SELINA Solution Class 9 Chapter 17 Circle Exercise 17D

Question 1

The radius of a circle is 13 cm and the length of one of its chords is 24 cm.
Find the distance of the chord from the center.

Sol:

To find: OM

Given that AB = 24 cm
Since OM ⊥ AB
⇒ OM bisects AB
So, AM = 12 cm
In right ⇒ OMA,
OA² = OM²  + AM²
OM² = OA² - AM²
OM² = 13² - 12²
OM² = 25
OM² = 5 cm
Hence, the distance of the chord from the centre is 5 cm.

Question 2

Prove that equal chords of congruent circles subtend equal angles at their center.

Sol:

Given: AB and CD are two equal chords of a congruent circle with centres O and O respectively.

To prove: ∠AOB = ∠CO'D

Proof:
In ΔOAB and ΔO'CD
OA = O'C                            ...[ ∵ Radii of congruent circles ]
OB = O'D                            ...[ ∵ Radii of congruent circles ]
AB = CD                              ...[ Given ] 
ΔOAB ≅ ΔO'CD                   ...[ By SSS congruence criterion ]
∠AOB = ∠CO'D                   ...[ c. p. c. t ]

Question 3

Draw two circles of different radii. How many points these circles can have in common? What is the maximum number of common points?

Sol:

So, the circle can have 0, 1 or 2 points in common.
The maximum number of common points is 2. 

Question 4

Suppose you are given a circle. Describe a method by which you can find the center of this circle.

Sol:

To draw the center of a given circle : 
1. Draw the circle.
2. Take any two different chords AB and CD of this circle and draw perpendicular bisector of these chords.
3. let these perpendicular bisectors meet at point O.

So, O will be the center of the given circle.

Question 5

Given two equal chords AB and CD of a circle with center O, intersecting each other at point P.
Prove that:
(i) AP = CP
(ii) BP = DP 

Sol:

In ΔOMP and ΔONP,
OP = OP                            ...( common sides )
∠OMP = ∠ONP                 ...( both are right angles )
OM = OM                          ...( side both the chords are equal, so the distance of the chords from the centre are also equal )
ΔOMP ≅ ΔONP                 ...( RHS congruence criterion )
⇒  MP = PN                       ...(c.p.c.t )  ....( a )

(i) Since AB = CD              ...( given )
⇒  AM = CN                     ...( drawn from the centre to the chord bisects the chord )
⇒  AM + MP = CN + NP  .....( from a )
⇒  AP = CP                       ....( b )

(ii) Since AB = CD
⇒ AP + BP = CP + DP
⇒  BP = DP                     ....( from  b )
Hence proved.

Question 6

In a circle of radius 10 cm, AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively.
Calculate the distance between the chords, if they are on:
(i) the same side of the center.
(ii) the opposite sides of the center.

Sol:

Given that AB = 16 cm and CD = 12 cm
So, AL = 8 cm and CM = 6 cm   ....( ⊥ from the center to the chord bisects the chord )

In right triangle OLA and OMC,
By Pythagoras theorem,
OA² = OL² + AL² and OC² = OM² + Cm² 
10² = OL² + 8² and 10² = OM² + 6²
OL² = 100 - 64 and OM² = 64
OL² = 6 cm and OM² = 8 cm

(i) In the first case, distance between AB and CD is
LM= OM - OL = 8 - 6 = 2 cm

(ii) In the second case , distance between AB and CD is
LM = OM + OL = 8 + 6 = 14 cm

Question 7

In the given figure, O is the center of the circle with radius 20 cm and OD is perpendicular to AB. If AB = 32 cm,
find the length of CD.

Sol:

To find: CD

Given : AB = 32 cm
⇒ AC = 16 cm( Since Perpendicular is drawn from the centre to the chord, bisects the chord )

In Right ΔOCA,
OA² = OC² + AC²        ....( By Pythagoras theorem )
⇒ OC² = OA² - AC²
⇒ OC² = 20² - 16²
⇒ OC²= 144
⇒ OC = 12cm
Since OD = 20 cm and OC = 12 cm
⇒ CD = OD - OC
          = 20 - 12 = 8 cm.

Question 8

In the given figure, AB and CD are two equal chords of a circle, with centre O. If P is the mid-point of chord AB, Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.

Sol:

It is given in the question that point.

