NCERT Class 9 Maths Solutions Chapter - 10 Circles, Ex - 10.6

Ex - 10.6

Question 1.  Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution


Let two circles having their centres as O and intersect each other at point A and B respectively. 
Construction: Let us join OO',

Now in AOO'  and BOO'
OA = OB                           (radius of circle 1)
O'A =  O'B                        (radius of circle 2)
OO'  = OO'                        (common)

 AOO'  BOO'         (by SSS congruence rule)
OAO'  = OBO'             (by CPCT)
So, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Question 2.  Two chords AB and CD of lengths 5 cm and 11cm 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

Draw OM  AB and ON  CD. Join OB and OD

                     (Perpendicular from centre bisects the chord)

Let ON be x, so OM will be 6 - x
In MOB

In NOD

We have OB = OD             (radii of same circle)
So, from equation (1) and (2)

From equation (2)

So, radius of circle is found to be  cm.

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 form the centre?

Solution

Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm

In OMB

In OND

OD=OB=5cm             (radii of same circle)

So, distance of bigger chord from centre is 3 cm.

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

In AOD and COE
    OA = OC             (radii of same circle)
    OD = OE             (radii of same circle)
    AD = CE            (given)
 AOD COE         (SSS congruence rule)

OAD = OCE         (by CPCT)        ... (1)
ODA = OEC         (by CPCT)        ... (2)
We also have
OAD = ODA        (As OA = OD)        ... (3)
From equations (1), (2) and (3), we have
OAD = OCE = ODA = OEC
Let OAD = OCE = ODA = OEC = x
In  OAC,
OA = OC
 OCA = OAC         (let a)

In  ODE,
OD = OE
OED = ODE         (let y)
ADEC is a cyclic quadrilateral
 CAD + DEC = 180^o         (opposite angles are supplementary)

x + a + x + y = 180^o
2x + a + y = 180^o
y = 180� - 2x - a                    ... (4)
But DOE = 180� - 2y
And AOC = 180� - 2a
Now, DOE - AOC = 2a - 2y = 2a - 2 (180� - 2x - a)
             = 4a + 4x - 360^o        ... (5)
Now, BAC + CAD = 180�    (Linear pair)
BAC = 180� - CAD = 180� - (a + x)
Similarly, ACB = 180� - (a + x)
Now, in ABC
ABC + BAC + ACB = 180�    (Angle sum property of a triangle)
ABC = 180� - BAC - ACB
= 180� - (180� - a - x) - (180� - a -x)
= 2a + 2x - 180�
=   [4a + 4x - 360^o]
ABC =  [DOE -  AOC]    [Using equation (5)] 

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

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 90^o.
 COD = 90^o

Also in rhombus the diagonals intersect each other at 90^o
AOB = BOC = COD = DOA = 90^o
So, point O has to lie on the circle. 

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 see that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral sum of opposite angles is 180^o
    AEC + CBA = 180^o 
    AEC + AED = 180^o        (linear pair)
    AED = CBA            ... (1) 
    For a parallelogram opposite angles are equal.
    ADE = CBA            ... (2)
    From (1) and (2)
   AED = ADE
    AD = AE            (angles opposite to equal sides of a triangle)

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

Let two chords AB and CD are intersecting each other at point O.
In AOB and COD
OA = OC                         (given)
OB = OD                         (given)
AOB = COD             (vertically opposite angles)
AOB COD          (SAS congruence rule)
AB = CD                        (by CPCT)  
Similarly, we can prove AOD COB
 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
 A = C

But A + C = 180^o      (ABCD is a cyclic quadrilateral)
A + A = 180^o
  A = 180^o
A = 90^o
As ACBD is a parallelogram and one of its interior angles is 90^o, so it is a rectangle.
A is the angle subtended by chord BD. And as A = 90^o, so BD should be diameter of circle. Similarly AC is diameter of circle.

Question 8.  Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90^o 

Solution

It is given that BE is the bisector of B
 ABE =  

But ADE = ABE             (angles in same segment for chord AE)
ADE =  
Similarly, ACF = ADF =      (angle in same segment for chord AF)
Now, D = ADE + ADF

               

Similarly we can prove that

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

AB is common chord in both congruent circles.
 APB = AQB  

Now in BPQ
APB = AQB
 BP = BQ            (angles opposite to equal sides of a triangle)

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

Solution

Let perpendicular bisector of side BC and angle bisector of A 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, BOC and BAC 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.
BOC = 2 BAC = 2A                 ... (1)
In BOE and COE

OE = OE                                   (common)
OB = OC                                  (radii of same circle)

OEB = OEC                       (Each 90^o as OD  BC)

 BOE  COE                (RHS congruence rule)

BOE = COE            (by CPCT)    ... (2)
But BOE + COE = BOC

 BOE +BOE = 2 A        [Using equations (1) and (2)]
BOE = 2A

BOE = A

 BOE = COE = A

The perpendicular bisector of side BC and angle bisector of A meet at point D.

 BOD = BOE = A                ... (3)

Since AD is the bisector of angle A

BAD =  
 2BAD = A                    ... (4)
From equations (3) and (4), we have

BOD = 2 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 A meet on the circum circle of triangle ABC.

Ex-10.5

Ex-11.1