NCERT Solutions for Class 9 Maths Chapter 9 Circles

NCERT Solutions for Class 9 Maths Chapter 9 Circles

Edited By Komal Miglani | Updated on May 02, 2025 07:58 AM IST

Have you ever tossed a coin, seen how beautiful the full moon looks in the night sky, or ordered a delicious pizza? So many things around us are round in shape; all of these round objects are examples of circles, an essential part of geometry. The chapter Circles of the NCERT Syllabus of Class 9 Maths includes the properties of circles, arcs, tangents, chords, and their distances from the centre, cyclic quadrilaterals, etc. These key concepts of circles will help the students understand more advanced geometry concepts effectively and enhance their problem-solving ability in real-world applications. The NCERT Solutions for this chapter also guide students in building strong conceptual clarity using step-by-step examples and theorems.

This Story also Contains
  1. Circles Class 9 Questions And Answers PDF Free Download
  2. Circles Class 9 Solutions - Important Formulae And Points
  3. NCERT Solutions for Class 9 Maths Chapter 9: Exercise Questions
  4. Class 9 Maths NCERT Chapter 9: Extra Question
  5. Approach to Solve Questions of Circles Class 9
  6. Circles Class 9 Solutions - Exercise Wise
  7. NCERT Solutions For Class 9 Maths - Chapter Wise
  8. NCERT Solutions For Class 9 - Subject Wise
  9. NCERT Class 9 Books and Syllabus
NCERT Solutions for Class 9 Maths Chapter 9 Circles
NCERT Solutions for Class 9 Maths Chapter 9 Circles

This article on NCERT solutions for class 9 Maths Chapter 9 Circles offers clear and step-by-step solutions for the exercise problems in the NCERT Books for class 9 Maths. Students who need the Circles class 9 solutions will find this article very useful. It covers all the important Class 9 Maths Chapter 9 solutions. According to the latest CBSE syllabus, these solutions are made by the Subject Matter Experts, ensuring that students can grasp the basic concepts effectively. NCERT Solutions for Class 9 and NCERT Solutions for other subjects and classes can be downloaded from the NCERT Solutions.

Circles Class 9 Questions And Answers PDF Free Download

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Circles Class 9 Solutions - Important Formulae And Points

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

Arc: An arc of a circle is a continuous portion of the circle.

Chord of a Circle: The chord of a circle is a line segment that connects any two points on the circle.

Some Important Properties Of Circle Chords:

  • The diameter of a circle is a chord that passes through its centre.

  • A circle's diameter divides it into two equal arcs, forming a semicircle.

  • Congruent arcs have the same degree measure.

  • Equal arcs have associated chords of the same length.

  • A perpendicular drawn from the centre to a chord bisects the chord, and vice versa.

  • Three non-collinear points define one and only one circle.

  • Chords equidistant from the centre are equal in length.

  • The line connecting the centres of two intersecting circles and their common chord is perpendicular.

  • The central angle of an arc is twice the angle it subtends on the circumference.

  • Any two angles in the same circle segment are equal.

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

  • The larger chord of a circle is closer to the centre than the smaller chord.

  • A semicircle contains a right angle.

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

Cyclic Quadrilateral:

  • A quadrilateral is termed cyclic if all of its vertices lie on the circumference of a circle.

  • The sum of opposite angles in a cyclic quadrilateral is 180°, and vice versa.

  • An exterior angle of a cyclic quadrilateral is equal to its opposite inner angle.

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Tangent and Radius:

  • The tangent and radius of a circle intersect at a right angle.


NCERT Solutions for Class 9 Maths Chapter 9: Exercise Questions

Circles class 9 NCERT solutions - Exercise: 9.1

Page number: 118, Total questions: 2

Question 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.

Answer:

1745317842320

In the given figure, two congruent circles are given and have centers O and O', and chords are PQ and XY, respectively.
In Δ POQ and XO'Y
PQ = XY (Given)
OP = O'X (Radius of congruent triangle)
OQ = O'Y (Radius of congruent triangle)
ΔPOQ ≅ ΔXO'Y (By SSS rule)
Therefore, ∠POQ = ∠XO'Y (By CPCT)

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

Answer:

Given: Two congruent circles have equal angles.
To Prove: Two congruent circles have equal angles, and then the chords are equal.

