The National Council of Educational Research and Training (NCERT) is an autonomous body of the Indian government that formulates the curricula for schools in India that are governed by the Central Board of Secondary Education (CBSE) and certain state boards. Therefore, students who will be taking the Class 10 tests administered by various boards should consult this NCERT Syllabus in order to prepare for those examinations, which in turn will assist those students get a passing score.

When working through the exercises in the NCERT textbook, if you run into any type of difficulty or uncertainty, you may use the swc NCERT Solutions for class 9 as a point of reference. While you are reading the theory form textbook, it is imperative that you always have notes prepared. You should make an effort to understand things from the very beginning so that you may create a solid foundation in the topic. Use the NCERT as your parent book to ensure that you have a strong foundation. After you have finished reading the theoretical section of the textbook, you should go to additional reference books.

NCERT Solutions for Class 9 Maths Exercise 10.1

Question 1. Fill in the blanks:

(i) The centre of a circle lies in _______________ of the circle.

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in _______________ of the circle.

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

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

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

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

Solution:
(i) Interior

(ii) Exterior

(iii) diameter

(iv) Semi-circle

(v) Chord

(vi) Three

Question 2. Write True or False:

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

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

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

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

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

(vi) A circle is a plane figure.

Solution:
(i) True

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

NCERT Solutions for Class 9 Maths Exercise 10.2

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
Solution:
Given: Two congruent circles with centres O and O’ and radii r, which have chords AB and CD respectively such that AB = CD.
To Prove: ∠AOB = ∠CO’D
Proof: In ∆AOB and ∆CO’D, we have
AB = CD [Given]
OA = O’C [Each equal to r]
OB = O’D [Each equal to r]
∴ ∆AOB ≅ ∆CO’D [By SSS congruence criteria]
⇒ ∠AOB = ∠CO’D [C.P.C.T.]
NCERT Solutions for Class 9 Maths Chapter-10 Circles/A1

Question 2. Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:
Given: Two congruent circles with centres O & O’ and radii r which have chords AB and CD respectively such that ∠AOB = ∠CO’D.
NCERT Solutions for Class 9 Maths Chapter-10 Circles/A2
To Prove: AB = CD
Proof: In ∆AOB and ∆CO’D, we have
OA = O’C [Each equal to r]
OB = O’D [Each equal to r]
∠AOB = ∠CO’D [Given]
∴ ∆AOB ≅ ∆CO’D [By SAS congruence criteria]
Hence, AB = CD [C.P.C.T.]

NCERT Solutions for Class 9 Maths Exercise 10.3

Question 1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
Let us draw different pairs of circles as shown below:
NCERT Solutions for Class 9 Maths Chapter-10 Circles/A1
We have

NCERT Solutions for Class 9 Maths Chapter-10 Circles

Thus, two circles can have at the most two points in common.

Question 2. Suppose you are given a circle. Give a construction to find its centre.
Solution:
Steps of construction :
Step I : Take any three points on the given circle. Let these points be A, B and C.
Step II : Join AB and BC.
Step III : Draw the perpendicular bisector, PQ of AB.
Step IV: Draw the perpendicular bisector, RS of BC such that it intersects PQ at O.
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image004.jpg
Thus, ‘O’ is the required centre of the given drcle.

Question 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution:
We have two circles with centres O and O’, intersecting at A and B.
∴ AB is the common chord of two circles and OO’ is the line segment joining their centres.
Let OO’ and AB intersect each other at M.
NCERT Solutions for Class 9 Maths Chapter-10 Circles/A3
∴ To prove that OO’ is the perpendicular bisector of AB,
we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’,
we have
OA = OB [Radii of the same circle]
O’A = O’B [Radii of the same circle]
OO’ = OO’ [Common]
∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence criteria]
⇒ ∠1 = ∠2 , [C.P.C.T.]
Now, in ∆AOM and ∆BOM, we have
OA = OB [Radii of the same circle]
OM = OM [Common]
∠1 = ∠2 [Proved above]
∴ ∆AOM = ∆BOM [By SAS congruence criteria]
⇒ ∠3 = ∠4 [C.P.C.T.]
But ∠3 + ∠4 = 180° [Linear pair]
∴∠3=∠4 = 90°
⇒ AM ⊥ OO’
Also, AM = BM [C.P.C.T.]
⇒ M is the mid-point of AB.
Thus, OO’ is the perpendicular bisector of AB.

