RS AGGARWAL CLASS 9 CHAPTER 12 CIRCLES EXERCISE 12A

  EXERCISE 12A

PAGE NO-436

Question 1:

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the circle.

Answer 1:

Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 cm

From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
BM = 162 cm=8 cm
In the right  ΔOMB, we have:
OB2 = OM2 + MB2   (Pythagoras theorem)
⇒ 102 = OM2 + 82
⇒ 100 = OM2 + 64
OM2 = (100 - 64) = 36
OM=36 cm=6 cm
Hence, the distance of the chord from the centre is 6 cm.

Question 2:

Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.

Answer 2:

Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm

From the right ΔOMB, we have:
OB2 = OM2 + MB2        (Pythagoras theorem)
⇒ 52 = 32 + MB2
⇒ 25 = 9 + MB2
MB2 = (25 9) = 16
MB=16 cm=4 cm
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = (2 × 4) cm = 8 cm
Hence, the required length of the chord is 8 cm.

Question 3:

A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle.

Answer 3:

Let AB be the chord of the given circle with centre O. The perpendicular distance from the centre of the circle to the chord is 8 cm.
Join OB.
Then OM = 8 cm and AB = 30 cm

We know that the perpendicular from the centre of a circle to a chord bisects the chord.
MB=AB2=302 cm=15 cm
From the right ΔOMB, we have:
OB2 = OM2 + MB2
OB2 = 82 + 152
OB2 = 64 + 225
OB2 = 289
OB=289 cm=17 cm
Hence, the  required length of the radius is 17 cm.

Question 4:

In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the distance between the chords if they are
(i) on the same side of the centre
(ii) on the opposite sides of the centre.

Answer 4:

We have:
(i)
Let AB and CD be two chords of a circle such that AB is parallel to CD on the same side of the circle.
Given: AB = 8 cm, CD = 6 cm and OB = OD = 5 cm
Join OL and OM.

The perpendicular from the centre of a circle to a chord bisects the chord.
∴ LB=AB2=82=4cm
Now, in right angled ΔBLO, we have:
OB2 = LB2 + LO2
LO2 = OB2 LB2
⇒ LO2 = 52 − 42  
⇒ LO2 = 25 − 16 = 9
LO = 3 cm

Similarly, MD=CD2=62=3cm
In right angled ΔDMO, we have:
OD2 = MD2 + MO2
MO2 = OD2 MD2
MO2   = 52 − 32
MO2 = 25 − 9 = 16
MO = 4 cm
∴ Distance between the chords = (MO LO) = (4 − 3) cm = 1 cm

(ii)
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the centre.
Given: AB = 8 cm and CD = 6 cm
Draw OL AB and OM CD.

Join OA and OC.
OA = OC = 5 cm (Radii of a circle)
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL=AB2=82=4cm
Now, in right angled ΔOLA, we have:
OA2 = AL2 + LO2
LO2 = OA2 − AL2
LO2 = 52 42  
 ⇒ LO2 = 25 16 = 9
LO = 3 cm
Similarly, CM=CD2=62=3cm
In right angled ΔCMO, we have:
OC2 = CM2 + MO2
MO2 = OC2 − CM2
MO2 = 52 32
MO2 = 25 9 = 16
MO = 4 cm
Hence, distance between the chords = (MO + LO) = (4 + 3) cm = 7 cm

Question 5:

Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. Find the distance between the chords.

Answer 5:

Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the centre.
Given: AB = 30 cm and CD = 16 cm
Draw OL AB and OM CD.

Join OA and OC.
OA = OC = 17 cm (Radii of a circle)
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL=AB2=302=15 cmAL=AB2=(82)=4cm
Now, in right angled ΔOLA, we have:
OA2 = AL2 + LO2
LO2 = OA2 AL2
LO2 = 172 − 152 
LO2 = 289 − 225 = 64
LO = 8 cm

Similarly, CM=CD2=162=8 cmCM=CD2=(62)=3cm
In right angled ΔCMO, we have:
OC2 = CM2 + MO2
MO2 = OC2 CM2
MO2 = 172 − 82
MO2  = 289 − 64 = 225
MO = 15 cm

Hence, distance between the chords = (LO + MO) = (8 + 15) cm = 23 cm

Question 6:

In the given figure, the diameter CD of a circle with centre O is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, calculate the radius of the circle.

