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NCERT Solutions for Class 9 Maths Chapter 10 Circles

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Edited By Ramraj Saini | Updated on May 11, 2023 03:25 PM IST

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Circles Class 9 Questions And Answers are provided here. These NCERT solutions are prepared by experts at Careers360 considering latest CBSE syllabus 2023. All NCERT problems are discussed in simple, easy to understand and comprehensive way. Thus, these will help students to score well in the exams. Many objects that we come across in our daily life is circular in shape, such as ring, bangle, wheels of the vehicle, clock, etc. NCERT solutions for class 9 maths chapter 10 Circles will help in solving all problems given in the book related to circular shapes.

The circle is an integral part of unit geometry. A circle divides the plane in which the circle lies into three parts as shown in figure 1, which are the interior of the circle, the exterior of the circle and the circle. In this particular chapter, you will learn the proof of theorems and problems based on those theorems. NCERT class 9 maths chapter 10 question answer are designed in such a way that a student can fetch 100% marks in a particular question. Here you will get NCERT solutions for class 9 Maths also.

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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 are 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.

Tangent and Radius:

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

Free download NCERT Solutions for Class 9 Maths Chapter 10 Circles for CBSE Exam.

Circles Class 9 NCERT Solutions (Intext Questions and Exercise)

Class 9 maths chapter 10 NCERT solutions - Exercise: 10.1

Fill in the blanks:

Q1 (i) The centre of a circle lies in _____________ of the circle. (exterior/ interior)

Answer:

The centre of a circle lies in the interior of the circle.

Fill in the blanks:

Q1 (ii) A point, whose distance from the centre of a circle is greater than its radius lies in_____________ of the circle. (exterior/ interior)

Answer:

A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.

Fill in the blanks:

Q1 (iii) The longest chord of a circle is a ___________ of the circle.

Answer:

The longest chord of a circle is a diameter of the circle.

Fill in the blanks:

Q1 (iv) An arc is a ________________ when its ends are the ends of a diameter.

Answer:

An arc is a semi- circle when its ends are the ends of a diameter.

Fill in the blanks:

Q1 (v) Segment of a circle is the region between an arc and ________________ of the circle.

Answer:

Segment of a circle is the region between an arc and chord of the circle.

Fill in the blanks:

Q1 (vi) A circle divides the plane, on which it lies, in _______________ parts.

Answer:

A circle divides the plane, on which it lies, in two parts.

Write True or False: Give reasons for your answers.

Q2 (i) Line segment joining the centre to any point on the circle is a radius of the circle.

Answer:

True. As line segment joining the centre to any point on the circle is a radius of the circle.

Write True or False: Give reasons for your answers.

Q2 (ii) A circle has only finite number of equal chords.

Answer:

False . As a circle has infinite number of equal chords.

Write True or False: Give reasons for your answers.

Q2 (iii) If a circle is divided into three equal arcs, each is a major arc.

Answer:

False. If a circle is divided into three equal arcs, each arc makes angle of 120 degrees whereas major arc makes angle greater than 180 degree at centre.

Write True or False: Give reasons for your answers.

Q2 (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

Answer:

True.A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

Write True or False: Give reasons for your answers.

Q2 (v) Sector is the region between the chord and its corresponding arc.

Answer:

False. As the sector is the region between the radii and arc.

Write True or False: Give reasons for your answers.

Q2 (vi) A circle is a plane figure.

Answer:

True. A circle is a plane figure.

Circles class 9 NCERT solutions - Exercise: 10.2

Q1 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:

Given: The two circles are congruent if they have the same radii.