P is the mid-point of the chord AB and Point Q is the mid-point of the CD.
⇒ ∠APO = 90°      ...( as the straight line drawn from the center of a  circle to bisect a chord, which is not a diameter, is at the right angle to the chord. )

As chords, AB and CD are equal therefore they are equidistant from the center i.e; PO = OQ      ...( ∵ Equal chords of a circle are equidistant from the center)

Now, the ΔPOQ is an isosceles triangle with OP = OQ as its two equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°.
⇒ ∠POQ + ∠OPQ + ∠PQO = 180°
⇒ ∠OPQ + ∠POQ + 150° = 180° ...( Given: ∠POQ = 150° )
⇒ 2∠OPQ = 180° - 150°  ...( As, ∠OPQ = ∠PQO ) 
⇒ 2∠OPQ = 30°
⇒ ∠OPQ = 15°

As ∠APO = 90°
⇒ ∠APQ + ∠OPQ = 90°
⇒ ∠APQ = 90° - 15°     ....( As, ∠OPQ = 15° )
⇒ ∠APQ = 75°.

Question 9

In the given figure, AOC is the diameter of the circle, with centre O. If arc AXB is half of arc BYC, find ∠BOC.

Sol:

Given :
1. AOC is the diameter.
2. Arc AXB = ${\displaystyle \frac{1}{2}}$ Arc BYC

From Arc AXB = ${\displaystyle \frac{1}{2}}$ Arc BYC We can see that
Arc AXB : Arc BYC = 1 : 2
⇒ ∠BOA : ∠BOC = 1 : 2

Since AOC is the diameter of the circle hence,
∠AOC = 180°
Now,
Assume that ∠BOA = x° and ∠BOC = 2x°
∠AOC = ∠BOA + ∠BOC = 180°
⇒ x + 2x = 180°
⇒ 3x = 180°
⇒ x = 60°
Hence, ∠BOA = 60° and ∠BOC = 120°.

Question 10

The circumference of a circle, with center O, is divided into three arcs APB, BQC, and CRA such that:
${\displaystyle \frac{\text{arc APB}}{2}=\frac{\text{arc BQC}}{3}=\frac{\text{arc CRA}}{4}}$

Find ∠BOC.

Sol:

From the given conditions given in the question
We can draw the circle with arc APB, arc BQC, and arc CRA

The given equation is
${\displaystyle \frac{\text{arc APB}}{2}=\frac{\text{arc BQC}}{3}=\frac{\text{arc CRA}}{4}}$
Let 
${\displaystyle \frac{\text{arc APB}}{2}=\frac{\text{arc BQC}}{3}=\frac{\text{arc CRA}}{4}}$ = k ( say )
then 
Arc APB = 2k, Arc BQC = 3k, Arc CRA = 4k
or
Arc APB : Arc BQC : Arc CRA = 2 : 3 : 4
⇒ ∠AOB : ∠BOC : ∠AOC = 2 : 3 : 4
and therefore,
and ∠AOB = 2k°, ∠BOC = 3k°, and ∠AOC = 4k°
Now,
Angle in a circle is 360°
So, 2k + 3k + 4k = 360°
⇒ 9k = 360°
⇒ k = 40°
Hence,
∠BOC = 3 x 40° = 120°.

SELINA Solution Class 9 Chapter 17 Circle Exercise 17C

Question 1

In the given figure, an equilateral triangle ABC is inscribed in a circle with center O.
Find: (i) ∠BOC
(ii) ∠OBC

Sol:

In the given figure, ABC is an equilateral triangle.

Hence all the three angles of the triangle will be equal to 60°
i.e. ∠A = ∠B = ∠C = 60°

As the triangle is an equilateral triangle, BO and CO will be the angle bisectors of B and C respectively.

Hence ∠OBC = ${\displaystyle \frac{\text{∠ABC}}{2}}$ 
                     = 30°
and as given in the figure we can see that OB and OC are the radii of the given circle.
Hence they are of equal length.

The ΔOBC is an isosceles triangle with OB = OC
In ΔOBC,
∠OBC = ∠OCB as they are angles opposite to the two equal sides of an isosceles triangle.
Hence, ∠OBC = 30° and ∠OCB = 30°
Since the sum of all angles of  a triangle is 180°

Hence in triangle OBC, ∠OCB + ∠OBC + ∠BOC + BOC = 180°
30° + 30° + ∠BOC= 180°
60° + BOC = 180°
∠BOC = 180° - 60°
∠BOC = 120°
Hence ∠BOC =120° and ∠OBC =30°

Question 2

In the given figure, a square is inscribed in a circle with center O.
Find:
(i) ∠BOC
(ii) ∠OCB
(iii) ∠COD
(iv) ∠BOD
Is BD a diameter of the circle?