1745317875792

In Δ POQ and XO'Y
∠POQ = ∠XO'Y (Given)
OP = O'X (Radius of congruent triangle)
OQ = O'Y (Radius of congruent triangle)
ΔPOQ ≅ ΔXO'Y (By SSS rule)
Therefore, PQ = XY (By CPCT)

NCERT Solutions for Class 9 Maths Chapter 9 Circles - Exercise: 9.2
Page number: 122, Total questions: 6

Question 1: Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points, and the distance between their centres is $\small 4\hspace{1mm}cm$. Find the length of the common chord.

Answer:

Given: Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points and the distance between their centres is $\small 4\hspace{1mm}cm$ .
To find the length of the common chord.
Construction: Join OP and ON draw $OM\perp AB\

1745296503432

Proof: AB is a chord of the circle, and OM is the bisector of chord AB.
$\therefore OM\perp AB$
$\angle OMA = 90 ^\circ$
Let, OM = x , so O'M = 4 - x
In $\triangle$ AOM, using Pythagoras' theorem
$AM^2=AO^2 - OM^2$ ...........................(1)
Also,
In $\triangle$ AO'M, using Pythagoras' theorem
$AM^2=AO'^2 - MO'^2$ ...........................(2)
From (1) and (2), we get
$AO^2 - OM^2 = AO'^2 - MO'^2$
$\Rightarrow 5^2 - x^2 = 3^2 - (4 - x)^2$
$\Rightarrow 25 - x^2 = 9 - 16 - x^2 + 8x$
$\Rightarrow 32 = 8x$
$\Rightarrow x = 4$
Put x = 4 in equation (1)
$AM^2 = 5^2 - 4^2 = 9$
$\Rightarrow AM = 3$
$\Rightarrow AB = 2AM = 6$

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

Answer:

Given: two equal chords of a circle intersect within the circle.
To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP = DP.
Construction: Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$
Proof:

1745296545668

In $\triangle$ OMP and $\triangle$ ONP,
OP = OP (Common)
OM = ON (Equal chords of a circle are equidistant from the centre)
$\angle$ OMP = $\angle$ ONP (Both are right angled)
Thus, $\triangle$ OMP $\cong$ $\triangle$ ONP (By SAS rule)
PM = PN..........................(1) (CPCT)
AB = CD ..........................(2)(Given)
$\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$
$\Rightarrow AM = CN$ .....................(3)
Adding equations (1) and (3), we have
AM + PM = CN + PN
$\Rightarrow AP = CP$ .................(4)
Subtract equation (4) from (2), we get
$ AB - AP = CD - CP$
$\Rightarrow PB = PD$

Question 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.

Answer:

Given: two equal chords of a circle intersect within the circle.
To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. $\angle$ OPM= $\angle$ OPN

1745296582641

Proof:
Construction: Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$
In $\triangle$ OMP and $\triangle$ ONP,
OP = OP (Common)
OM = ON (Equal chords of a circle are equidistant from the centre)
$\angle$ OMP = $\angle$ ONP (Both are right-angled)
Thus, $\triangle$ OMP $\cong$ $\triangle$ ONP (By RHS rule)
$\angle$ OPM= $\angle$ OPN (CPCT)

Question 4: If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that $\small AB=CD$ (see Fig. $\small 10.25$ ).