NCERT Solutions for Class 9 Maths Exercise 10.4

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

Solution:
Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.jpg

Radius OA = 5 cm, Radius O’A = 3 cm,

Distance between their centers OO’ = 4 cm

In triangle AOO’,

52 = 42 + 32

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png25 = 16 + 9

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png 25 = 25

Hence AOO’ is a right triangle, right angled at O’.

Since, perpendicular drawn from the center of the circle bisects the chord.

Hence O’ is the mid-point of the chord AB. Also O’ is the centre of the circle II.

Therefore length of chord AB = Diameter of circle II

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png Length of chord AB = 2 x 3 = 6 cm.

Question 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.

Solution:
Given: Let AB and CD are two equal chords of a circle of centers O intersecting each other at point E within the circle.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image004.jpg

To prove: (a) AE = CE

(b) BE = DE

Construction: Draw OM NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png AB, ON NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png CD. Also join OE.

Proof: In right triangles OME and ONE,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngOME = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngONE = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image007.png

OM = ON

[Equal chords are equidistance from the centre]

OE = OE [Common]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngOMEONE [RHS rule of congruency]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png ME = NE [By CPCT] ……….(i)

Now, O is the centre of circle and OM NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png AB

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png AM = AB

[Perpendicular from the centre bisects the chord] …..(ii)

Similarly, NC = CD ……….(iii)

But AB = CD [Given]

From eq. (ii) and (iii), AM = NC ……….(iv)

Also MB = DN …….…(v)

Adding (i) and (iv), we get,

AM + ME = NC + NE

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.pngAE = CE [Proved part (a)]

Now AB = CD [Given]

AE = CE [Proved]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.pngAB – AE = CD – CE

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.pngBE = DE [Proved part (b)]

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 chord.

Solution:
Given: AB and CD be two equal chords of a circle with centre O intersecting each other with in the circle at point E. OE is joined.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image011.jpg

To prove: NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngOEM = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngOEN

Construction: Draw OM NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png AB and

ON NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png CD.

Proof: In right angled triangles OME and ONE,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngOME = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngONE [Each NCERT Solutions for Class 9 Maths Chapter-10 Circles/image007.png]

OM = ON [Equal chords are equidistant from the centre]

OE = OE [Common]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngOMENCERT Solutions for Class 9 Maths Chapter-10 Circles/image009.pngONE [RHS rule of congruency]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngOEM = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngOEN [By 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 AB = CD. (See figure)

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.jpg

Solution:
Given: Line NCERT Solutions for Class 9 Maths Chapter-10 Circles/image013.png intersects two concentric circles with centre O at points A, B, C and D.

To prove: AB = CD

Construction: Draw OLNCERT Solutions for Class 9 Maths Chapter-10 Circles/image014.png

Proof: AD is a chord of outer circle and OLNCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png AD.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png AL = LD ………(i) [Perpendicular drawn from the centre bisects the chord]

Now,   BC is a chord of inner circle and

OLNCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png BC

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png BL = LC …(ii) [Perpendicular drawn from the centre bisects the chord]

Subtracting (ii) from (i), we get,

AL – BL = LD – LC

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.pngAB = CD

Question 5. Three girls Reshma, Salma and Mandip are standing on a circle of radius 5 m 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 6 m each, what is the distance between Reshma and Mandip?

Solution:
Let Reshma, Salma and Mandip takes the position C, A and B on the circle. Since AB = AC

The centre lies on the bisector of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngBAC.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image015.jpg

Let M be the point of intersection of BC and OA.