Sol:

In the given figure we can extend the straight line OB to BD and CO to CA

Then we get the diagonals of the square which intersect each other at 90 by the property of Square.

From the above statement, we can see that
∠COD = 90°

The sum of the angle ∠BOC and ∠COD is 180° as BD is a straight line.
Hence ∠BOC + ∠OCD = ∠BOD = 180°
∠BOC + 90° = 180°
∠BOC + 180° - 90°
∠BOC = 90°

We can see that the OCB is an isosceles triangle with sides OB and OC of Equal length as they are the radii of the same are.

In ΔOCB,
∠OBC = ∠OCB as they are opposite angles to the two equal sides of an isosceles triangle.

Sum of all the angles of a triangle is 180°
so, ∠OBC + ∠OCB + ∠BOC =180°
∠OBC + ∠OBC + 90° = 180° as, ∠OBC = ∠OCB
2∠OBC = 180° - 90°
2∠OBC = 90°
2∠OBC = 45°
as ∠OBC = ∠OCB So,
∠OBC = OCB = 45°

Yes BD is the diameter of the order.

Question 3

In the given figure, AB is a side of regular pentagon and BC is a side of regular hexagon.
(i) ∠AOB
(ii) ∠BOC
(iii) ∠AOC
(iv) ∠OBA
(v) ∠OBC
(vi) ∠ABC

Sol:

As given that AB is the side of a pentagon the angle subtended by each arm of the pentagon at the center of the circle is = ${\displaystyle \frac{360°}{5}}$ = 72°

Thus angle ∠AOB = 72°

Similarly, as BC is the side of a hexagon hence the angle subtended by BC at the center is = ${\displaystyle \frac{360°}{6}}$ i.e. 60°
∠BOC = 60°

Now ∠AOC = ∠AOB + ∠BOC =72° + 60° = 132°

The triangle thus formed, ΔAOB is an isosceles triangle with OA = OB as they are radii of the same circle.

Thus ∠OBA = ∠BAO as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°
so, ∠AOB + ∠OBA + ∠BAO = 180°
⇒ 2∠OBA + 72° = 180° as, ∠OBA = ∠BAO
⇒ 2∠OBA = 180° - 72°
⇒ 2∠OBA = 180°
⇒ 2∠OBA =54° 
as, ∠OBA = ∠BAO So,
∠OBA = ∠BAO = 54°

The triangle thus formed, ΔBOC is an isosceles triangle with OB = OC as they are radii of the same are.

Thus ∠OBC = ∠OCB as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°
so, ∠BOC + ∠OBC + ∠OCB = 180°
2∠OBC + 60° = 180° as , ∠OBC = ∠OCB
2∠OBC = 180° - 60°
2∠OBC = 120°
∠OBC = 60°
as ∠OBC = ∠OCB
So, ∠OBC = ∠OCB = 60°
∠ABC = ∠OBA + ∠OBC = 54° + 60°= 114°

Question 4

In the given figure, arc AB and arc BC are equal in length. If ∠AOB = 48°, find:
(i) ∠BOC
(ii) ∠OBC
(iii) ∠AOC
(iv) ∠OAC

Sol:

We know that the arc of equal lengths subtends equal angles at the center.

hence ∠AOB = ∠BOC = 48°
Then ∠AOC = ∠AOB + ∠BOC = 48° + 48° = 96°
The triangle thus formed, ΔBOC is an isosceles triangle with OB = OC as they are radii of the same circle.

Thus ∠OBC = ∠OCB as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°.
So, ∠BOC + ∠OBC + ∠OCB = 180°   
 2∠OBC + 48° = 180°              as ∠OBC = ∠OCB
 2∠OBC = 180° - 48°
 2∠OBC = 132°
∠OBC = 66°
as ∠OBC = ∠OCB 
So, ∠OBC = ∠OCB = 66°

The triangle thus formed, ΔAOC is an isosceles triangle with OA = OC as they are radii of the same circle.
Thus ∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°.
So, ∠COA + ∠OAC + ∠OCA = 180°
2∠OAC + 96° = 180°                 as, ∠OAC = ∠OCA 
2∠OAC = 180° - 96°
2∠OAC = 84° 
∠OAC = 42°
as ∠OCA = ∠OAC
So, ∠OCA = ∠OAC = 42°.

