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# Class 9 NCERT Solutions Maths Chapter 9 - Circles

## Circles Exercise Ex. 9.1

### Solution 1

A circle is a collection of points which are equidistant from a fix point. This fix point is called as the centre of circle and this equal distance is called as radius of circle. And thus shape of a circle depends on the radius of the circle.

So, if we try to superimpose two circles of equal radius, one each other both circles will cover each other.
So, two circles are congruent if they have equal radius.

Now consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths

Now in AOB and CO'D
AB = CD            (chords of same length)
OA = O'C            (radii of congruent circles)
OB = O'D            (radii of congruent circles)
AOB  CO'D        (SSS congruence rule)

AOB = CO'D            (by CPCT)
Hence equal chords of congruent circles subtend equal angles at their centres.

### Solution 2

Let us consider two congruent circles (circles of same radius) with centres as O and O'.

In AOB and CO'D
AOB = CO'D        (given)
OA = O'C            (radii of congruent circles)
OB = O'D            (radii of congruent circles)
AOB CO'D        (SSS congruence rule)
AB = CD            (by CPCT)
Hence, if chords of congruent circles subtend equal angles at their centres then chords are equal.

## Circles Exercise Ex. 9.2

### Solution 1

Let radius of circle centered at O and O' be 5 cm and 3 cm respectively.
OA = OB = 5 cm
O'A = O'B = 3 cm
OO' will be the perpendicular bisector of chord AB.
AC = CB

Given that OO' = 4 cm
Let OC be x. so, O'C will be 4 - x
In OAC
OA2 = AC2 + OC2
52 = AC2 + x2
25 - x2 = AC2            ... (1)
In O'AC
O'A2 = AC2 + O'C2
32 = AC2 + (4 - x)2
9 = AC2 + 16 + x2 - 8x
AC2 = - x2 - 7 + 8x        ... (2)
From equations (1) and (2), we have
25 - x2 = - x2 - 7 + 8x
8x = 32
x = 4
So, the common chord will pass through the centre of smaller circle i.e. O'. and hence it will be diameter of smaller circle.

Now, AC2 = 25 - x2 = 25 - 42 = 25 - 16 = 9
AC = 3 m

The length of the common chord AB = 2 AC = (2  3) m = 6 m

### Solution 2

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.

Draw perpendiculars OV and OU on these chords.
In OVT and OUT
OV = OU                                (Equal chords of a circle are equidistant from the centre)
OVT = OUT                     (Each 90o)
OT = OT                                (common)
OVT  OUT              (RHS congruence rule)
VT = UT                             (by CPCT)        ... (1)

It is given that
PQ = RS                                           ... ... ... ... (2)
PV = RU                                    ... ... ...  ... (3)
On adding equations (1) and (3), we have
PV + VT = RU + UT
PT = RT                                    ... ... ...   ... (4)
On subtracting equation (4) from equation (2), we have
PQ - PT = RS - RT
QT = ST                                      ... ... ... ... (5)
Equations (4) and (5) shows that the corresponding segments of
chords PQ and RS are congruent to each other.

### Solution 3

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In OVT and OUT
OV = OU                           (Equal chords of a circle are equidistant from the centre)
OVT = OUT                 (Each 90o)
OT = OT                            (common)
OVT OUT            (RHS congruence rule)
OTV = OTU                 (by CPCT)
Hence, the line joining the point of intersection to the centre makes equal angles with the chords.

### Solution 4

Let us draw a perpendicular OM on line AD.

Here, BC is chord of smaller circle and AD is chord of bigger circle.
We know that the perpendicular drawn from centre of circle bisects the chord.
BM = MC             ... (1)

And AM = MD      ... (2)
Subtracting equations (2) from (1), we have
AM - BM = MD - MC
AB = CD

### Solution 5

Draw perpendiculars OA and OB on RS and SM respectively.
Let R, S and M be the position of Reshma, Salma and Mandip respectively.

AR = AS = = 3cm
OR = OS = OM = 5 m     (radii of circle)
In OAR
OA2 + AR2 = OR2
OA2 + (3 m)2 = (5 m)2
OA2 = (25 - 9) m2 = 16 m2
OA = 4 m
We know that in an isosceles triangle altitude divides the base, so in RSM
RCS will be of 90o and RC = CM
Area of ORS =   OARS

RC = 4.8
RM = 2RC = 2(4.8)= 9.6
So, distance between Reshma and Mandip is 9.6 m.

### Solution 6

Given that AS = SD = DA
So, ASD is a equilateral triangle
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.
We also know that median intersect each other at the 2: 1. As AB is the median of equilateral triangle ABC, we can write

AB = OA + OB = (20 + 10) m = 30 m.