To prove: The equal chords of congruent circles subtend equal angles at their centres i.e. \angle BAC= \angle QPR

Proof :

1640237118228

In \triangle ABC and \triangle PQR,

BC = QR (Given)

AB = PQ (Radii of congruent circle)

AC = PR (Radii of congruent circle)

Thus, \triangle ABC \cong \triangle PQR (By SSS rule)

\angle BAC= \angle QPR (CPCT)

Q2 Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer:

Given : chords of congruent circles subtend equal angles at their centres,

To prove : BC = QR

Proof : 1651649779761

In \triangle ABC and \triangle PQR,

\angle BAC= \angle QPR (Given)

AB = PQ (Radii of congruent circle)

AC = PR (Radii of congruent circle)

Thus, \triangle ABC \cong \triangle PQR (By SAS rule)

BC = QR (CPCT)

Class 9 circles ncert solutions - Exercise: 10.3

Q1 Draw different pairs of circles. How many points does each pair have in common? What ii the maximum number of common points?

Answer:

1640237144301

In (i) we do not have any common point.

In (ii) we have 1 common point.

In (iii) we have 1 common point.

In (iv) we have 2 common points.

The maximum number of common points is 2.

Q2 Suppose you are given a circle. Give a construction to find its centre.
Answer:

1640237176161

Given : Points P,Q,R lies on circle.

Construction :

1. Join PR and QR

2. Draw perpendicular bisector of PR and QR which intersects at point O.

3. Taking O as centre and OP as radius draw a circle.

4. The circle obtained is required.

Q3 If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Answer:

Given: Two circles intersect at two points.

To prove: their centres lie on the perpendicular bisector of the common chord.

1640237204682

Construction: Joinpoint P and Q to midpoint M of chord AB.

Proof: AB is a chord of circle C(Q,r) and QM is the bisector of chord AB.

\therefore PM\perp AB

Hence, \angle PMA =90 \degree

Similarly, AB is a chord of circle(Q,r' ) and QM is the bisector of chord AB.

\therefore QM\perp AB

Hence, \angle QMA =90 \degree

Now, \angle QMA +\angle PMA=90 \degree+90 \degree= 180 \degree

\angle PMA and \angle QMA are forming linear pairs so PMQ is a straight line.

Hence, P and Q lie on the perpendicular bisector of common chord AB.

NCERT Solutions for Class 9 Maths Chapter 10 Circles - Exercise: 10.4

Q1 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 draw OM\perp AB\, \, and \, \, \, ON\perp CD.

1640237228523

Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.

\therefore PM\perp AB

\angle PMA=90 \degree

Let, PM = x , so QM=4-x

In \triangle APM, using Pythagoras theorem

AM^2=AP^2-PM^2 ...........................1

Also,

In \triangle AQM, using Pythagoras theorem

AM^2=AQ^2-MQ^2 ...........................2

From 1 and 2, we get

AP^2-PM^2=AQ^2-MQ^2

\Rightarrow 3^2-x^2=5^2-(4-x)^2

\Rightarrow 9-x^2=25-16-x^2+8x

\Rightarrow 9=9+8x

\Rightarrow 8x=0

\Rightarrow x=0

Put,x=0 in equation 1

AM^2=3^2-0^2=9

\Rightarrow AM=3

\Rightarrow AB=2AM=6

Q2 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.
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 :

1640237267424

In \triangle OMP and \triangle ONP,

AP = AP (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 1 and 3, we have

AM + PM = CN + PN

\Rightarrow AP = CP

Subtract 4 from 2, we get

AB-AP = CD - CP

\Rightarrow PB = PD

Q3 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

Proof :

Construction: Join OP and draw OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.

In \triangle OMP and \triangle ONP,

AP = AP (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)

Q4 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

Proof :

1640237314606

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 1 from 2, we get

AM-BM = DM - CM

\Rightarrow AB = CD


Q5 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, M are the position of Reshma, Salma, Mandip respectively.

So, RS = SM = 6 cm

Construction : Join OR,OS,RS,RM and OM.Draw OL\perp RS .