Question 5

In the given figure, the lengths of arcs AB and BC are in the ratio 3:2. If ∠AOB = 96°,
find: (i) ∠BOC (ii) ∠ABC

Sol:

We know that for two arcs are in ratio 3: 2 then
∠AOB: ∠BOC = 3: 2
As give ∠AOC = 96°
So, 3x = 96
       x = 32
There ∠BOC = 2 x 32 = 64°
The triangle thus formed, ΔAOB is an isosceles triangle with OA = OB as they are radii of the same circle.

Thus ∠OBA = ∠BAO as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOB + ∠OBA + ∠BAO = 180°
2∠OBA + 96° = 180°       as,  ∠OBA = ∠BAO
2∠OBA = 180° - 96° 
2∠OBA = 84°
∠OBA = 42°
as, ∠OBA = ∠BAO So,
∠OBA = ∠BAO = 42°

The triangle thus formed, ΔBOC is an isosceles triangle with OB = OC as they are radii of the same circle.
Thus ∠OBC = ∠OCB as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠BOC + ∠OBC + ∠OCB = 180°
2∠OBC + 64° = 180°          as, ∠OBC = ∠OCB 
2∠OBC = 180° - 64° 
2∠OBC =  116° 
∠OBC = 58°
As ∠OBC = ∠OCB   So,
∠OBC = ∠OCB = 58° 
∠ABC = ∠BOA + ∠OBC = 42°+ 58° = 100°

Question 6

In the given figure, AB = BC = DC and ∠AOB = 50°.
(i) ∠AOC
(ii) ∠AOD
(iii) ∠BOD
(iv) ∠OAC
(v) ∠ODA

Sol:

Since arc AB and BC are equal.
So, ∠AOB = ∠BOC = 50°
Now,
∠AOC = ∠AOB + ∠BOC = 50° + 50° = 100°
As arc AB, arc BC and arc CD so, 
∠AOB = ∠BOC = ∠COD = 50°
∠AOD = ∠AOB + ∠BOC + ∠COD = 50° + 50° + 50° = 150°
Now, ∠BOD = ∠BOC + ∠COD 
∠BOD = 50° + 50° 
∠BOD = 100°

The triangle thus formed, ΔAOC is an isosceles triangle with OA = OC as they are radii of the same circle.
 Thus ∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOC + ∠OAC + ∠OCA = 180°
2∠OAC + 100° = 180°     as, ∠OAC = ∠OCA
2∠OAC = 180° - 100° 
2∠OAC = 80°
∠OAC = 40°

as ∠OCA = ∠OAC So,
∠OCA = ∠OAC = 40°

The triangle thus formed, ΔAOD is an isosceles triangle with OA = OD as they are radii of the same circle.
Thus, ∠OAD = ∠ODA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOD + ∠OAD + ∠ODA = 180°
2∠OAD + 150° = 180°         as, ∠OAD = ∠ODA
2∠OAD = 180° - 150° 
∠OAD = 30°

as ∠OAD = ∠ODA    So,
∠OAD = ∠ODA  = 15°.

Question 7

In the given figure, AB is a side of a regular hexagon and AC is a side of a regular eight-sided polygon.
Find:
(i) ∠AOB
(ii) ∠AOC
(iii) ∠BOC 
(iv) ∠OBC

Sol:

As AB is the side of a hexagon so the
∠AOB = ${\displaystyle \frac{360°}{6}}$ = 60°

AC is the side of an eight-sided polygon so,
∠AOC = ${\displaystyle \frac{360°}{8}}$ = 45°

From the given figure we can see that:
∠BOC = ∠AOB + ∠AOC 
⇒ 60° + 45° = 105°
Again, from the figure, we can see that ∠BOC is an isosceles triangle with sides BO = OC as they are the radii of the same circle.
Angles ∠OBC = ∠OCB    as they are opposite angles to the equal sides of an isosceles triangle.

Sum of all the angles of a triangle is 180°
∠OBC + ∠OCB + ∠BOC = 180°
2∠OBC + 105° = 180°                    as, ∠OBC = ∠BOC
2∠OBC = 180° - 105°
2∠OBC = 75°
∠OBC = 37.5° = 37°30'
As, ∠OBC = ∠BOC
∠OBC = ∠BOC = 37.5° = 37°30'.

