ML Aggarwal Solution Class 9 Chapter 15 Circle Test

 Test

Question 1

In the given figure, a chord PQ of a circle with centre O and radius 15 cm is bisected at M by a diameter AB. If OM = 9 cm, find the lengths of :

(i) PQ

(ii) AP

(iii) BP













Sol :

Given : radius=15 cm

⇒OA=OB=OP=OQ=15 cm

Also, OM=9 cm








∴MB=OB-OM=15-9=6=6 cm

AM=OA+OM=15+9 cm=24 cm


In ΔOMP , by using Pythagoras Theorem

⇒OP2=OM2+PM2

⇒152=92+PM2

=PM2=√144=12 cm

Also, In ΔOMQ,

by using Pythagoras theorem

⇒OQ2=OM2+QM2

⇒152=OM2+QM2

⇒152=92+QM2

⇒QM2=225-81

⇒QM=√144=12 cm

∴PQ=PM+QM

(as radius is bisected at M)

⇒PQ=12+12 cm=24 cm


(ii) Now in ΔAPM

⇒AP2=AM2+OM2

⇒AP2=242+122

⇒AP2=242+122

⇒AP2=576+144

⇒AP=√720=12√5 cm


(iii) Now in ΔBMP

⇒BP2=BM2+PM2

⇒BP2=62+122

⇒BP2=36+144

⇒BP=√180=6√5 cm


Question 2

The radii of two concentric circles are 17 cm and 10 cm ; a line PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, calculate PQ.

Sol :
A line PQRS intersects the outer circle at P and S and inner circle at Q and R. Radius of outer circle OP=17 cm and radius of inner circle OQ=10 cm











QR=12 cm

From O , draw OM⟂PS
∴QM$=\frac{1}{2}$QR$=\frac{1}{2}\times 12$=6 cm

In right ΔOQM

⇒OQ2=OM2+QM2

⇒(10)2=OM2+(6)2

⇒OM2=102-62

⇒OM2=100-36=64=82

∴OM=8 cm

Now in right ΔOPM,

⇒OP2=OM2+PM2

⇒(17)2=(8)2+PM2

⇒PM2=172-82

=289-64=225=(15)2
∴PM=15 cm
∴PQ=PM-QM=15-6=9 cm

Question 3

A chord of length 48 cm is at a distance of 10 cm from the centre of a circle. If another chord of length 20 cm is drawn in the same circle, find its distance from the centre of the circle.

Sol :
O is the centre of the circle
Length of chord AB=48 cm and chord CD=20 cm










OL⟂AB and OM⟂CD are drawn 
∴AL=LB$=\frac{48}{2}=24$ cm
and CM=MD$=\frac{20}{2}=10$ cm
OL=10 cm

Now in right ΔAOL

⇒OA2=AL2+OL2 (Pythagoras theorem)

⇒OA2=(24)2+(10)2

⇒OA2=576+100=676=262
∴OA=26 cm

But OC=OA (radii of the same circle)
∴OC=26 cm

Now in right ΔOCM

⇒OC2=OM2+CM2 (Pythagoras theorem)

⇒(26)2=OM2+(10)2 
=676=OM2+100
⇒OM2=676-100
⇒OM2=576=(24)2 

∴OM=24 cm


Question 4

(a) In the figure (i) given below, two circles with centres C, D intersect in points P, Q. If length of common chord is 6 cm and CP = 5 cm, DP = 4 cm, calculate the distance CD correct to two decimal places.

(b) In the figure (ii) given below, P is a point of intersection of two circles with centres C and D. If the st. line APB is parallel to CD, Prove that AB = 2 CD.
Sol :







∴PM=MQ$=\frac{6}{2}$=3 cm

Now in right ΔCPM

⇒CP2=CM2+PM2 
⇒52=CM2+32 
⇒25=CM2+9 
⇒CM2=25-9=16=42 

∴CM=4 cm

and in right ΔPDM

⇒PD2=PM2+MD2 
⇒(4)2=(3)2+MD2 
⇒16=9+MD2 
⇒MD2 =16-9=7

∴MD=√7=2.65 cm

∴CD=CM+MD=4+2.65=6.65 cm


(b) Given : Two circles with centre C and D intersects each other at P and Q. A straight line APB is drawn parallel to CD
To prove: AB=2CD

Construction : Draw CM and DN perpendicular to AB from  C and D

Proof : ∵CM⟂AP

∴AM=MP or AP=2 MP and DN⟂PB






∴BN=PN or PB=2PN

Adding 

⇒AP+PB=2MP+2PN

⇒AB=2(MP+PN)=2MN

⇒AB=2CD  (Q.E.D)


Question 5

(a) In the figure (i) given below, C and D are centres of two intersecting circles. The line APQB is perpendicular to the line of centres CD.Prove that:

(i) AP=QB

(ii) AQ = BP.

(b) In the figure 

(ii) given below, two equal chords AB and CD of a circle with centre O intersect at right angles at P. If M and N are mid-points of the chords AB and CD respectively, Prove that NOMP is a square.

Sol :






(a) Given : Two circles with centre C and D intersecting each other . A line APQB is drawn perpendicular to CD at M

To prove :

(i) AP=QB
(ii) AQ=BP

Construction : Join AC and BC, DP and DQ

Proof : 

(i) In right ΔACM and ΔBCM

Hyp. AC=BC (radii of same circle)

Side CM=CM (common)

∴ΔACM≅ΔBCM

(R.H.S axiom of congruency)

∴AM=BM ..(ii)

Again in right ΔPDM and ΔQDM

Hypo. PD=QD (radii of the same circle)

Side DM=DM (common)

∴ΔPD≅ΔQDM  (R.H.S axiom of congruency)

∴PM=QM ...(ii)

Subtracting (ii) from (i)

⇒AM-PM=BM-QM

⇒AP=QB


(ii) Adding PQ both sides

⇒AP+PQ=PQ+QB

⇒AQ=PB  (Q.E.D)


(b) Given : Two chords AB and CD intersects each other at P at right angle in the circle. M and N are mid points of the chord AB and CD

To prove : NOMP is a square

Proof : ∵M and N are the mid points of AB and CD respectively
∴OM⟂AB and ON⟂CD
and OM=ON
(∵equal chords are at equal distance from the centre)
∵AB⟂CD
∴OM⟂ON
Hence NOMP is a square

Question 6

In the given figure, AD is diameter of a circle. If the chord AB and AC are equidistant from its centre O, prove that AD bisects ∠BAC and ∠BDC.
Sol :












Given : AB and AC are equidistant from its centre O
So, AB=AC
In ΔABD and ΔACD
AB=AC (given)
∠B=∠C (∵Angle is a semicircle is 90°)
AD=AD (common)
∴ΔABD≅ΔACD  (SSS rule of congruency)
∴AD bisects ∠BAC and ∠BDC

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