In ABD

So, length of string of each phone will be  m.

## Circles Exercise Ex. 9.3

### Solution 1

We may observe that
AOC = AOB + BOC
= 60o + 30o
= 90o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

### Solution 2

In OAB
AB = OA = OB = radius
OAB is an equilateral triangle.

So, each interior angle of this triangle will be of 60o

AOB = 60o

Now,

ADB = 180o - 30o = 150o
So, angle subtended by this chord at a point on major arc and minor arc are 30o and 150o respectively.

### Solution 3

Consider PR as a chord of circle.
Take any point S on major arc of circle.
Now PQRS is a cyclic quadrilateral.
PQR + PSR = 180o                    (Opposite angles of cyclic quadrilateral)
PSR = 180o - 100o = 80o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.
POR = 2PSR = 2 (80o) = 160o
In POR
OP = OR                                        (radii of same circle)
OPR = ORP                         (Angles opposite equal sides of a triangle)
OPR + ORP + POR = 180o    (Angle sum property of a triangle)
2 OPR + 160o= 180o
2 OPR = 180o - 160o = 20o
OPR = 10o

### Solution 4

In ABC
BAC + ABC + ACB = 180o     (Angle sum property of a triangle)
BAC + 69o + 31o = 180o
BAC = 180o - 100º
BAC = 80o
BDC = BAC = 80o                    (Angles in same segment of circle are equal)

### Solution 5

In CDE
CDE + DCE = CEB        (Exterior angle)
CDE + 20o = 130o
CDE = 110o
But BAC = CDE               (Angles in same segment of circle)
BAC = 110o

### Solution 6

For chord CD
CBD = CAD                    (Angles in same segment)
BCD + 100o = 180o
BCD = 80o
In ABC
AB = BC                               (given)
BCA = CAB               (Angles opposite to equal sides of a triangle)
BCA = 30o
We have BCD = 80o
BCA + ACD = 80o
30oACD = 80o
ACD = 50o
ECD = 50o

### Solution 7

Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.
(Consider BD as a chord)

BCD = 180o - 90o = 90o
(Considering AC as a chord)

90o + ABC = 180o
ABC = 90o
Here, each interior angle of cyclic quadrilateral is of 90o. Hence it is a rectangle.

### Solution 8

Consider a trapezium ABCD with AB | |CD and BC = AD Draw AM  CD and BN CD
In AMD and BNC
AMD = BNC                      (By construction each is 90o)
AM = BM    (Perpendicular distance between two parallel lines is same)
AMD  BNC              (RHS congruence rule)
ADC = BCD                   (CPCT)    ... (1)
BAD + BCD = 180o          [Using equation (1)]
This equation shows that the opposite angles are supplementary.
So, ABCD is a cyclic quadrilateral.

### Solution 9

Join chords AP and DQ
For chord AP
PBA = ACP         (Angles in same segment)        ... (1)
For chord DQ
DBQ = QCD         (Angles in same segment)        ... (2)
ABD and PBQ are line segments intersecting at B.
PBA = DBQ         (Vertically opposite angles)        ... (3)
From equations (1), (2) and (3), we have
ACP = QCD

### Solution 10

Consider a ABC
Two circles are drawn while taking AB and AC as diameter.
Let they intersect each other at D and let D does not lie on BC.
ADB = 90o            (Angle subtend by semicircle)
ADC = 90o            (Angle subtend by semicircle)
Hence BDC is straight line and our assumption was wrong.
Thus, Point D lies on third side BC of ABC

### Solution 11

In ABC
ABC + BCA + CAB = 180o    (Angle sum property of a triangle)
90o + BCA + CAB = 180o
BCA + CAB = 90o        ... (1)
CDA + ACD + DAC = 180o    (Angle sum property of a triangle)
90o + ACD + DAC = 180o
ACD + DAC = 90o        ... (2)
Adding equations (1) and (2), we have
BCA + CAB + ACD + DAC = 180o
(BCA + ACD) + (CAB + DAC) = 180o BCD + DAB = 180o        ... (3)
But it is given that
B + D = 90o + 90o = 180o        ... (4)
From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180o.
So, it is a cyclic quadrilateral.
Consider chord CD.
Now, CAD = CBD                      (Angles in same segment)

### Solution 12

Let ABCD be a cyclic parallelogram.
A + C = 180o     (Opposite angle of cyclic quadrilateral)    ... (1)
We know that opposite angles of a parallelogram are equal
A = C and B = D
From equation (1)
A + C = 180o
A + A = 180o
A = 180o
A = 90o
Parallelogram ABCD is having its one of interior angles as 90o, so, it is a rectangle.