Question 8

In the given figure, O is the center of the circle and the length of arc AB is twice the length of arc BC. If ∠AOB = 100°,
find: (i) ∠BOC (ii) ∠OAC

Sol:

We know that when two arcs are in ratio 2: 1 then the subtended by them is also in ratio 2: 1
As given arc AB is twice the length of arc BC.
Therefore, arc AB: arc BC = 2: 1
Hence, ∠AOB: ∠BOC = 2: 1

Now given that ∠AOB = 100°.
So, ∠BOC = ${\displaystyle \frac{1}{2}\angle \text{AOB}=\frac{1}{2}\times 100°=50°}$

Now, ∠AOC = ∠AOB + ∠BOC = 100° + 50° = 150°.
The triangle thus formed, ∠AOC is an isosceles triangle with OA = OC as they are radii of the same circle.
Thus,
∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°.
So, ∠COA + ∠OAC + ∠OCA = 180°
2∠OAC + 150° = 180° as, ∠OAC = ∠OCA 
2∠OAC = 180° - 150°
2∠OAC = 30°
∠OAC = 15°
as ∠OCA = ∠OAC So,
∠OCA = ∠OAC = 15°.

SELINA Solution Class 9 Chapter 17 Circle Exercise 17B

Question 1

The figure shows two concentric circles and AD is a chord of a larger circle.
Prove that: AB = CD.

Sol:

Drop OP ⊥ AD
∴ OP bisects AD.     ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
⇒ AP = PD               .....(i)

Now, BC is a chord for the inner circle and OP ⊥ BC.
∴ OP bisects BC      ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
⇒ BP = PC               .....(ii)

Subtracting (ii) from (i),
AP - BP = PD - PC 
⇒ AB = CD.

Question 2

A straight line is drawn cutting two equal circles and passing through the mid-point M of the line joining their centers O and O'. Prove that the chords AB and CD, which are intercepted by the two circles, are equal.

Sol:

Given: A straight line AD intersects two circles of equal radii at A, B, C and D.
The line joining the centers OO' intersect AD at M and M is the midpoint of OO'.

To Prove: AB = CD.

Construction: From O, draw OP ⊥ AB and from O', draw O'Q ⊥ CD.

Proof:  
In ΔOMP and ΔO'MQ,
∠OMP = ∠O'MQ            ...( Vertically Opposite angles )
∠OPM = ∠O'QM            ...( each = 90° ) 
OM = O'M                      ...( Given )

By Angle-Angle-Side criterion of congruence,
∴ ΔOMP ≅ ΔO'MQ,       ...( by AAS )

The corresponding parts of the congruent triangles are congruent.
∴ OP = O'Q                     ...( c.p.c.t. )

We know that two chords of a circle or equal circles which are equidistant from the center are equal.
∴ AB = CD.

Question 3

M and N are the mid-points of two equal chords AB and CD respectively of a circle with center O.
Prove that: (i) ∠BMN = ∠DNM
                  (ii) ∠AMN = ∠CNM

Sol:

Drop OM ⊥ AB and ON ⊥ CD.
∴ OM bisects AB and ON bisects CD.    ...( Perpendicular drawn from the centre of a circle to a chord bisects it. )

⇒ BM = ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}\text{CD}}$ = DN  ....(1)

Applying Pythagoras theorem,
OM² = OB² - BM²
         = OD² - DN²                         ....( By 1 )
         = ON²
∴ OM = ON
⇒  ∠OMN = ∠ONM                   ....(2)
( Angles opp to equal sides are equal. )  

(i) ∠OMB = ∠OND                  .....( both 90° )
Subtracting (2) from above,
∠BMN = ∠DNM

(ii) ∠OMA = ∠ONC                  .....( both 90° )
Adding (2) to above,
∠AMN = ∠CNM.

Question 4

In the following figure; P and Q are the points of intersection of two circles with centers O and O'. If straight lines APB and CQD are parallel to OO';
prove that: (i) OO' = ${\displaystyle \frac{1}{2}}$AB ; (ii) AB = CD

Sol:

Drop OM and O'N perpendicular on AB and OM' and O'N' perpendicular on CD.

∴ OM, O'N, OM' and O'N' bisect AP, PB, CQ and QD respectively.
( Perpendicular is drawn from the center of a circle to a chord bisects it. )

∴ MP = ${\displaystyle \frac{1}{2}\text{AP},\text{PN}=\frac{1}{2}\text{BP},\text{M'Q}=\frac{1}{2}\text{CQ},\text{QN}\prime =\frac{1}{2}\text{QD}}$ 

Now, OO' = MN = MP + PN = ${\displaystyle \frac{1}{2}\left(\text{AP + BP}\right)=\frac{1}{2}\text{AB}}$   ...(i)

and OO' = M'N' = M'Q + QN' = ${\displaystyle \frac{1}{2}\left(\text{CQ + QD}\right)=\frac{1}{2}\text{CD}}$  ...(ii)

By (i) and (ii),
AB = CD.