Answer:

Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.
To prove: AB = CD
Construction: Draw $OM\perp AD$

1745296629919

BC is a chord of the inner circle and $OM\perp BC$
So, BM = CM .................(1)
(Perpendicular OM bisect BC)
Similarly,
AD is a chord of the outer circle and $OM\perp AD$
So, AM = DM .................(2) (Perpendicular OM bisect AD)
Subtracting equation (1) from (2), we get
$ AM-BM = DM - CM$
$\Rightarrow AB = CD$

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

Answer:

Given: From the figure, R, S, and M are the positions of Reshma, Salma, and Mandip, respectively.
So, RS = SM = 6 cm
Construction: Join ON, OS, OR and OM. Draw $OL\perp RS$.
Proof:

1745296667139

In $\triangle$ ORS,
OS = OR and $OL\perp RS$ (by construction)
So, RL = LS = 3cm (RS = 6 cm)
In $\triangle$ OLS, by Pythagoras' theorem,
$OL^2 = OS^2 - SL^2$
$\Rightarrow OL^2 = 5^2 - 3^2 = 25 - 9 = 16$
$\Rightarrow OL = 4$
In $\triangle$ ORN and $\triangle$ OMN,
OR = OM (Radii)
$\angle$ RON = $\angle$ MON (Equal chords subtend equal angles at the centre)
ON = ON (Common)
$\triangle$ ORN $\cong$ $\triangle$ OMN (By SAS)
RN = MN (CPCT)
Thus, $ON\perp RM$
Area of $\triangle$ ORS = $\frac{1}{2}\times RS\times OL$ ..................(1)
Area of $\triangle$ ORS = $\frac{1}{2}\times OS\times NR$ ..................(2)
From 1 and 2, we get
$\frac{1}{2}\times RS\times OL$ $ = \frac{1}{2}\times OS\times NR$
$\Rightarrow RS\times OL= OS\times KR$
$\Rightarrow 6\times 4 = 5\times KR$
$\Rightarrow NR = 4.8 cm$
Thus, $RM =2 NR = 2\times 4.8 cm = 9.6 cm$

Question 6: A circular park of radius $\small 20m$ 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 each other. Find the length of the string of each phone.

Answer:

Given: In the figure, A, B, and C are positioned as Ankur, Syed and David, respectively.
So, AB = BC = CD
Radius of circular park = 20m
So, AO = OB = OC = 20m
Construction: AF $\perp$ BC
Proof:

1745296698281

Let AC = CB = AB = 2x cm
In $\triangle$ ABC,
AC = AB and AF $\perp$ BC
So, CF = FB = x cm
In $\triangle$ OFB, by Pythagoras,
$OF^2=OB^2-FB^2$
$\Rightarrow OF^2 = 20^2-x^2 = 400-x^2$
$\Rightarrow OF = \sqrt{400-x^2}$
In $\triangle$ AFB, by Pythagoras,
$AF^2 = AB^2-FB^2$
$\Rightarrow (AO+OF)^2 + x^2 = (2x)^2$
$\Rightarrow (20+\sqrt{400 - x^2})^2 + x^2 = 4x^2$
$\Rightarrow 400 + 400 - x^2 + 40\sqrt{400 - x^2} + x^2 = 4x^2$
$\Rightarrow 800 + 40\sqrt{400 - x^2} = 4x^2$
$\Rightarrow 200 + 10\sqrt{400 - x^2} = x^2$
$\Rightarrow 10\sqrt{400 - x^2} = x^2 - 200$
Squaring both sides,
$\Rightarrow 100(400 - x^2) = (x^2 - 200)^2$
$\Rightarrow 40000 - 100x^2 = x^4-40000 - 400x^2$
$\Rightarrow x^4 - 300x^2 = 0$
$\Rightarrow x^2(x^2 - 300) = 0$
$\Rightarrow x^2 = 300$
$\Rightarrow x = 10\sqrt{3}$
Hence, the length of the string of each phone $= 2x = 20\sqrt{3}$ m

NCERT Solutions for Class 9 Maths Chapter 9 Circles - Exercise: 9.3
Page number: 127-129, Total questions: 12

Question 1: In Fig. $\small 10.36$ , A,B and C are three points on a circle with centre O such that $\small \angle BOC=30^{\circ}$ and $\small \angle AOB=60^{\circ}$ . If D is a point on the circle other than the arc ABC, find $\small \angle ADC$.