Answer 6:

CD is the diameter of the circle with centre O and is perpendicular to chord AB.
Join OA.

Given: AB = 12 cm and CE = 3 cm
Let OA = OC = r cm   (Radii of a circle)
Then OE = (r - 3) cm
Since the perpendicular from the centre of the circle to a chord bisects the chord, we have:
AE=AB2=122 cm=6 cm
Now, in right angled ΔOEA, we have:
⇒ OA2 = OE2 + AE2
⇒  r2 = (r − 3)2 + 62 
⇒  r2 = r2 − 6r + 9 + 36
r2r2 + 6r = 45
⇒ 6r = 45
r=456 cm=7.5 cm
r = 7.5 cm
Hence, the required radius of the circle is 7.5 cm.

Question 7:

In the given figure, a circle with centre O is given in which a diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. Find the radius of the circle.

Answer 7:

AB is the diameter of the circle with centre O, which bisects the chord CD at point E.
Given: CE = ED = 8 cm and EB = 4 cm
Join OC.

Let OC = OB = r cm   (Radii of a circle)
Then OE = (r − 4) cm
Now, in right angled ΔOEC, we have:
OC2 = OE2 + EC2      (Pythagoras theorem)
⇒  r2 = (r − 4)2 + 82 
⇒  r2 = r2 − 8r + 16 + 64
r2r2 + 8r = 80
⇒ 8r = 80
r=808 cm=10 cm
r = 10 cm
Hence, the required radius of the circle is 10 cm.

PAGE NO-437

Question 8:

In the adjoining figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show that AC || CD and AC = 2 × OD.

Answer 8:


Given:
BC is a diameter of a circle with centre O and OD AB.
To prove: AC parallel to OD and AC = 2 × OD
Construction: Join AC.
Proof:
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
Here, OD AB
D is the mid point of AB.
i.e., AD = BD
Also, O is the mid point of BC.
i.e., OC = OB
Now, in ΔABC, we have:
D is the mid point of AB and O is the mid point of BC.
According to the mid point theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.

AC = 2 × OD

Hence, proved.

Question 9:

In the given figure, O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects ∠BPD. Prove that AB = CD.

Answer 9:


Given:
O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects ∠BPD.
To prove: AB = CD
Construction: Draw OE AB and OF CD
Proof: In ΔOEP and ΔOFP, we have:
∠OEP = ∠OFP         (90° each)
OP = OP                   (Common)
OPE = ∠OPF         (∵ OP bisects ∠BPD )
Thus, ΔOEP ≅ ΔOFP      (AAS criterion)
OE = OF               
Thus, chords AB and CD are equidistant from the centre O.
⇒  AB = CD         (∵ Chords equidistant from the centre are equal)
AB =  CD

Question 10:

Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.

Answer 10:


Given:
AB and CD are two parallel chords of a circle with centre O.
POQ is a diameter which is perpendicular to AB.
To prove: PF CD and CF = FD
Proof:
AB || CD and POQ is a diameter.
PEB = 90°    (Given)
∠PFD = ∠PEB          (∵ AB || CD, Corresponding angles)
Thus, PF CD 
OF CD
We know that the perpendicular from the centre to a chord bisects the chord.
i.e., CF = FD
Hence, POQ is perpendicular to CD and bisects it.

Question 11:

Prove that two different circles cannot intersect each other at more than two points.

Answer 11:

Given: Two distinct circles
To prove: Two distinct circles cannot intersect each other in more than two points.
Proof: Suppose that two distinct circles intersect each other in more than two points.
∴ These points are non-collinear points.
Three non-collinear points determine one and only one circle.
∴ There should be only one circle.
This contradicts the given, which shows that our assumption is wrong.
Hence, two distinct circles cannot intersect each other in more than two points.