Question 5

Two equal chords AB and CD of a circle with center O, intersect each other at point P inside the circle.
Prove that: (i) AP = CP ; (ii) BP = DP

Sol:

Drop OM and ON perpendicular on AB and CD.
Join OP, OB, and OD.

∴ OM and ON bisect AB and CD respectively.     ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
∴ MB = ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}\text{CD}=\text{ND}}$....(i)

In right ΔOMB,
OM² = OB² - MB²                            ....(ii)
In right ΔOND,
ON² = OD² - ND²                           ....(iii)

From (i), (ii), and (iii),
OM = ON

In ΔOPM and ΔOPN,
∠OMP = ∠ONP          ....( both 90° )
OP = OP                     ....( common )
OM = ON                   ....( proved above )
By Right Angle-Hypotenuse-Side criterion of congruence,
∴ ΔOPM ≅ ΔOPN      ....( by RHS )

The corresponding parts of the congruent triangles are congruent.
∴ PM = PN                ....( c.p.c.t. )

Adding (i) to both sides,
MB + PM = ND + PN
⇒ BP = DP
Now, AB = CD
∴  AB - BP = CD - DP     ...( ∵ BP = DP )
⇒ AP = CP.

Question 6

In the following figure, OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q.
Prove that:
( i ) ΔOPA ≅ ΔOQC 
( ii ) ΔBPC ≅ ΔBQA

Sol:

(i) In ΔOPA and ΔOQC,
 OP = OQ                          ....[ radii of same circle ]
∠AOP = ∠COQ                  ... [ both 90° ] 
OA = OC                           ... [ sides of the square ]

By Side- Angle - Side criterion of congruence.
∴ ΔOPA ≅ ΔOQC             ...[ by SAS ]

(ii) Now, OP = OQ           ...[ radii ]
 and  OC = OA                 ...[ sides of the square ]
∴ OC - OP = OA - OQ  
⇒ CP = AQ                      ....(i)

In ΔBPC and ΔBQA,
BC = BA                           ...[ sides of the square ]
∠PCB = ∠QAB                 ...[ both 90° ]
 PC = QA                         ...[ by ( i ) ]

By Side- Angle-Side criterion of congruence,
∴ ΔBPC ≅ ΔBQA                ...[ by SAS ]

Question 7

The length of the common chord of two intersecting circles is 30 cm. If the diameters of these two circles are 50 cm and 34 cm, calculate the distance between their centers.

Sol:

OA = 25 cm and AB = 30 cm

∴    AD = ${\displaystyle \frac{1}{2}\times \text{AB} =(\frac{1}{2}\times 30)}$ cm = 15 cm 

Now in right angled ADO
OA² + AD² + OD²  
⇒  OD² = OA² - OD²  = 25² - 15² 
             = 625 - 225 = 400
∴ OD = ${\displaystyle \sqrt{400}}$ = 20 cm
Again, we have  O'A = 17 cm.

In right-angle ADO'
O'A² = A'D² + O'D²  
⇒  O'D² = O'A² - AD² 
= 17² - 15² 
= 289 - 225 = 64

∴ O'D = 8 cm
∴ OO' = ( OD + O'D )
          = ( 20 + 8 ) = 28 cm

∴ the distance between their centres is 28 cm.

Question 8

The line joining the midpoints of two chords of a circle passes through its center.
Prove that the chords are parallel.

Sol:

Given: AB and CD are the two chords of a circle with center O.
L and M are the mid-points of AB and CD and O lies in the line joining ML.

To prove : AB || CD.

Proof:
AB and CD are two chords of a circle with center O.
Line LOM bisects them at L and M.
Then,                OL ⊥ AB
and,                OM ⊥ CD
∴               ∠ALM = ∠LMO = 90°
But they are alternate angles
∴                 AB || CD.

Question 9

In the following figure, the line ABCD is perpendicular to PQ; where P and Q are the centers of the circles.
Show that:
(i) AB = CD ;
(ii) AC = BD.

Sol:

In the circle with center Q, QO ⊥ AD
∴ OA = OD                            ....(i)   ...[ perpendicular drawn the center of a circle to a chord bisects it ]

In circle with center P, PO ⊥ BC
∴ OB  = OC                          ....(ii)  ....[  perpendicular drawn the center of a circle to a chord bisects it ]

(i) (i) - (ii) gives,
AB = CD                              ....(iii)

(ii) Adding BC to both sides of equation (iii) 
 AB + BC + CD + BC
⇒ AC = BC

Question 10

AB and CD are two equal chords of a circle with center O which intersect each other at a right angle at point P.
If OM ⊥ AB and ON ⊥ CD;
show that OMPN is a square.