1745297591812

Answer:

$\angle$ AOC = $\angle$ AOB + $\angle$ BOC = $60 ^\circ+30 ^\circ=90 ^\circ$

$\angle$ AOC = 2 $\angle$ ADC (angle subtended by an arc at the centre is double the angle subtended by it at any)

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

$\Rightarrow \angle ADC=\frac{1}{2}90 ^\circ = 45 ^\circ$

Question 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.

Answer:

Given: A chord of a circle is equal to the radius of the circle i.e. OA = AB.
To find: $\angle$ADB and $\angle$ ACB.

1745297632182

In $\triangle$ OAB,OA = AB (Given)

OA = OB (Radii of circle)

So, OA = OB = AB

$\Rightarrow$ AOB is an equilateral triangle.

So, $\angle$ AOB = $60 ^\circ$

$\angle$ AOB = 2 $\angle$ ADB

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

$\Rightarrow \angle ADB = \frac{1}{2}60 ^\circ=30$

ACBD is a cyclic quadrilateral.

So, $\angle$ ACB+ $\angle$ ADB = $180 ^\circ$

$\Rightarrow \angle ACB + 30 ^\circ= 180 ^\circ$

$\Rightarrow \angle ACB = 180 ^\circ-30 ^\circ=150 ^\circ$

Question 3: In Fig. $\small 10.37$ , $\small \angle PQR=100^{\circ}$ , where P, Q and R are points on a circle with centre O. Find $\small \angle OPR$.

1745321888736

Answer:

1745321929523

Construction: Join PO and OR.
PQSR is a cyclic quadrilateral.

So, $\angle$ PSR + $\angle$ PQR = $180 ^\circ$

$\Rightarrow \angle PSR + 100 ^\circ = 180 ^\circ$

$\Rightarrow \angle PSR = 180 ^\circ - 100 ^\circ = 80 ^\circ$

Here, $\angle$ POR = 2 $\angle$ PSR

$\Rightarrow \angle POR = 2\times 80 ^\circ = 160 ^\circ$

In $\triangle$ OPR,

OP = OR (Radii)

$\angle$ ORP = $\angle$ OPR (the angles opposite to equal sides)

In $\triangle$ OPR,

$\angle$ OPR+ $\angle$ ORP + $\angle$ POR = $180 ^\circ$

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

$\Rightarrow 2\angle OPR = 180 ^\circ - 160 ^\circ$

$\Rightarrow 2\angle OPR = 20 ^\circ$

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

Question 4: In Fig. $\small 10.38$ , $\small \angle ABC=69^{\circ}, \angle ACB=31^{\circ},$ find $\small \angle BDC$

1745297886746

Answer:

In $\triangle$ ABC,

$\angle$ A+ $\angle$ ABC+ $\angle$ ACB= $180^\circ$

$\Rightarrow \angle A+69 ^\circ+31 ^\circ=180^\circ$

$\Rightarrow \angle A+100 ^\circ=180^\circ$

$\Rightarrow \angle A=180 ^\circ-100^\circ$

$\Rightarrow \angle A=80 ^\circ$

$\angle$ A = $\angle$ BDC = $80 ^\circ$ (Angles in same segment)

Question 5: In Fig. $\small 10.39$ , A, B, C and D are four points on a circle. AC and BD intersect at a point E such that $\small \angle BEC=130^{\circ}$ and $\small \angle ECD=20^{\circ}$ . Find $\small \angle BAC$

1745297908720

Answer:

$\angle$ DEC+ $\angle$ BEC = $180 ^\circ$ (linear pairs)

$\Rightarrow$ $\angle$ DEC+ $130 ^\circ$ = $180 ^\circ$ ( $\angle$ BEC = $130 ^\circ$ )

$\Rightarrow$ $\angle$ DEC = $180 ^\circ$ - $130 ^\circ$

$\Rightarrow$ $\angle$ DEC = $50 ^\circ$

In $\triangle$ DEC,

$\angle$ D+ $\angle$ DEC+ $\angle$ DCE = $180 ^\circ$

$\Rightarrow \angle D+50 ^\circ+20 ^\circ= 180 ^\circ$

$\Rightarrow \angle D+70 ^\circ= 180 ^\circ$

$\Rightarrow \angle D= 180 ^\circ-70 ^\circ=110 ^\circ$

$\angle$ D = $\angle$ BAC (angles in same segment are equal )

