NCERT Solutions for Class 9 Maths Chapter 10 - Circles

Chapter 10 - Circles Exercise Ex. 10.1

Solution 1
(i)   The centre of a circle lies in interior of the circle. (exterior/interior)
(ii)  A point, whose distance from the centre of a circle is greater than its radius lies in exterior of
      the circle. (exterior/interior)
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

Solution 2
(i)  True, all the points on circle are at equal distance from the centre of circle, and this equal distance
 it called as radius of circle.
(ii) False, on a circle there are infinite points. So, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords.
(iii) False, consider three arcs of same length as AB, BC and CA. Now we may observe that for minor arc BDC. CAB is major arc. So AB, BC and CA are minor arcs of circle.
                                 Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
(iv) True, let AB be a chord which is twice as long as its radius. In this situation our chord will be passing through centre of circle. So it will be the diameter of circle.
 
                                  Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
(v) False, sector is the region between an arc and two radii joining the centre to the end points of the arc as in the given figure OAB is the sector of circle.
 
                                       Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
(vi) True, A circle is a two dimensional figure and it can also be referred as plane figure.

Chapter 10 - Circles Exercise Ex. 10.2

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
 
                         Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCO'D
AB = CD            (chords of same length)
OA = O'C            (radii of congruent circles)
OB = O'D            (radii of congruent circles)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCO'D        (SSS congruence rule)

Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCO'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'.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCO'D
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCO'D        (given)
OA = O'C            (radii of congruent circles)
OB = O'D            (radii of congruent circles)
 Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCO'D        (SSS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles AB = CD            (by CPCT)
Hence, if chords of congruent circles subtend equal angles at their centres then chords are equal.

Chapter 10 - Circles Exercise Ex. 10.3

Solution 1
Consider the following pair of circles.
(i) circles don't intersect each other at any point, so circles are not having any point in common.
                    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
(ii) Circles touch each other only at one point P so there is only 1 point in common.
                                 
                     Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles     
 
(iii) Circles touch each other at 1 point X only. So the circles have 1 point in common.     

                                   Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles                   
 

(iv) These circles intersect each other at two points P and Q. So the circles have two points in common. We may observe that there can be maximum 2 points in common.     

                         Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles

We can have a situation in which two congruent circles are superimposed on each other, this situation can be referred as if we are drawing circle two times.

Solution 2
Following are the steps of construction:  

Step1. Take the given circle centered at point O.
Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these
          chords.
Step3. Let these perpendicular bisectors meet at point O. Now, O is the centre of given circle.
 
                    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Solution 3
Consider two circles centered at point O and O' intersect each other at point A and B respectively.
Join AB. AB is the chord for circle centered at O, so perpendicular bisector of AB will pass through O.
Again AB is also chord of circle centered at O', so, perpendicular bisector of AB will also pass through O'.
Clearly centres of these circles lie on the perpendicular bisector of common chord.

Chapter 10 - Circles Exercise Ex. 10.4

Solution 1
 
                      Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles

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.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles AC = CB

    Given that OO' = 4 cm
    Let OC be x. so, O'C will be 4 - x
    In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAC
    OA2 = AC2 + OC2  
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 52 = AC2 + x2
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 25 - x2 = AC2            ... (1)
    In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesO'AC
    O'A2 = AC2 + O'C2
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 32 = AC2 + (4 - x)2
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 9 = AC2 + 16 + x2 - 8x
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 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.   
 
                      Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Now, AC2 = 25 - x2 = 25 - 42 = 25 - 16 = 9
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles AC = 3 m

The length of the common chord AB = 2 AC = (2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 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.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
 
    Draw perpendiculars OV and OU on these chords.
    In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOVT and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOUT
    OV = OU                                (Equal chords of a circle are equidistant from the centre)
   Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOVT = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOUT                     (Each 90o)
   OT = OT                                (common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesOVT Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOUT              (RHS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles VT = UT                             (by CPCT)        ... (1)

    It is given that 
    PQ = RS                                           ... ... ... ... (2)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles PV = RU                                    ... ... ...  ... (3)
    On adding equations (1) and (3), we have
    PV + VT = RU + UT
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles PT = RT                                    ... ... ...   ... (4)
    On subtracting equation (4) from equation (2), we have
    PQ - PT = RS - RT
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles QT = ST                                      ... ... ... ... (5)
    Equations (4) and (5) shows that the corresponding segments of
    chords PQ and RS are congruent to each other.