Proof:

1640237339300 In \triangle ORS,

OS = OR and OL\perp RS (by construction )

So, RL = LS = 3cm (RS = 6 cm )

In \triangle OLS, by pytagoras theorem,

OL^2=OS^2-SL^2

\Rightarrow OL^2=5^2-3^2=25-9=16

\Rightarrow OL=4

In \triangle ORK and \triangle OMK,

OR = OM (Radii)

\angle ROK = \angle MOK (Equal chords subtend equal angle at centre)

OK = OK (Common)

\triangle ORK \cong \triangle OMK (By SAS)

RK = MK (CPCT)

Thus, OK\perp RM

area of \triangle ORS = \frac{1}{2}\times RS\times OL ...............................1

area of \triangle ORS = \frac{1}{2}\times OS\times KR .............................2

From 1 and 2, we get

\frac{1}{2}\times RS\times OL =\frac{1}{2}\times OS\times KR

\Rightarrow RS\times OL=OS\times KR

\Rightarrow 6\times 4=5\times KR

\Rightarrow KR=4.8 cm

Thus, RM =2 KR=2\times 4.8 cm=9.6 cm

Q6 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, S, D are positioned Ankur, Syed and David respectively.

So, AS = SD = AD

Radius of circular park = 20 m

so, AO=SO=DO=20 m

Construction: AP \perp SD

Proof :

1640237370579

Let AS = SD = AD = 2x cm

In \triangle ASD,

AS = AD and AP \perp SD

So, SP = PD = x cm

In \triangle OPD, by Pythagoras,

OP^2=OD^2-PD^2

\Rightarrow OP^2=20^2-x^2=400-x^2

\Rightarrow OP=\sqrt{400-x^2}

In \triangle APD, by Pythagoras,

AP^2=AD^2-PD^2

\Rightarrow (AO+OP)^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, length of string of each phone = 2x=20\sqrt{3} m

NCERT Solutions for Class 9 Maths Chapter 10 Circles - Exercise: 10.5

Q1 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 .

1640237406418

Answer:

\angle AOC = \angle AOB + \angle BOC= 60 \degree+30 \degree=90 \degree

\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 \degree= 45 \degree

Q2 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=OB.

To find: ADB and \angle ACB.

Solution :

1640237429779 In \triangle OAB,

OA = AB (Given )

OA = OB (Radii of circle)

So, OA=OB=AB

\Rightarrow ABC is a equilateral triangle.

So, \angle AOB = 60 \degree

\angle AOB = 2 \angle ADB

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

\Rightarrow \angle ADB=\frac{1}{2}60 \degree=30

ACBD is a cyclic quadrilateral .

So, \angle ACB+ \angle ADB = 180 \degree

\Rightarrow \angle ACB+30 \degree= 180 \degree

\Rightarrow \angle ACB= 180 \degree-30 \degree=150 \degree

Q3 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 .

1640237467378

Answer:

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, \angle PSR + \angle PQR = 180 \degree

\Rightarrow \angle PSR+100 \degree=180 \degree

\Rightarrow \angle PSR=180 \degree-100 \degree=80 \degree

Here, \angle POR = 2 \angle PSR

\Rightarrow \angle POR=2\times 80 \degree=160 \degree

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 \degree

\Rightarrow 2\angle OPR+160 \degree= 180 \degree

\Rightarrow 2\angle OPR=180 \degree- 160 \degree

\Rightarrow 2\angle OPR=20 \degree

\Rightarrow \angle OPR=10 \degree


Q4 In Fig. \small 10.38 , \small \angle ABC=69^{\circ}, \angle ACB=31^{\circ}, find \small \angle BDC


1640237504699

Answer:

In \triangle ABC,

\angle A+ \angle ABC+ \angle ACB= 180\degree

\Rightarrow \angle A+69 \degree+31 \degree=180\degree

\Rightarrow \angle A+100 \degree=180\degree

\Rightarrow \angle A=180 \degree-100\degree

\Rightarrow \angle A=80 \degree

\angle A = \angle BDC = 80 \degree (Angles in same segment)

Q5 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

1640237534860 .