Sol:

Clearly , all the angles of OMPN are 90°.
OM ⊥ AB and ON ⊥ CD

∴ BM = ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}}$CD = CN      ....(i) ...[ perpendicular drawn from the center of a circle to a chord bisects it ]

As the two equal chords, AB and CD intersect at point P inside the circle,

∴ AP = DP and CP = BP                 .....(ii)
Now, CN - CP = BM - BP                ...[ by (i) and (ii) ]
⇒ PN = MP

∴ Quadrilateral OMPN is A square.

SELINA Solution Class 9 Chapter 17 Circle Exercise 17A

Question 1

A chord of length 6 cm is drawn in a circle of radius 5 cm.
Calculate its distance from the center of the circle.

Sol:

Let AB be the chord and O be the center of the circle.

Let OC be the perpendicular drawn from O to AB.

We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.

∴ AC = CB = 3 cm

In ΔOCA,
OA² = OC² + AC²                    ...( By Pythagoras theorem )
⇒ OC² = ( 5 )² - ( 3 )³ = 16 
⇒ OC = 4 cm 

Question 2

A chord of length 8 cm is drawn at a distance of 3 cm from the center of the circle.
Calculate the radius of the circle.

Sol:

Let AB be the chord and O be the center of the circle.

Let OC be the perpendicular drawn from O to AB.

We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.
∴ AB = 8 cm
⇒ AC = CB = ${\displaystyle \frac{\text{AB}}{2}}$

⇒ AC = CB = ${\displaystyle \frac{8}{2}}$

⇒ AC = CB =  4 cm

In OCA,
OA² = OC² + AC²            ...( By Pythagoras theorem )
⇒ OA² = ( 4 )² + ( 3 )² = 25
⇒ OA = 5 cm

Hence, radius of the circle is 5 cm. 

Question 3

The radius of a circle is 17.0 cm and the length of the perpendicular drawn from its center to a chord is 8.0 cm.
Calculate the length of the chord.

Sol:

Let AB be the chord and O be the center of the circle.

Let OC be the perpendicular drawn from O to AB.

We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.

∴ AC = CB

In ΔOCA,
OA² = OC² + AC²                       ...( By Pythagoras theorem )

⇒ AC² = (17)² - (8)² = 225

⇒ Ac = 15 cm

∴ AB = 2 AC = 2 x 15 = 30 cm. 

Question 4

A chord of length 24 cm is at a distance of 5 cm from the center of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the center.

Sol:

Let AB be the chord of length 24 cm and O be the center of the circle.

Let OC be the perpendicular drawn from O to AB.

We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.

∴ AC = CB = 12 cm

In OCA,
OA² = OC² + AC²  ....( By Pythagoras theorem )

=(5)² + ( 12 )² = 169

⇒ OA = 13 cm

∴  radius of the circle = 13 cm.
Let A ' B ' be the new chord at a distance of 12 cm from the center.

∴  ( OA' )² = ( OC' )² + ( A'C' )²

⇒ ( A'C' )² = ( 13 )² - ( 12 )²   = 25

∴  A'C' = 5 cm

Hence, length of the new chord = 2 x 5 = 10 cm.

Question 5

In the following figure, AD is a straight line, OP ⊥ AD and O is the centre of both circles. If OA = 34cm, OB = 20 cm and OP = 16 cm;
find the length of AB.

Sol:

For the inner circle, BC is a chord and OP ⊥ BC.

We know that the perpendicular to a chord, from the center of a circle, bisects the chord.

∴ BP = PC
By Pythagoras theorem,
OB² = OP² + BP²
⇒ BP² = 20² - 16² = 144
∴ BP = 12 cm

For the outer circle, AD is the chord and OP ⊥ AD.

We know that the perpendicular to a chord, from the center of a circle, bisects the chord.

∴ AP = PD
By Pythagoras Theorem,
OA² = OP² + AP²
⇒ AP² = (34)² - (16)² = 900
⇒ AP = 30 cm

AB = AP - BP = 30 - 12 = 18 cm

Question 6

In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn. Find the distance between the chords,
if both the chords are:
(i) on the opposite sides of the centre;
(ii) on the same side of the centre.

Sol:

Let O be the center of the circle and AB and CD be the two parallel chords of length 30 cm and 16 cm respectively.

Drop OE and OF perpendicular on AB and CD from the center O.