Again, since AB = AC and AM bisects

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.pngCAB.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png AM NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngCB and M is the mid-point of CB.

Let OM = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image016.png then MA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image017.png

From right angled triangle OMB,

OB2 = OM2 + MB2

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png52 = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image018.png + MB……….(i)

Again, in right angled triangle AMB,

AB2 = AM2 + MB2

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png62 = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image019.png + MB2 ……….(ii)

Equating the value of MB2 from eq. (i) and (ii),

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image020.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image021.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image022.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image023.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image024.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image025.png

Hence, from eq. (i),

MB2 = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image026.png = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image027.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image028.png = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image029.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png MB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image030.png = 4.8 cm

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png BC = 2MB = 2 x 4.8 = 9.6 cm

Question 6. A circular park of radius 20 m 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.

Solution:
Let position of three boys Ankur, Syed and David are denoted by the points A, B and C respectively.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image031.jpg

A = B = C = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image032.png [say]

Since equal sides of equilateral triangle are as equal chords and perpendicular distances of equal chords of a circle are equidistant from the centre.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png OD = OE = OF = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image033.png cm [say]

Join OA, OB and OC.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngAOB

= Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngBOC = Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngAOC

And Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngABC

= Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngAOB + Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngBOC + Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngAOC

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.pngAnd Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngABC = 3 x Area of NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngBOC

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png NCERT Solutions for Class 9 Maths Chapter-10 Circles/image034.png = 3 (NCERT Solutions for Class 9 Maths Chapter-10 Circles/image010.pngBC x OE)

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png NCERT Solutions for Class 9 Maths Chapter-10 Circles/image034.png = 3 (NCERT Solutions for Class 9 Maths Chapter-10 Circles/image010.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image035.png )

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image036.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image037.png ……….(i)

Now, CE NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png BC

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image003.png BE = EC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image010.pngBC    [NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png Perpendicular drawn from the centre bisects the chord]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.pngBE = EC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image039.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png BE = EC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image040.png [Using eq. (i)]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.pngBE = EC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image041.png

Now in right angled triangle BEO,

OE2 + BE2 = OB2 [Using Pythagoras theorem]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image042.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image043.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image044.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image045.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image002.png NCERT Solutions for Class 9 Maths Chapter-10 Circles/image033.png = 10 m

And NCERT Solutions for Class 9 Maths Chapter-10 Circles/image037.png = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image046.png = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image047.png m

Thus distance between any two boys is NCERT Solutions for Class 9 Maths Chapter-10 Circles/image047.pngm.

NCERT Solutions for Class 9 Maths Exercise 10.5

Question 1. In figure, A, B, C are three points on a circle with centre O such that BOC = AOB = If D is a point on the circle other than the arc ABC, find ADC.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image004.jpg

Solution:
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOB + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBOC

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image006.png

Now NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOC = 2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngADC

[NCERT Solutions for Class 9 Maths Chapter-10 Circles/image007.pngAngled subtended by an arc, at the centre of the circle is double the angle subtended by the same arc at any point in the remaining part of the circle]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngADC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOC

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngADC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image009.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image010.png

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

Solution:
Let AB be the minor arc of circle.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image011.jpg

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngChord AB = Radius OA = Radius OB

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image013.pngAOB is an equilateral triangle.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image014.png

Now NCERT Solutions for Class 9 Maths Chapter-10 Circles/image015.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOB + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBOA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image016.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBOA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image016.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image017.jpg

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBOA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image018.png

D is a point in the minor arc.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image019.png= 2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBDA

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBOA = 2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBDA

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBDA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBOA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image020.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBDA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image021.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image022.jpg

Thus angle subtended by major arc, NCERT Solutions for Class 9 Maths Chapter-10 Circles/image023.pngat any point D in the minor arc is NCERT Solutions for Class 9 Maths Chapter-10 Circles/image024.png

Let E be a point in the major arc NCERT Solutions for Class 9 Maths Chapter-10 Circles/image025.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image026.png= 2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAEB

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOB = 2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAEB

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAEB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image008.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAOB

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngAEB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image027.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image028.png

Question 3. In figure, PQR = where P, Q, R are points on a circle with centre O. Find OPR.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image030.jpg

Solution: 
In the figure, Q is a point in the minor arc NCERT Solutions for Class 9 Maths Chapter-10 Circles/image031.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image032.png= 2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPQR

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngROP = 2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPQR

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngROP = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image033.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image034.png

Now NCERT Solutions for Class 9 Maths Chapter-10 Circles/image035.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPOR + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngROP = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image016.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPOR + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image034.png=

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPOR = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image036.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image037.png…..(i)

Now NCERT Solutions for Class 9 Maths Chapter-10 Circles/image013.pngOPR is an isosceles triangle.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngOP = OR [radii of the circle]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngOPR = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngORP [angles opposite to equal sides are equal] …..(ii)

Now in isosceles triangle OPR,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngOPR + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngORP + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPOR = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngOPR + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngORP + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image037.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngOPR =NCERT Solutions for Class 9 Maths Chapter-10 Circles/image039.png[Using (i) & (ii)]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.png2NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngOPR = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image040.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngOPR = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image041.png

Question 4. In figure, ABC = ACB = find BDC.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image044.jpg

Solution: 
In triangle ABC,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBAC + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngABC + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngACB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBAC + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image045.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBAC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image046.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBAC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image047.png…….(i)

Since, A and D are the points in the same segment of the circle.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBDC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBAC

[Angles subtended by the same arc at any points in the alternate segment of a circle are equal]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBDC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image047.png[Using (i)]

Question 5. In figure, A, B, C, D are four points on a circle. AC and BD intersect at a point E such that BEC = and

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngECD = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image048.pngFind NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBAC.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image050.jpg

Solution: 
Given: NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBEC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image048.pngand NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngECD = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image040.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDEC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image051.pngBEC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image052.png[Linear pair]

Now in NCERT Solutions for Class 9 Maths Chapter-10 Circles/image013.pngDEC,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDEC + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDCE + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngEDC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png[Angle sum property]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image053.pngEDC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngEDC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image054.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngBAC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngEDC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image054.png[Angles in same segment]

Question 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. DBC = BAC is findBCD. Further if AB = BC, find ECD.

Solution:
For chord CD

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image056.jpg

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image057.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image058.png(Angles in same segment)

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image059.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image060.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image061.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image062.png(Opposite angles of a cyclic quadrilateral)
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image057.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image057.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image064.png

AB = BC (given)
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image065.png(Angles opposite to equal sides of a triangle)
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image066.png
We haveNCERT Solutions for Class 9 Maths Chapter-10 Circles/image057.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image067.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image068.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image069.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image070.png

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

Solution: 
Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image071.jpg
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image072.png

(Consider BD as a chord)NCERT Solutions for Class 9 Maths Chapter-10 Circles/image073.png(Cyclic quadrilateral)

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image074.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image075.png

(Considering AC as a chord)

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image076.png(Cyclic quadrilateral)
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image077.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image078.png
Here, each interior angle of cyclic quadrilateral is ofNCERT Solutions for Class 9 Maths Chapter-10 Circles/image079.png. Hence it is a rectangle.

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

Solution:
Given: A trapezium ABCD in which ABNCERT Solutions for Class 9 Maths Chapter-10 Circles/image080.pngCD and AD = BC.

To prove: The points A, B, C, D are concyclic.

Construction: Draw DENCERT Solutions for Class 9 Maths Chapter-10 Circles/image080.pngCB.

Proof: Since DENCERT Solutions for Class 9 Maths Chapter-10 Circles/image080.pngCB and EBNCERT Solutions for Class 9 Maths Chapter-10 Circles/image080.pngDC.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image081.jpg

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngEBCD is a parallelogram.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngDE = CB and NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDEB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDCB

Now AD = BC and DA = DE

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDAE = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDEB

But NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDEA + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDEB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDAE + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDCB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

[NCERT Solutions for Class 9 Maths Chapter-10 Circles/image007.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDEA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDAE and NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDEB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDCB]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDAB + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngDCB = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngA + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngC = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

Hence, ABCD is a cyclic trapezium.