Answer:

\angle DEC+ \angle BEC = 180 \degree (linear pairs)

\Rightarrow \angle DEC+ 130 \degree = 180 \degree ( \angle BEC = 130 \degree )

\Rightarrow \angle DEC = 180 \degree - 130 \degree

\Rightarrow \angle DEC = 50 \degree

In \triangle DEC,

\angle D+ \angle DEC+ \angle DCE = 180 \degree

\Rightarrow \angle D+50 \degree+20 \degree= 180 \degree

\Rightarrow \angle D+70 \degree= 180 \degree

\Rightarrow \angle D= 180 \degree-70 \degree=110 \degree

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

\angle BAC = 110 \degree


Q6 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:

1640237557595

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

\angle BDC= 30 \degree

In \triangle BDC,

\angle BCD+\angle BDC+\angle DBC= 180 \degree

\Rightarrow \angle BCD+30 \degree+70 \degree= 180 \degree

\Rightarrow \angle BCD+100 \degree= 180 \degree

\Rightarrow \angle BCD=180 \degree- 100 \degree=80 \degree

If AB = BC ,then

\angle BCA=\angle BAC

\Rightarrow \angle BCA=30 \degree

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

\Rightarrow \angle ECD+30 \degree=80 \degree

\Rightarrow \angle ECD=80 \degree-30 \degree=50 \degree


Q7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer:

1640237583763

AC is the diameter of the circle.

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

Similarly, BD is the diameter of the circle.

Thus, \angle BAD=90 \degree and \angle BCD=90 \degree ............................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 \degree

Hence, ABCD is a rectangle.

Q8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer:

1640237665409

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 \degree (linear pair)

\Rightarrow \angle A+\angle C=180 \degree

Thus, ABED is a cyclic quadrilateral.


Q9 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 .

1640237697041

Answer:

1640237709420

\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

Q10 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.

Answer:

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

Construction: Join AD.

1640237735195

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

\angle ADB = 90 \degree ........................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 \degree ........................2(angle in a semi-circle)

From 1 and 2, we have

\angle ADB+ \angle ADC= 90 \degree + 90 \degree = 180 \degree

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

Hence, point D lies on this side.


Q11 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 :

1640237758812

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 same segment are equal)


Q12 Prove that a cyclic parallelogram is a rectangle.
Answer:

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

1640237806403

In cyclic quadrilateral ABCD.

\angle A + \angle C = 180 \degree .......................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 \degree

\Rightarrow 2\angle A = 180 \degree

\Rightarrow \angle A = 90 \degree

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

Hence, ABCD is a rectangle.

NCERT maths chapter 10 class 9 - Exercise: 10.6

Q1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Answer:

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : \angle PAQ = \angle PBQ

Proof : In \triangle APQ and \triangle BPQ,

PA = PB (radii of same circle)

PQ = PQ (Common)

QA = QB (radii of same circle)

So, \triangle APQ \cong \triangle BPQ (By SSS)

\angle PAQ = \angle PBQ (CPCT)


Q2 Two chords AB and CD of lengths \small 5\hspace {1mm}cm and \small 11\hspace {1mm}cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is \small 6\hspace {1mm}cm , find the radius of the circle.

Answer:

Given : AB = 5 cm, CD = 11 cm and AB || CD.

To find Radius (OA).

Construction: Draw OM \perp CD \, \, and \, \, \, ON\perp AB

Proof :

1640237834315

Proof: CD is a chord of circle and OM \perp CD

Thus, CM = MD = 5.5 cm (perpendicular from centre bisects chord)

and AN = NB = 2.5 cm

Let OM be x.