$\angle$ BAC = $110 ^\circ$

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

Answer:

1745297946422

$\angle BDC=\angle BAC$ (angles in the same segment are equal )

$\angle BDC= 30 ^\circ$

In $\triangle BDC,$

$\angle BCD+\angle BDC+\angle DBC= 180 ^\circ$

$\Rightarrow \angle BCD+30 ^\circ+70 ^\circ= 180 ^\circ$

$\Rightarrow \angle BCD+100 ^\circ= 180 ^\circ$

$\Rightarrow \angle BCD=180 ^\circ- 100 ^\circ=80 ^\circ$

If AB = BC ,then

$\angle BCA=\angle BAC$

$\Rightarrow \angle BCA=30 ^\circ$

Here, $\angle ECD+\angle BCE=\angle BCD$

$\Rightarrow \angle ECD+30 ^\circ=80 ^\circ$

$\Rightarrow \angle ECD=80 ^\circ-30 ^\circ=50 ^\circ$

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

Answer:

1745298019285

AC is the diameter of the circle.

Thus, $\angle ADC=90 ^\circ$ and $\angle ABC=90 ^\circ$ ............................1(Angle in a semi-circle is a right angle)

Similarly, BD is the diameter of the circle.

Thus, $\angle BAD=90 ^\circ$ and $\angle BCD=90 ^\circ$ ............................2(Angle in a semi-circle is a right angle)

From 1 and 2, we get

$\angle BCD=\angle ADC=\angle ABC=\angle BAD =90 ^\circ$

Hence, ABCD is a rectangle.

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

Answer:

1745298063019

Given: ABCD is a trapezium.

Construction: Draw AD || BE.

Proof: In quadrilateral ABED,

AB || DE (Given )

AD || BE ( By construction )

Thus, ABED is a parallelogram.

AD = BE (Opposite sides of parallelogram )

AD = BC (Given )

so, BE = BC

In $\triangle$ EBC,

BE = BC (Proved above )

Thus, $\angle C = \angle 2$ ...........1(angles opposite to equal sides )

$\angle A= \angle 1$ ...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

$\angle 1+\angle 2=180 ^\circ$ (linear pair)

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

Thus, ABED is a cyclic quadrilateral.

Question 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 (see Fig. $\small 10.40$ ). Prove that $\small \angle ACP=\angle QCD$.

1745298090691

Answer:

$\angle ABP=\angle QBD$ ................1(vertically opposite angles)

$\angle ACP=\angle ABP$ ..................2(Angles in the same segment are equal)

$\angle QBD=\angle QCD$ .................3(angles in the same segment are equal)

From 1,2,3 ,we get

$\angle ACP=\angle QCD$

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

Answer:

Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

1745298167595

Proof: AB is the diameter of the circle and $\angle$ ADB is formed in a semi-circle.

$\angle$ ADB = $90 ^\circ$ ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and $\angle$ ADC is formed in a semi-circle.

$\angle$ ADC = $90 ^\circ$ ........................2(angle in a semi-circle)

From 1 and 2, we have

$\angle$ ADB+ $\angle$ ADC= $90 ^\circ$ + $90 ^\circ$ = $180 ^\circ$

$\angle$ ADB and $\angle$ ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

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

Answer:

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove : $\small \angle CAD =\angle CBD$

Proof :

1745298195372

Triangle ABC and ADC are on common base BC and $\angle$ BAC = $\angle$ BDC.

Thus, point A, B, C, D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)

$\angle$ CAD = $\angle$ CBD (Angles in the same segment are equal)

Question 12: Prove that a cyclic parallelogram is a rectangle.

Answer:

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

1745298231341

In cyclic quadrilateral ABCD.