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, P, Q respectively (see figure). Prove that ACP = QCD.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image082.jpg

Solution:;
In triangles ACD and QCP,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngA = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngP and NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngQ = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngD [Angles in same segment]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngACD = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngQCP [Third angles] ……….(i)

Subtracting NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPCD from both the sides of eq. (i), we get,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngACD – NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPCD = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngQCP – NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPCD

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngACPO = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngQCD

Hence proved.

Question 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.

Solution:
Given: Two circles intersect each other at points A and B. AP and AQ be their respective diameters.

To prove: Point B lies on the third side PQ.

Construction : Join A and B.

Proof: AP is a diameter.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image083.jpg

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.png1 = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image084.png

[Angle in semicircle]

AlsoAQ is a diameter.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.png2 = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image084.png

[Angle in semicircle]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.png1 + NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.png2 = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image085.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngPBQ = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image038.png

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngPBQ is a line.

Thus point B. i.e. point of intersection of these circles lies on the third side i.e., on PQ.

Question 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that CAD = ABD.

Solution: 
We have ABC and ADC two right triangles, right angled at B and D respectively.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image086.jpg

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image005.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngABC = ADC [Each NCERT Solutions for Class 9 Maths Chapter-10 Circles/image084.png]

If we draw a circle with AC (the common hypotenuse) as diameter, this circle will definitely passes through of an arc AC, Because B and D are the points in the alternate segment of an arc AC.

Now we have NCERT Solutions for Class 9 Maths Chapter-10 Circles/image087.pngsubtending NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngCBD and NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngCAD in the same segment.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image012.pngNCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngCAD = NCERT Solutions for Class 9 Maths Chapter-10 Circles/image001.pngCBD

Hence proved.

NCERT Solutions for Class 9 Maths Exercise 10.6

Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given : Two circles with centres O and O’ respectively such that they intersect each other at P and Q.
To Prove: ∠OPO’ = ∠OQO’.
Construction : Join OP, O’P, OQ, O’Q and OO’.
Proof: In ∆OPO’ and ∆OQO’, we have
NCERT Solutions for Class 9 Maths Chapter-10 Circles/ A1
OP = OQ [Radii of the same circle]
O’P = O’Q [Radii of the same circle]
OO’ = OO’ [Common]
∴ AOPO’ = AOQO’ [By SSS congruence criteria]
⇒ ∠OPO’ = ∠OQO’ [C.P.C.T.]

Question 2. Two chords AB and CD of lengths 5 cm and 11 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 6 cm, find the radius of the circle.
Solution:
We have a circle with centre O.
AB || CD and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm.
NCERT Solutions for Class 9 Maths Chapter-10 Circles/ A2
Let r cm be the radius of the circle.
Let us draw OP ⊥ AB and OQ ⊥ CD such that
PQ = 6 cm
Join OA and OC.
Let OQ = x cm
∴ OP = (6 – x) cm
∵ The perpendicular drawn from the centre of a circle to chord bisects the chord.
NCERT Solutions for Class 9 Maths Chapter-10 Circles/ A2a