So, ON = 6 - x (MN = 6 cm )

In \triangle OCM , using Pythagoras,

OC ^2=CM^2+OM^2 .............................1

and

In \triangle OAN , using Pythagoras,

OA ^2=AN^2+ON^2 .............................2

From 1 and 2,

CM ^2+OM^2=AN^2+ON^2 (OC=OA =radii)

5.5 ^2+x^2=2.5^2+(6-x)^2

\Rightarrow 30.25+x^2=6.25+36+x^2-12x

\Rightarrow 30.25-42.25=-12x

\Rightarrow -12=-12x

\Rightarrow x=1

From 2, we get

OC^2=5.5^2+1^2=30.25+1=31.25

\Rightarrow OC=\frac{5}{2}\sqrt{5} cm

OA = OC

Thus, the radius of the circle is \frac{5}{2}\sqrt{5} cm

Q3 The lengths of two parallel chords of a circle are \small 6\hspace {1mm}cm and \small 8\hspace {1mm}cm . If the smaller chord is at distance \small 4\hspace {1mm}cm from the centre, what is the distance of the other chord from the centre?

Answer:

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw OM \perp CD \, \, and \, \, \, ON\perp AB

Proof :

1640237861219

Proof: CD is a chord of circle and OM \perp CD

Thus, CM = MD = 3 cm (perpendicular from centre bisects chord)

and AN = NB = 4 cm

Let MN be x.

So, ON = 4 - x (MN = 4 cm )

In \triangle OCM , using Pythagoras,

OC ^2=CM^2+OM^2 .............................1

and

In \triangle OAN , using Pythagoras,

OA ^2=AN^2+ON^2 .............................2

From 1 and 2,

CM ^2+OM^2=AN^2+ON^2 (OC=OA =radii)

\Rightarrow 3 ^2+4^2=4^2+(4-x)^2

\Rightarrow 9+16=16+16+x^2-8x

\Rightarrow 9=16+x^2-8x

\Rightarrow x^2-8x+7=0

\Rightarrow x^2-7x-x+7=0

\Rightarrow x(x-7)-1(x-7)=0

\Rightarrow (x-1)(x-7)=0

\Rightarrow x=1,7

So, x=1 (since x\neq 7> OM )

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

Q4 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that \small \angle ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer:

Given : AD = CE

To prove : \angle ABC = \frac{1}{2}(\angle AOC-\angle DOE)

Construction: Join AC and DE.

Proof :

1640237883312

Let \angle ADC = x , \angle DOE = y and \angle AOD = z

So, \angle EOC = z (each chord subtends equal angle at centre)

\angle AOC + \angle DOE + \angle AOD + \angle EOC = 360 \degree

\Rightarrow x+y+z+z=360 \degree

\Rightarrow x+y+2z=360 \degree .........................................1

In \triangle OAD ,

OA = OD (Radii of the circle)

\angle OAD = \angle ODA (angles opposite to equal sides )

\angle OAD + \angle ODA + \angle AOD = 180 \degree

\Rightarrow 2\angle OAD+z=180 \degree

\Rightarrow 2\angle OAD=180 \degree-z

\Rightarrow \angle OAD=\frac{180 \degree-z}{2}

\Rightarrow \angle OAD=90 \degree-\frac{z}{2} .............................................................2

Similarly,

\Rightarrow \angle OCE=90 \degree-\frac{x}{2} .............................................................3

\Rightarrow \angle OED=90 \degree-\frac{y}{2} ..............................................................4

\angle ODB is exterior of triangle OAD . So,

\angle ODB = \angle OAD + \angle ODA

\Rightarrow \angle ODB=90 \degree-\frac{z}{2}+z (from 2)

\Rightarrow \angle ODB=90 \degree+\frac{z}{2} .................................................................5

similarly,

\angle OBE is exterior of triangle OCE . So,

\angle OBE = \angle OCE + \angle OEC

\Rightarrow \angle OEB=90 \degree-\frac{z}{2}+z (from 3)

\Rightarrow \angle OEB=90 \degree+\frac{z}{2} .................................................................6

From 4,5,6 ;we get

\angle BDE = \angle BED = \angle OEB - \angle OED

\Rightarrow \angle BDE=\angle BED=90 \degree+\frac{z}{2}-(90-\frac{y}{2})=\frac{y+z}{2}

\Rightarrow \angle BDE+\angle BED=y+z ..................................................7

