the Centre of a Chord

• Circles Through Three Points

• Chords and Their Distance From the Centre

• Angle Subtended by the Area of Circle

6. How do I solve the problems of Class 9 Maths Chapter 9 Circles?

Before attempting to solve any problems, it is important that the core concepts are clear in a student’s mind. In order to correctly solve the problems and understand the concepts, students can refer to NCERT Solutions for Class 9 Maths Chapter 9 Circles. These solutions can also be accessed from the Vedantu app. These solutions attempt to solve the problems in the most logical and straightforward way as prepared by Vedantu’s Mathematics expert teachers.

7. What are the most important theorems that come in Class 9 Chapter 9 Circles?

Class 9 Maths Chapter 9 Circles has a lot of theorems. All the theorems mentioned in the chapter are important, and students should practice them well. The main theorems include:

• Equal chords of a circle subtend equal angles at the centre.

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

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

In addition to these, all theorems are important and must be practised well.

8. What is the most important theorem in circles for Class 9?

The most important theorem in Class 9 circles is Thales' Theorem (also known as the Angle in a Semi-Circle Theorem). Thales' Theorem is a key concept in Class 9 circles, stating that if you draw a straight line (a diameter) across the middle of a circle and then make an angle from any point on the circle's edge, that angle will always be 90 degrees, or a right angle. Essentially, this theorem helps to explain why any triangle formed with the endpoints of the diameter and a point on the circle is always a right triangle, showcasing a fundamental property of circles in geometry.

9. What is special about circles in Circles Class 9?

In Circles Class 9 , circles are special because they introduce fundamental concepts and properties such as:

• Chord properties: Equal chords are equidistant from the centre.

• Angle properties: Angles subtended by the same arc are equal, and the angle in a semicircle is a right angle.

• Tangent properties: A tangent to a circle is perpendicular to the radius at the point of tangency.

10. How to prove a circle in Ch 9 Maths Class 9?

To prove a circle in Ch 9 Maths Class 9, you can demonstrate that all points on the circle are equidistant from a fixed point called the centre. This involves showing that every line segment drawn from the centre to the circumference (the radius) is of equal length. Additionally, you can verify that any point on the circumference maintains a constant distance from the centre, ensuring that this distance remains the same for all points around the circle. These properties confirm that the shape in question is indeed a circle, as they align with the fundamental definition of a circle in geometry.

11. How many types of circles are there in Ch 9 Maths Class 9?

While there is fundamentally one type of geometric figure called a circle, circles can be categorized based on their relative positions and relationships with each other:

• Concentric Circles: Circles that share the same centre but have different radii.

• Tangent Circles: Circles that touch each other at exactly one point. They can be internally tangent (one circle inside another) or externally tangent (touching from outside).

• Congruent Circles: Circles with the same radius.

12. What is a circle in Class 9 Maths Ch 9?

A circle is essentially a shape where every point along its edge is the exact same distance from a central point. This constant distance is known as the radius. Imagine drawing a perfect loop around a point, making sure every part of the loop is equally far from the centre — that's your circle.

13. What is a chord in a circle Class 9 Maths Ch 9?

A chord in a circle is like a straight road that cuts straight across the circle, touching it at two points on the perimeter. This line doesn't go all the way around; instead, it just stretches from one side of the circle to the other, connecting those two points directly.