Question 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ?
Solution:
We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre.
NCERT Solutions for Class 9 Maths Chapter-10 Circles/ A3
Let r cm be the radius of the circle and draw OP ⊥ AB and join OA and OC.
∵ OP ⊥ AB
∴ P is the mid-point of AB.
⇒ AP = NCERT Solutions for Class 9 Maths Chapter-10 CirclesAB\frac { 1 }{ 2 }(6 cm)= 3 cm
Similarly, CQ = NCERT Solutions for Class 9 Maths Chapter-10 CirclesNCERT Solutions for Class 9 Maths Chapter-10 Circles(8 cm)= 4 cm
Now in ∆OPA, we have OA2 = OP2 + AP2
⇒ r2 = 42 + 32
⇒ r2 = 16 + 9 = 25
⇒ r = NCERT Solutions for Class 9 Maths Chapter-10 Circles=5
Again, in ∆CQO, we have OC2 = OQ2 + CQ2
⇒ r2 = OQ2 + 42
⇒ OQ2 = r2 – 42
⇒ OQ2 = 52 – 42 = 25 – 16 = 9 [∵ r = 5]
⇒ OQ
⇒ √9 = 3
The distance of the other chord (CD) from the centre is 3 cm.
Note: In case if we take the two parallel chords on either side of the centre, then
NCERT Solutions for Class 9 Maths Chapter-10 Circles/ A3A
In ∆POA, OA= OP2 + PA2
⇒ r2 = 42 + 32 = 52
⇒ r = 5
In ∆QOC, OC2 = CQ2 + OQ2
⇒ OQ2 = 42 + OQ2
⇒ OQ= 52 – 42 = 9
⇒ OQ = 3

Question 4. 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 ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
Given : ∠ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE.
To prove: ∠ABC = NCERT Solutions for Class 9 Maths Chapter-10 Circles[∠DOE – ∠AOC]
Construction : Join AE.
Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles.
∴ In ∆BAE, we have
∠DAE = ∠ABC + ∠AEC ……(i)
The chord DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.
NCERT Solutions for Class 9 Maths Chapter-10 Circles/ A4
⇒ ∠ABC = NCERT Solutions for Class 9 Maths Chapter-10 Circles[(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)]
⇒ ∠ABC = NCERT Solutions for Class 9 Maths Chapter-10 Circles[Difference of the angles subtended by the chords DE and AC at the centre]

Question 5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Solution:
We have a rhombus ABCD such that its diagonals AC and BD intersect at O.
Taking AB as diameter, a circle is drawn. Let us draw PQ || DA and RS || AB, both are passing through O.
P, Q, R and S are the mid-points of DC, AB, AD and BC respectively,
∵ Q is the mid-point of AB.
⇒ AQ = QB …(i)
Since AD = BC [ ∵ ABCD is a rhombus]
∴ NCERT Solutions for Class 9 Maths Chapter-10 CirclesAD = NCERT Solutions for Class 9 Maths Chapter-10 CirclesBC
⇒ RA = SB
⇒ RA = OQ …(ii)
NCERT Solutions for Class 9 Maths Chapter-10 Circles/ A5
[ ∵ PQ is drawn parallel to AD and AD = BC]
We have, AB = AD [Sides of rhombus are equal]
⇒ NCERT Solutions for Class 9 Maths Chapter-10 CirclesAB = NCERT Solutions for Class 9 Maths Chapter-10 CirclesAD
⇒ AQ = AR …(iii)
From (i), (ii) and (iii), we have AQ = QB = OQ
i.e. A circle drawn with Q as centre, will pass through A, B and O.
Thus, the circle passes through the of intersection ‘O’ of the diagonals rhombus ABCD.

Question 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
We have a circle passing through A, B and C is drawn such that it intersects CD at E.
ABCE is a cyclic quadrilateral.
∴∠AEC + ∠B = 180° …(i)
[Opposite angles of a cyclic quadrilateral are supplementary] But ABCD is a parallelogram. [Given]
∴∠D = ∠B …(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
∠AEC + ∠D = 180° …(iii)
But ∠AEC + ∠AED = 180° [Linear pair] …(iv)
NCERT Solutions for Class 9 Maths Chapter-10 Circles/ A6
From (iii) and (iv), we have ∠D = ∠AED
i.e., The base angles of AADE are equal.
∴ Opposite sides must be equal.
⇒ AE = AD

Question 7. AC and BD are chords of a circle which bisect each other. Prove that:

(i) AC and BD are diameters.

(ii) ABCD is a rectangle.