$\angle A + \angle C = 180 ^\circ$ .......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

$\angle A = \angle C$ ........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

$\angle A + \angle A = 180 ^\circ$

$\Rightarrow 2\angle A = 180 ^\circ$

$\Rightarrow \angle A = 90 ^\circ$

We know that a parallelogram with one angle a right angle is a rectangle.

Hence, ABCD is a rectangle.


Class 9 Maths NCERT Chapter 9: Extra Question

Question: The radius of a circle is 5 cm, and the length of the chord is 6 cm. Find the distance from the centre to its chord.

Answer:

Let AB = 6 cm be the chord of the circle with radius AO = 5 cm. Draw OP perpendicular to AB.

Now, apply Pythagoras' theorem

(AO)2 = (OP)2 + (AP)2

Here, AP = $\frac{1}{2}$ AB

So, AP = 3 cm

Now, putting values, we get:

52 = (OP)2 + 32

25 = (OP)2 + 9

(OP)2 = 16

Therefore, OP = 4 cm

Thus, the distance from the centre to its chord is 4 cm.

Approach to Solve Questions of Circles Class 9

1. Learn key circle terms: Students must learn fundamental circle vocabulary that includes radius, diameter, chord, arc and tangent. These form the base of all problems in this chapter.

2. Apply the perpendicular theorem: The perpendicular line drawn from the circle centre splits any chord exactly in two parts, and this property serves as a fundamental concept in many of our questions.

3. Use equal chord concepts: An object which reaches the centre of a circle at the same distance from the centre holds an equal length with the object. The concepts apply to both numerical tasks and proof-based assignments.

4. Understand angle subtended properties: Resolving angle problems requires knowing that the centre-point arc angle equals two times the angle derived at any circle-perimeter location.

5. Practice cyclic quadrilateral concepts: Students should employ opposite angles of cyclic quadrilaterals, adding 180 degrees as a frequently examined concept.

6. Draw neat diagrams and use logical steps: The ability to perform geometrical reasoning improves through diagrams which are correctly drawn, together with points clearly labelled, particularly when solving proof-based problems.

Circles Class 9 Solutions - Exercise Wise

Interested students can practice these Class 9 Maths Chapter 9 question answers using the exercise solutions provided below.

NCERT Solutions For Class 9 Maths - Chapter Wise


NCERT Solutions For Class 9 - Subject Wise

Here are the subject-wise links for the NCERT solutions of class 9:

NCERT Class 9 Books and Syllabus

Given below are some useful links for NCERT books and the NCERT syllabus for class 9:

Keep working hard & happy learning!

Frequently Asked Questions (FAQs)

1. What are the important topics in chapter ch 10 maths class 9 Circles ?

Circles and the related terms, angle subtended by a chord at a point, perpendicular from the centre to a chord, circle through three points, equal chords and their distances from the centre, angle subtended by an arc of a circle are the important topics of this chapter.

2. What benefits do Class 9 students derive from studying NCERT Solutions for Maths Chapter 10?

Students in Class 9 can benefit from class 9 chapter 10 maths by gaining a thorough understanding of all the concepts within the subject, which can serve as the basis for their future academic pursuits. The solutions provided by Careers360 experts are designed in a clear and comprehensive manner, making it easier for students to solve complex problems with greater efficiency. By mastering these solutions, CBSE Class 9 students can establish a strong foundation in the fundamentals and achieve excellent scores in their final exams.

3. What is the rationale behind adhering to circles class 9 NCERT solutions?

The use of class 9th circles NCERT solutions  is a reliable and effective approach to help students develop mastery of the subject's concepts. Reviewing these solutions, in conjunction with the textbooks, can aid in solving any problems that may arise on the board exams. These solutions also enhance students' problem-solving skills and logical reasoning abilities. They are among the most widely used study materials for CBSE exams. Consistent practice with these solutions can improve students' performance and help them excel in the subject.

4. Where can I find the complete solutions of NCERT for class 9 ?

Here you will get the detailed NCERT solutions for class 9. you can practice these maths ch 10 class 9 solutions and problems to command the concepts discussed in the chapter. after practicing you will get confidence to solve any kind of problem related to circle given in class 9th.

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0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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