Question 12:

Two circles of radii 10 cm and 8 cm intersect each other, and the length of the common chord is 12 cm. Find the distance between their centres.

Answer 12:



Given: OA = 10 cm, O'A = 8 cm and AB = 12 cm

Now, in right angled ΔADO, we have:
OA2 = AD2 + OD2
OD2 = OA2 - AD2
             = 102 - 62
             = 100 - 36 = 64
OD = 8 cm

Similarly, in right angled ΔADO', we have:
O'A2 = AD2 + O'D2
O'D2 = O'A2 - AD2
              = 82 - 62
              = 64 - 36
              = 28
cm
Thus, OO' = (OD + O'D)
                = 
Hence, the distance between their centres is .

Question 13:

Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA = QB.

Answer 13:

Given: Two equal circles intersect at point P and Q.
A straight line passes through P and meets the circle at points A and B.
To prove: QA = QB
Construction: Join PQ.

Proof:
Two circles will be congruent if and only if they have equal radii.
Here, PQ is the common chord to both the circles.
Thus, their corresponding arcs are equal (if two chords of a circle are equal, then their corresponding arcs are congruent).
So, arc PCQ = arc PDQ
∴ ∠QAP = ∠QBP (Congruent arcs have the same degree in measure)
Hence,  QA = QB     (In isosceles triangle, base angles are equal)

Question 14:

If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.

Answer 14:

Given: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at points L and M.
To prove: AB || CD
Proof: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at L and M.

Then OL AB
Also, OM CD
∴ ∠ ALM = ∠ LMD = 90o
Since alternate angles are equal, we have:
AB|| CD     

Question 15:

In the adjoining figure, two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

Answer 15:

Two circles with centres A and B of respective radii 5 cm and 3 cm touch each other internally.
The perpendicular bisector of AB meets the bigger circle at P and Q.
Join AP.

Let PQ intersect AB at point L.
Here, AP = 5 cm
Then AB = (5 - 3) cm = 2 cm
Since PQ is the perpendicular bisector of AB, we have:
AL=AB2=22=1 cm
Now, in right angled ΔPLA, we have:
AP2 = AL2 + PL2
PL2 = AP2 - AL2
            =  52 - 12
            = 25 - 1 = 24
PL=24=26 cm
Thus PQ = 2 × PL
              =  2×26=46cm
Hence, the required length of PQ is 46cm.

PAGE NO-438

Question 16:

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ACD = y° and ∠AOD = x°, prove that x = 3y.

Answer 16:

We have:
OB = OC, ∠BOC = ∠BCO = y
External ∠OBA = ∠BOC + ∠BCO = (2y)
Again, OA = OB, ∠OAB = ∠OBA = (2y)
External ∠AOD = ∠OAC + ∠ACO
Or x = ∠OAB + ∠BCO
Or x = (2y) + y = 3y
Hence, x = 3y

Question 17:

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre then prove that 4q2 = p2 + 3r2.

Answer 17:

Let AC = a.
Since, AB = 2AC, ∴ AB = 2a.


From centre O, perpendicular is drawn to the chords AB and AC at points M and N, respectively.

It is given that OM = p and ON = q.

We know that perpendicular drawn from the centre to the chord, bisects the chord.

AM = MB = a                     ...(1)
and AN = NC = a2                   ...(2)

In ∆OAN,
(AN)2 + (NO)2 = (OA)2          (Pythagoras theorem)
a22+q2=r2a24+q2=r2a2+4q24=r2a2+4q2=4r2a2=4r2-4q2        ....3

In ∆OAM,
(AM)2 + (MO)2 = (OA)2          (Pythagoras theorem)
a2+p2=r2a2=r2-p2       ....4

From eq. (3) and (4),
4r2-4q2=r2-p24r2-r2+p2=4q23r2+p2=4q2

Hence, 4q2 = p2 + 3r2.