Solution 3
            Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
     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 Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOVT and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOUT
     OV = OU                           (Equal chords of a circle are equidistant from the centre)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOVT = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOUT                 (Each 90o)
    OT = OT                            (common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesOVT Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOUT            (RHS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesOTV = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOTU                 (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.

                     Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles


    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.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles BM = MC             ... (1)

     And AM = MD      ... (2)
     Subtracting equations (2) from (1), we have
     AM - BM = MD - MC
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 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.
 
 
                                    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
 
   AR = AS = Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles = 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesRSM
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesRCS will be of 90o and RC = CM             
    Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesORS =   Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesOANcert Solutions Cbse Class 9 Mathematics Chapter - CirclesRS    
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
                      RC = 4.8
    RM = 2RC = 2(4.8)= 9.6   
    So, distance between Reshma and Mandip is 9.6 m.
Solution 6
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
 
    Given that AS = SD = DA
             So, ASD is a equilateral triangle
             OA (radius) = 20 m.
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
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
   Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles AB = OA + OB = (20 + 10) m = 30 m.

    In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABD

    AD2 = AB2 + BD2
    AD2 = (30)2 +   Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles

   Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles    
    So, length of string of each phone will be Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles m.

Chapter 10 - Circles Exercise Ex. 10.5

Solution 1
We may observe that
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOC = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOC
        = 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.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Solution 2
                       Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
 
Consider a Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC
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.
 Join AD
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADB = 90o            (Angle subtend by semicircle)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADC = 90o            (Angle subtend by semicircle)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBDC = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADB + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADC = 90o + 90o = 180o
 Hence BDC is straight line and our assumption was wrong.
 Thus, Point D lies on third side BC of Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC

 
 
                              Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Solution 3
                     Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAB = 180o    (Angle sum property of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 90o + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAB = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAB = 90o        ... (1)
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADC
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCDA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDAC = 180o    (Angle sum property of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 90o + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDAC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDAC = 90o        ... (2)
Adding equations (1) and (2), we have
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAB + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDAC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles (Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACD) + (Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAB + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDAC) = 180o Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDAB = 180o        ... (3)
    But it is given that
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesB + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesD = 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, Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCBD                      (Angles in same segment)
 
                                          Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Solution 4
                                       Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Let ABCD be a cyclic parallelogram.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesC = 180o     (Opposite angle of cyclic quadrilateral)    ... (1)
    We know that opposite angles of a parallelogram are equal
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesC and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesB = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesD
    From equation (1)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesA = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA = 90o
Parallelogram ABCD is having its one of interior angles as 90o, so, it is a rectangle.
Solution 5
                           Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAB
    AB = OA = OB = radius
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAB is an equilateral triangle.

So, each interior angle of this triangle will be of 60o
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB = 60o
 
Now,  Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
In cyclic quadrilateral ACBD
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACB + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADB = 180o        (Opposite angle in cyclic quadrilateral)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADB = 180o - 30o = 150o
So, angle subtended by this chord at a point on major arc and minor arc are 30o and 150o respectively.