In \triangle BDE ,

\angle DBE + \angle BDE + \angle BED = 180 \degree

\Rightarrow \angle DBE +y+z=180 \degree

\Rightarrow \angle DBE =180 \degree-(y+z)

\Rightarrow \angle ABC =180 \degree-(y+z) ...................................................8

Here, from equation 1,

\frac{x-y}{2}=\frac{360 \degree-y-2x-y}{2}

\Rightarrow \frac{x-y}{2}=\frac{360 \degree-2y-2x}{2}

\Rightarrow \frac{x-y}{2}=180 \degree-y-x ...................................9

From 8 and 9,we have

\angle ABC=\frac{x-y}{2}=\frac{1}{2}(\angle AOC-\angle DOE)

Q5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Answer:

Given : ABCD is a rhombus.

To prove: the circle drawn with AB as diameter passes through the point O.

Proof :

1640237931704

ABCD is rhombus.

Thus, \angle AOC = 90 \degree (diagonals of a rhombus bisect each other at 90 \degree )

So, a circle drawn AB as diameter will pass through point O.

Thus, the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Q6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that \small AE=AD .

Answer:

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove : AE = AD

Proof :

1640237957576

\angle ADC = \angle 3 , \angle ABC = \angle 4, \angle ADE = \angle 1 and \angle AED = \angle 2

\angle 3+\angle 1=180 \degree .................1(linear pair)

\angle 2+\angle 4=180 \degree ....................2(sum of opposite angles of cyclic quadrilateral)

\angle 3 = \angle 4 (oppsoite angles of parallelogram )

From 1 and 2,

\angle 3+ \angle 1 = \angle 2 + \angle 4

From 3, \angle 1 = \angle 2

From 4, \triangle AQB, \angle 1 = \angle 2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

Q7 (i) AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA.

Proof :

1640237982446

In \triangle ABD and \triangle CDO,

AO = OC (Given )

\angle AOB = \angle COD (Vertically opposite angles )

BO = DO (Given )

So, \triangle ABD \cong \triangle CDO (By SAS)

\angle BAO = \angle DCO (CPCT)

\angle BAO and \angle DCO are alternate angle and are equal .

So, AB || DC ..............1

Also AD || BC ...............2

From 1 and 2,

\angle A+\angle C=180 \degree ......................3(sum of opposite angles)

\angle A = \angle C ................................4(Opposite angles of the parallelogram )

From 3 and 4,

\angle A+\angle A=180 \degree

\Rightarrow 2\angle A=180 \degree

\Rightarrow \angle A=90 \degree

BD is a diameter of the circle.

Similarly, AC is a diameter.

Q7 (ii) AC and BD are chords of a circle which bisect each other. Prove that ABCD is a rectangle.

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: ABCD is a rectangle.

Construction : Join AB,BC,CD,DA.

Proof :

1651650039652

ABCD is a parallelogram. (proved in (i))

\angle A=90 \degree (proved in (i))

A parallelogram with one angle 90 \degree , is a rectangle )

Thus, ABCD is rectangle.

Q8 Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are \small 90^{\circ}-\frac{1}{2}C , \small 90^{\circ}-\frac{1}{2}B and \small 90^{\circ}-\frac{1}{2}A

Answer:

Given : Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove : the angles of the triangle DEF are \small 90^{\circ}-\frac{1}{2}C , \small 90^{\circ}-\frac{1}{2}B and \small 90^{\circ}-\frac{1}{2}A

Proof :

1640238007328

\angle 1 and \angle 3 are angles in same segment.therefore,

\angle 1 = \angle 3 ................1(angles in same segment are equal )

and \angle 2 = \angle 4 ..................2

Adding 1 and 2,we have

\angle 1+ \angle 2= \angle 3+ \angle 4

\Rightarrow \angle D=\frac{1}{2}\angle B+\frac{1}{2}\angle C ,

\Rightarrow \angle D=\frac{1}{2}(\angle B+\angle C)

\Rightarrow \angle D=\frac{1}{2}(180 \degree+\angle C)

and \Rightarrow \angle D=\frac{1}{2}(180 \degree-\angle A)

\Rightarrow \angle D=90 \degree-\frac{1}{2}\angle A

Similarly, \Rightarrow \angle E=90 \degree-\frac{1}{2}\angle B and \angle F=90 \degree-\frac{1}{2}\angle C

Q9 Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that \small BP=BQ .