Question 18:

In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, OP ⊥  AB and OQAC, prove that PB = QC.

Answer 18:

Given: AB and AC are chords of the circle with centre O. AB = AC, OP ⊥ AB and OQ ⊥ AC


To prove: PB = QC
Proof:
AB = AC      (Given)
12AB=12AC
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ MB = NC            ...(i)
Also, OM = ON    (Equal chords of a circle are equidistant from the centre)
 and OP = OQ (Radii)
⇒ OP - OM = OQ - ON
∴ PM = QN          ...(ii)
Now, in ΔMPB and ΔNQC, we have:
MB = NC                [From (i)]
∠PMB = ∠QNC     [90° each]
PM = QN                [From (ii)]
i.e., ΔMPB ≅ ΔNQC    (SAS criterion)
∴ PB = QC        (CPCT)

Question 19:

In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.

Answer 19:

Given: BC is a diameter of a circle with centre O. AB and CD are two chords such that AB || CD.
TO prove: AB = CD
Construction: Draw OL AB and OM CD.

Proof:
In ΔOLB and ΔOMC, we have:
∠OLB = ∠OMC        [90° each]
∠OBL = ∠OCD          [Alternate angles as AB || CD]
OB = OC                      [Radii of a circle]
∴ ΔOLB ≅ ΔOMC   (AAS criterion)
Thus, OL = OM   (CPCT)
We know that chords equidistant from the centre are equal.
Hence, AB = CD

Question 20:

An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

Answer 20:

Let ΔABC be an equilateral triangle of side 9 cm.
Let AD be one of its median.

Then, AD BC     (ΔABC is an equilateral triangle)
Also, BD=BC2=92=4.5cm
In right angled ΔADB, we have:
AB2 = AD2 + BD2
AD2 = AB2 - BD2
AD=AB2-BD2
         =92-922cm
         =932cm
In the equilateral triangle, the centroid and circumcentre coincide and AG : GD = 2 : 1.
Now, radius = AG=23AD
AG=23×932=33cm
∴ The radius of the circle is 33cm.

Question 21:

In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisector of ∠BAC.

Answer 21:

Given: AB and AC are two equal chords of a circle with centre O.


To prove: ∠OAB = ∠OAC
Construction: Join OA, OB and OC.
Proof:
In ΔOAB and ΔOAC, we have:
AB = AC         (Given)
OA = OA        (Common)
OB = OC         (Radii of a circle)
∴ Δ OAB Δ OAC  (By SSS congruency rule)
∠OAB = ∠OAC    (CPCT)
Hence, point O lies on the bisector of ∠BAC.

Question 22:

In the adjoining figure, OPQR is a square. A circle drawn with centre O cuts the square at X and Y. Prove that QX = QY.

Answer 22:

Given: OPQR is a square. A circle with centre O cuts the square at X and Y.
To prove: QX = QY
Construction: Join OX and OY.

Proof:
In ΔOXP and ΔOYR, we have:
∠OPX = ∠ORY      (90° each)
OX = OY                (Radii of a circle)
OP = OR                (Sides of a square)
∴ ΔOXPΔOYR    (BY RHS congruency rule)
PX = RY              (By CPCT)
PQ - PX = QR - RY   (PQ and QR are sides of a square)
QX = QY
Hence, proved.

PAGE NO-439

Question 23:

Two circles with centres O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through A or B, intersecting the circles at P and Q. Prove that PQ = 2OO'.

Answer 23:

Given: Two circles with centres O and O' intersect at two points A and B.



Draw a line PQ parallel to OO' through B, OX perpendicular to PQ, O'Y perpendicular to PQ, join all.

We know that perpendicular drawn from the centre to the chord, bisects the chord.

PX = XB and YQ = BY

PX + YQ = XB + BY

On adding XB + BY on both sides, we get

PX + YQ + XB + BY = 2(XB + BY)
PQ = 2(XY)
PQ = 2(OO')

Hence, PQ = 2OO'.

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