Solution 6
                               Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 

Consider PR as a chord of circle.
Take any point S on major arc of circle.
Now PQRS is a cyclic quadrilateral.
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesPQR + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesPSR = 180o                    (Opposite angles of cyclic quadrilateral)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesPSR = 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.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesPOR = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesPSR = 2 (80o) = 160o
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesPOR
OP = OR                                        (radii of same circle) 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOPR = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesORP                         (Angles opposite equal sides of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOPR + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesORP + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesPOR = 180o    (Angle sum property of a triangle)
2 Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOPR + 160o= 180o
2 Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOPR = 180o - 160o = 20o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOPR = 10o

Solution 7
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACB = 180o     (Angle sum property of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC + 69o + 31o = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC = 180o - 100º
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC = 80o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBDC = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC = 80o                    (Angles in same segment of circle are equal)
    

Solution 8
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCDE
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCDE + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDCE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCEB        (Exterior angle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCDE + 20o = 130o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCDE = 110o
But Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCDE               (Angles in same segment of circle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC = 110o

Solution 9
                                Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
For chord CD
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCBD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAD                    (Angles in same segment)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAD = 70o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAD = 30o + 70o = 100o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAD = 180o        (Opposite angles of a cyclic quadrilateral)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD + 100o = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD = 80o
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC
AB = BC                               (given)
 Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAB               (Angles opposite to equal sides of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCA = 30o
We have Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD = 80o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACD = 80o
30oNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesACD = 80o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACD = 50o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesECD = 50o
Solution 10
                          Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles    (Consider BD as a chord)
 
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAD = 180o            (Cyclic quadrilateral)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD = 180o - 90o = 90o
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles       (Considering AC as a chord)
 
   Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC = 180o            (Cyclic quadrilateral)
    90o + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC = 180o
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC = 90o
    Here, each interior angle of cyclic quadrilateral is of 90o. Hence it is a rectangle.

Solution 11
                            Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Consider a trapezium ABCD with AB | |CD and BC = AD Draw AM Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles CD and BN Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles CD
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAMD and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBNC
 AD = BC                                 (Given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAMD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBNC                      (By construction each is 90o)
AM = BM    (Perpendicular distance between two parallel lines is same)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesAMD Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles BNC              (RHS congruence rule) 
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesADC = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD                   (CPCT)    ... (1)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAD and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADC are on same side of transversal AD
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADC = 180o                ... (2)    
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBCD = 180o          [Using equation (1)]
This equation shows that the opposite angles are supplementary.
So, ABCD is a cyclic quadrilateral.
Solution 12
 
                                  Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
    Join chords AP and DQ
    For chord AP
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesPBA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACP         (Angles in same segment)        ... (1)  
    For chord DQ
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDBQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesQCD         (Angles in same segment)        ... (2)   
    ABD and PBQ are line segments intersecting at B.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesPBA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDBQ         (Vertically opposite angles)        ... (3)    
    From equations (1), (2) and (3), we have
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACP = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesQCD

Chapter 10 - Circles Exercise Ex. 10.6

Solution 1
                     Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Let two circles having their centres as O and intersect each other at point A and B respectively.
Construction: Let us join OO',
 
                         Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOO'  and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOO'
OA = OB                           (radius of circle 1)
O'A =  O'B                        (radius of circle 2)
OO'  = OO'                        (common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles AOO'  Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOO'         (by SSS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAO'  = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOBO'             (by CPCT)
So, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution 2
                                        Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Let perpendicular bisector of side BC and angle bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA meet at point D.
 Let perpendicular bisector of side BC intersects it at E.

Perpendicular bisector of side BC will pass through circum centre O of circle. Now, Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOC and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively.
We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOC = 2 Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA                 ... (1)
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOE
OE = OE                                   (common)
OB = OC                                  (radii of same circle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOEB = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOEC                       (Each 90o as OD Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles BC)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles COE                (RHS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOE            (by CPCT)    ... (2)
But Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOC
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles BOE +Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles A        [Using equations (1) and (2)]
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA
 Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNcert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA
The perpendicular bisector of side BC and angle bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA meet at point D.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA                ... (3)
Since AD is the bisector of angle Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAD =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles 2Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA                    ... (4)
From equations (3) and (4), we have
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOD = 2Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles BAD
It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.  
Therefore, the perpendicular bisector of side BC and angle bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA meet on the circum circle of triangle ABC.
    