OE ⊥ AB and OF ⊥ CD.
∴ OE bisects AB and OF bisects CD.   ...( Perpendicular is drawn from the centre of a circle to a chord bisects it. )
⇒ AE = ${\displaystyle \frac{30}{2}}$ = 15 cm;
    CF = ${\displaystyle \frac{16}{2}}$ = 8 cm

In right ΔOAE,
OA² = OE² + AE²
⇒ OE² = OA² - AE² = 17² - 15² = 64
∴ OE = 8 cm

In right ΔOCF,
OC² = OF² + CF²
⇒ OF² = OC² - CF² = 17² - 8² = 225
∴ OF = 15 cm

(i) The chord are on the opposite sides of the centre :
∴ EF = EO + OF = 8 + 15 = 23cm

(ii) The chord are on the same side of the centre :
∴ EF = OF - OE = 15 - 8 = 7 cm.

Question 7

Two parallel chords are drawn in a circle of diameter 30.0 cm. The length of one chord is 24.0 cm and the distance between the two chords is 21.0 cm;
find the length of another chord.

Sol:

Since the distance between the chords is greater than the radius of the circle (15 cm), so the chords will be on the opposite sides of the center.

Let O be the center of the circle and AB and CD be the two parallel chords such that AB = 24 cm.
Let the length of the CD be 2x cm.

Drop OE and OF perpendicular on AB and CD from the center O.
OE ⊥ AB and OF ⊥ CD

∴ OE bisects AB and OF bisects CD.      ....( Perpendicular drawn from the center of a circle to a chord bisects it.)

⇒ AE = ${\displaystyle \frac{24}{2}}$ = 12 cm ;

    CF = ${\displaystyle \frac{2x}{2}}$ = x cm

In right ΔOAE,
OA² = OE² + AE²
⇒ OE² = OA² - AE² = 15² - 12² = 81
∴ OE = 9 cm
∴ OF = EF - OE = 21 - 9 = 12 cm

In right ΔOCF,
OC² = OF² + CF²
⇒ x² = OC² - OF² = 15² - 12² = 81
∴ x = 9 cm
Hence, length of chord CD = 2x = 2 x 9 = 18 cm.

Question 8

A chord CD of a circle whose center is O is bisected at P by a diameter AB. Given OA = OB = 15 cm and OP = 9 cm.
Calculate the lengths of: (i) CD ; (ii) AD ; (iii) CB.

Sol:

(i) OP ⊥ CD
∴ OP bisects CD.   ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
⇒ CP = ${\displaystyle \frac{\text{CD}}{2}}$

In right ΔOPC,
OC² = OP² + CP²
⇒ CP² = OC² - OP² 
⇒ 15² - 9² = 144
∴ CP = 12 cm
∴ CD = 12 x 2 = 24 cm

(ii) Join BD,
∴ BP = OB - OP = 15 - 9 = 6 cm.
In right ΔBPD,
BD² = BP² + PD² 
       = 6² + 12² = 180
In ΔADB,
∠ADB = 90°     ...( Angle in a semi-circle is a right angle )
∴ AB² = AD² + BD²
⇒ AD² = AB² - BD²
= 30² - 180 = 720
∴ AD = ${\displaystyle \sqrt{720}}$ = 26.83 cm

(iii) Also, BC = BD = ${\displaystyle \sqrt{180}}$ = 13.42 cm.

Question 9

The figure given below shows a circle with center O in which diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4 cm,
find the radius of the circle.

Sol:

Let the radius of the circle be r cm.
∴ OE = OB - EB = r - 4
Join OC.
In right ΔOEC,
OC² = OE² + CE²
⇒ r² = ( r - 4 )² + (8)² 
⇒ r² = r² - 8r + 16 + 64
⇒  8r = 80
∴ r = 10 cm
Hence, radius of the circle is 10 cm.

Question 10

In the given figure, O is the center of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm,

Find the :
(i) the radius of the circle
(ii) length of chord CD.

Sol:

(i) AB is the chord of the circle and OM is perpendicular to AB.
So, AM = MB = 12 cm ....( Since ⊥ bisects the chord )
In right ΔOMA,
OA² = OM² + AM²
⇒ OA² = 5² + 12^2
⇒  OA = 13 cm
So, radius of the circle is 13 cm.

(ii) So, OA = OC = 13 cm  ....( radii of the same circle )
In right ΔONC,
NC² = OC² - ON²
⇒ NC² = 13² - 12²
⇒ NC = 5 cm
So, CD = 2NC = 10 cm.