NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 - Circles

Q1 Two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm . Find the length of the common chord.

Solution:
Given: Two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm .
To find the length of the common chord.
Construction: Join OP and ON draw $OM\perp AB\

Proof: AB is a chord of the circle, and OM is the bisector of chord AB.
∴OM⊥AB
∠OMA=90∘
Let, OM = x , so O'M = 4 - x
In △ AOM, using Pythagoras' theorem
AM2=AO2−OM2 ...........................(1)
Also,
In △ AO'M, using Pythagoras' theorem
AM2=AO′2−MO′2 ...........................(2)
From (1) and (2), we get
AO2−OM2=AO′2−MO′2
⇒52−x2=32−(4−x)2
⇒25−x2=9−16−x2+8x
⇒32=8x
⇒x=4
Put x = 4 in equation (1)
AM2=52−42=9
⇒AM=3
⇒AB=2AM=6

Q2 If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution:
Given: two equal chords of a circle intersect within the circle.
To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP = DP.
Construction: Join OP and draw OM⊥ABandON⊥CD.
Proof:

In △ OMP and △ ONP,
OP = OP (Common)
OM = ON (Equal chords of a circle are equidistant from the centre)
∠ OMP = ∠ ONP (Both are right angled)
Thus, △ OMP ≅ △ ONP (By SAS rule)
PM = PN..........................(1) (CPCT)
AB = CD ..........................(2)(Given)
⇒12AB=12CD
⇒AM=CN .....................(3)
Adding equations (1) and (3), we have
AM + PM = CN + PN
⇒AP=CP .................(4)
Subtract equation (4) from (2), we get
AB−AP=CD−CP
⇒PB=PD

Q3 If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Solution:
Given: two equal chords of a circle intersect within the circle.
To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. ∠ OPM= ∠ OPN


Proof:
Construction: Join OP and draw OM⊥ABandON⊥CD.
In △ OMP and △ ONP,
OP = OP (Common)
OM = ON (Equal chords of a circle are equidistant from the centre)
∠ OMP = ∠ ONP (Both are right-angled)
Thus, △ OMP ≅ △ ONP (By RHS rule)
∠ OPM= ∠ OPN (CPCT)

Q4 If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB=CD (see Fig. 10.25 ).

Solution:
Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.
To prove: AB = CD
Construction: Draw OM⊥AD
Proof:

BC is a chord of the inner circle and OM⊥BC
So, BM = CM .................(1)
(Perpendicular OM bisect BC)
Similarly,
AD is a chord of the outer circle and OM⊥AD
So, AM = DM .................(2) (Perpendicular OM bisect AD)
Subtracting equation (1) from (2), we get
AM−BM=DM−CM
⇒AB=CD

Q5 Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Solution:
Given: From the figure, R, S, and M are the positions of Reshma, Salma, and Mandip, respectively.
So, RS = SM = 6 cm
Construction: Join ON, OS, OR and OM. Draw OL⊥RS.
Proof:

In △ ORS,
OS = OR and OL⊥RS (by construction)
So, RL = LS = 3cm (RS = 6 cm)
In △ OLS, by Pythagoras' theorem,
OL2=OS2−SL2
⇒OL2=52−32=25−9=16
⇒OL=4
In △ ORN and △ OMN,
OR = OM (Radii)
∠ RON = ∠ MON (Equal chords subtend equal angles at the centre)
ON = ON (Common)
△ ORN ≅ △ OMN (By SAS)
RN = MN (CPCT)
Thus, ON⊥RM
Area of △ ORS = 12×RS×OL ..................(1)
Area of △ ORS = 12×OS×NR ..................(2)
From 1 and 2, we get
12×RS×OL =12×OS×NR
⇒RS×OL=OS×KR
⇒6×4=5×KR
⇒NR=4.8cm
Thus, RM=2NR=2×4.8cm=9.6cm

Q6 A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Solution:
Given: In the figure, A, B, and C are positioned as Ankur, Syed and David, respectively.
So, AB = BC = CD
Radius of circular park = 20m
So, AO = OB = OC = 20m
Construction: AF ⊥ BC
Proof:

Let AC = CB = AB = 2x cm
In △ ABC,
AC = AB and AF ⊥ BC
So, CF = FB = x cm
In △ OFB, by Pythagoras,
OF2=OB2−FB2
⇒OF2=202−x2=400−x2
⇒OF=400−x2
In △ AFB, by Pythagoras,
AF2=AB2−FB2
⇒(AO+OF)2+x2=(2x)2
⇒(20+400−x2)2+x2=4x2
⇒400+400−x2+40400−x2+x2=4x2
⇒800+40400−x2=4x2
⇒200+10400−x2=x2
⇒10400−x2=x2−200
Squaring both sides,
⇒100(400−x2)=(x2−200)2
⇒40000−100x2=x4−40000−400x2
⇒x4−300x2=0
⇒x2(x2−300)=0
⇒x2=300
⇒x=103
Hence, the length of the string of each phone =2x=203 m