Solution 3
Draw OM Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles AB and ON Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles CD. Join OB and OD
 
                      Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles                     (Perpendicular from centre bisects the chord)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Let ON be x, so OM will be 6 - x
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesMOB
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesNOD
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
We have OB = OD             (radii of same circle)
So, from equation (1) and (2)
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
From equation (2)
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
So, radius of circle is found to be Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles cm.
Solution 4
                                             Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm

Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOMB
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOND
OD=OB=5cm             (radii of same circle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
So, distance of bigger chord from centre is 3 cm.
Solution 5
                                 Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOD and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOE
    OA = OC             (radii of same circle)
    OD = OE             (radii of same circle)
    AD = CE            (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOD Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOE         (SSS congruence rule)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOCE         (by CPCT)        ... (1)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesODA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOEC         (by CPCT)        ... (2)
We also have
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesODA        (As OA = OD)        ... (3)
From equations (1), (2) and (3), we have
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOCE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesODA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOEC
Let Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOCE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesODA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOEC = x
In Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles OAC,
OA = OC
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOCA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOAC         (let a)
In Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles ODE,
OD = OE
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesOED = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesODE         (let y)
ADEC is a cyclic quadrilateral
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAD + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDEC = 180o         (opposite angles are supplementary)
x + a + x + y = 180o
2x + a + y = 180o
y = 180 - 2x - a                    ... (4)
But Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDOE = 180 - 2y
And Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOC = 180 - 2a
Now, Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDOE - Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOC = 2a - 2y = 2a - 2 (180 - 2x - a)
             = 4a + 4x - 360o        ... (5)
Now, Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAD = 180    (Linear pair)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC = 180 - Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCAD = 180 - (a + x)
Similarly, Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACB = 180 - (a + x)
Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACB = 180    (Angle sum property of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC = 180 - Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBAC - Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACB
= 180 - (180 - a - x) - (180 - a -x)
= 2a + 2x - 180
= Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles  [4a + 4x - 360o]
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABC = Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles [Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDOE -  Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOC]    [Using equation (5)]

Solution 6
                              Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn taking side CD as its diameter.
We know that angle in a semicircle is of 90o.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOD = 90o

Also in rhombus the diagonals intersect each other at 90o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBOC = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesDOA = 90o
So, point O has to lie on the circle.  

Solution 7
                               Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
We see that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral sum of opposite angles is 180o
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAEC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCBA = 180o
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAEC + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAED = 180o        (linear pair)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAED = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCBA            ... (1)
    For a parallelogram opposite angles are equal.
    Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCBA            ... (2)
    From (1) and (2)
   Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAED = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADE
    AD = AE            (angles opposite to equal sides of a triangle)

Solution 8
                                      Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Let two chords AB and CD are intersecting each other at point O.
In Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB and Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOD
OA = OC                         (given)
OB = OD                         (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOD             (vertically opposite angles)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOB Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOD          (SAS congruence rule)
AB = CD                        (by CPCT)    
Similarly, we can prove Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAOD Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesCOB
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles AD = CB                     (by CPCT)
 
Since in quadrilateral ACBD opposite sides are equal in length.
Hence, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesC
But Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesC = 180o      (ABCD is a cyclic quadrilateral)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA = 180o
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles A = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA = 90o
As ACBD is a parallelogram and one of its interior angles is 90o, so it is a rectangle.
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA is the angle subtended by chord BD. And as Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesA = 90o, so BD should be diameter of circle. Similarly AC is diameter of circle.

Solution 9
                              Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 

It is given that BE is the bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesB
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABE =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles

But Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADE = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesABE             (angles in same segment for chord AE)
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADE =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Similarly, Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesACF = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADF =  Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles    (angle in same segment for chord AF)
Now, Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesD = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADE + Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesADF
               Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Similarly we can prove that
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
Solution 10
                                  Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles
 
AB is common chord in both congruent circles.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAPB = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAQB  

Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesBPQ
Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAPB = Ncert Solutions Cbse Class 9 Mathematics Chapter - CirclesAQB  
Ncert Solutions Cbse Class 9 Mathematics Chapter - Circles BP = BQ            (angles opposite to equal sides of a triangle)