Solution:
Given: AC and BD of a circle bisect each other at O.

Then OA = OC and OB = OD

To prove: (i) AC and BD are the diameters. In other words, O is the centre of the circle.

(ii) ABCD is a rectangle.

Proof: (i) In triangles AOD and BOC,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image039.jpg

AO = OC [given]

AOD = BOC [Vertically opp.]

OD = OB [given]

AODCOB [SAS congruency]

AD = CB [By CPCT]

Similarly AOBCOD

AB = CD

[Arcs opposite to equal chords]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image007.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image041.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image007.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image042.png

AC = BD [Chords opposites to equal arcs]

AC and BD are the diameters as only diameters can bisects each other as the chords of the circle.

(ii) Ac is the diameter. [Proved in (i)]

B = D = ….(i) [Angle in semi-circle]

Similarly BD is the diameter.

A = C = …(ii) [Angle in semi-circle]

Now diameters AC = BD

[Arcs opposite to equal chords]

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image007.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image044.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image007.png
NCERT Solutions for Class 9 Maths Chapter-10 Circles/image045.png

AD = BC [Chords corresponding to the equal arcs] ….(iii)

Similarly AB = DC ….(iv)

From eq. (i), (ii), (iii) and (iv), we observe that each angle of the quadrilateral is and opposite sides are equal.

Hence ABCD is a rectangle.

Question 8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that angles of the triangle are and respectively.

Solution:
According to question, AD is bisector of A.

1 = 2 = 

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image049.jpg

And BE is the bisector of B.

3 = 4 = 

Also CF is the bisector of C.

5 = 6 = 

Since the angles in the same segment of a circle are equal.

9 = 3 [angles subtended by ] ….(i)

And 8 = 5 [angles subtended by ] ….(ii)

Adding both equations,

9 + 8 = 3 + 5

D = 

Similarly E = 

And F = 

In triangle DEF,

D + E + F = 

D = (E + F )

D = 

D = 

D = [A + B + C = ]

D = 

Similarly, we can prove that

E = and F = 

Question 9. 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 BP = BQ.

Solution: 
Given: Two equal circles intersect in A and B.

A straight line through A meets the circles in P and Q.

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image065.jpg

To prove: BP = BQ

Construction: Join A and B.

Proof: AB is a common chord and the circles are equal.

Arc about the common chord are equal, i.e.,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image066.png

Since equal arcs of two equal circles subtend equal angles at any point on the remaining part of the circle, then we have,

1 = 2

In triangle PBQ,

1 = 2 [proved]

Sides opposite to equal angles of a triangle are equal.

Then we have, BP = BQ

Question 10. In any triangle ABC, if the angle bisector of and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.

Solution:
Given: ABC is a triangle and a circle passes through its vertices.

Angle bisector of A and the perpendicular bisector (say ) of its opposite side BC intersect each other at a point P.

To prove: Circumcircle of triangle ABC also passes through point P.

Proof: Since any point on the perpendicular bisector is equidistant from the end points of the corresponding side,

NCERT Solutions for Class 9 Maths Chapter-10 Circles/image069.jpg

BP = PC ….(i)

Also we have 1 = 2 [AP is the bisector of A (given)] ….(ii)

From eq. (i) and (ii) we observe that equal line segments are subtending equal angles in the same segment i.e., at point A of circumcircle of ABC. Therefore BP and PC acts as chords of circumcircle of ABC and the corresponding congruent arcs and acts as parts of circumcircle. Hence point P lies on the circumcircle. In other words, points A, B, P and C are concyclic (proved).

Conclusions for NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 10 CIRCLES

SWC academic staff has developed NCERT answers for this chapter of the ninth grade mathematics curriculum. We have solutions prepared for all of the exercises of this chapter. The answers, broken down into steps, to all of the questions included in the NCERT textbook’s chapter are provided here. Read this chapter on theory. Be certain that you have read the theory section of this chapter of the NCERT textbook and that you have learnt the formulas for the chapter that you are studying. 

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