Answer:

Given: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.

To prove : BP = BQ

Proof :

1640238028954

AB is a common chord in both congruent circles.

\therefore \angle APB = \angle AQB

In \triangle BPQ,

\angle APB = \angle AQB

\therefore BQ = BP (Sides opposite to equal of the triangle are equal )

Q10 In any triangle ABC, if the angle bisector of \small \angle A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answer:

Given :In any triangle ABC, if the angle bisector of \small \angle A and perpendicular bisector of BC intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

1640238078439

Let \angle ABD = \angle 1 , \angle ADC = \angle 2 , \angle DCB = \angle 3 , \angle CBD = \angle 4

\angle 1 and \angle 3 lies in same segment.So,

\angle 1 = \angle 3 ..........................1(angles in same segment)

similarly, \angle 2 = \angle 4 ......................2

also, \angle 1= \angle 2 ..............3(given)

From 1,2,3 , we get

\angle 3 = \angle 4

Hence, BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.

Summary Of NCERT Maths Circles Class 9

  • A circle is a closed figure in which all points on the boundary are equidistant from a fixed point called the center of the circle.
  • The distance between the center and any point on the circle is called the radius of the circle.
  • The diameter of a circle is twice the radius.
  • The circumference of a circle is the distance around its boundary and is given by the formula C = 2πr, where r is the radius of the circle and π is the mathematical constant pi (approximately equal to 3.14).
  • The area of a circle is given by the formula A = πr², where r is the radius of the circle.
  • The chord of a circle is a line segment joining any two points on the circle.
  • The diameter is the longest chord of a circle and passes through the center.
  • The tangent to a circle is a line that touches the circle at only one point.
  • A line drawn perpendicular to a tangent at the point of contact is called the normal to the circle at that point.
  • The angle between a tangent and a chord drawn from the point of contact is equal to the angle in the alternate segment.
  • If a line intersects two chords of a circle, then the product of the segments of one chord is equal to the product of the segments of the other chord.
  • The angle subtended by an arc at the center of a circle is double the angle subtended by it at any point on the remaining part of the circle.
  • The angle in a semicircle is a right angle.

Interested students can practice these class 9 maths ch 10 question answer using the exercise solutions provided below.

NCERT Solutions For Class 9 Maths Chapter Wise

Chapter No. Chapter Name
Chapter 1 Number Systems
Chapter 2 Polynomials
Chapter 3 Coordinate Geometry
Chapter 4 Linear Equations In Two Variables
Chapter 5 Introduction to Euclid's Geometry
Chapter 6 Lines And Angles
Chapter 7 Triangles
Chapter 8 Quadrilaterals
Chapter 9 Areas of Parallelograms and Triangles
Chapter 10 Circles
Chapter 11 Constructions
Chapter 12 Heron’s Formula
Chapter 13 Surface Area and Volumes
Chapter 14 Statistics
Chapter 15 Probability

NCERT Solutions For Class 9 Subject Wise

How To Use NCERT Solutions For Class 9 Maths Chapter 10 Circles

  • Learn and memorize some theorems related to circles
  • Learn the application of those theorems in the problems
  • Start applying the theorems and concepts on the practice exercises.
  • During the practice exercises, you can take the help of NCERT solutions for class 9 maths chapter 10 Circles.
  • After doing all the above-written things, you can practice more using the previous year questions papers.

NCERT Books and NCERT Syllabus

Keep working hard & happy learning!

Frequently Asked Question (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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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

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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