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.