question bank maths class 10 circles
TRANSCRIPT
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CIRCLES
A. SUMMATIVE ASSESSMENT
10.1 TANGENT TO A CIRCLE
1.A circle may be regarded as a collection of
points in a plane at a fixed distance from a fixed
point. The fixed point is called the centre of the
circle. The fixed distance between the centre of
the circle and the circumference, is calledradius.
2. The perimeter of the circle is referred to as
the circumference of the circle.
3.Achordof a circle is a line segment joining
any two points on the circumference.
4. An arc o f a c i r c l e i s a p a r t o f t h e
circumference.
5. A diameter of a circle is a chord which
passes through the centre of the circle.6. A line, which intersects the circle in two
distinct points, is called a secant.
7.A line which has only one point common to
the circle is called a tangent to the circle.
TEXTBOOKS EXERCISE 10.1
Q.1. How many tangents can a circle have?
Sol. A circle can have infinitely many tangents.
Since there are infinitely many points on a circle and at
each point of it, it has a unique tangent.
Q.2. Fill in the blanks :
(i) A tangent to a circle intersects it in
_______ point(s).
(ii) A line intersecting a circle in two points
is called a _______.
(iii) A circle can have _______ parallel
tangents at the most.
(iv) The common point of a tangent to a
circle and the circle is called _______.
Sol. (i) A tangent to a circle intersects it in one
point(s).
(ii) A line intersecting a circle in two points is
called a secant.
(iii) A circle can have twoparallel tangents at the
most.
(iv) The common point of a tangent to a circle
and the circle is called point of contact.
Q.3.A tangent PQ at a point P of a circle of
radius 5 cm meets a line through the centre O at
a point Q so that OQ = 12 cm. Length PQ is :
(a) 12 cm (b) 13 cm
(c) 8.5 cm (d) 119 cm
Sol. PQ is the tangent and OP is the radius.
OPQ = 90In right angled OPQ,
OQ2 = OP2 + PQ2 (By Pythagoras Theorem)
(12)2 = (5)2 + PQ2
144 = 25 + PQ2
PQ
2
= 144 25 PQ2 = 119
PQ 119 cm
Hence, the length PQ is 119 cm.
So, the correct choice is (d).
Q.4. Draw a circle and two lines parallel to
a given line such that one is a tangent and other,
a secant to the circle.
Sol. According to the given information we draw a
circle having O as centre andlis the given line.
10
Question Bank In Mathematics Class X (TermII)
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Now,mandnbe two lines parallel to a given line l
and that m is tangent as well as parallel to land n is
secant to the circle as well as parallel to l.
TEXTBOOKS EXERCISE 10.2
In Q.1 to 3 choose the correct option and give
justification.
Q.1. From a point Q, the length of the
tangent to a circle is 24 cm and the distance of
Q from the centre is 25 cm. The radius of thecircle is :
(a) 7 cm (b) 12 cm
(c) 15 cm (d) 24.5 cm
Sol.QPO = 90 [Angle between tangent andradius through the point of contact]
In right angledQPO, OQ2 = OP2 + PQ2
[By Pythagoras Theorem]
(25)2 = OP2 + (24)2
625 = OP2 + 576
OP2 = 625 576
OP2 = 49
OP 49 7 cm
So, radius of the circle is 7 cm.
Hence, the correct option is (a).
Q.2. In figure, if TP and TQ are the two
tangents to a circle with centre O so thatPOQ
= 110, then PTQ is equal to :
(a) 60 (b) 70
(c) 80 (d) 90
Sol.POQ = 110 (Given)
OPT = 90 andOQT = 90
[Angle between tangent and radius through
the point of contact is 90]
In quadrilateral OPTQ,POQ +OPT +OQT +PTQ = 360
[The sum of all the angles of a quadrilateral
is 360]
110 + 90 + 90 +PTQ = 360
290 +PTQ = 360
PTQ = 360 290 = 70
Hence, the correct choice is (b).
Q.3. If tangents PA and PB from a point P to
a circle with centre O are inclined to each other
10.2 NUMBER OF TANGENTS FROM A
POINT ON A CIRCLE
1. There is one and only one tangent at a
point of the circle.
2. The tangent at any point of a circle isperpendicular to the radius through the
point of contact.
3. No tangent can be drawn from a point
inside the circle.
4. The lengths of tangents drawn from an
external point to a circle are equal.
5. The perpendicular at the point of contact to
the tangent to a circle passes through thecentre of the circle.
6. Tangents drawn at the end points of a
diameter of a circle are parallel.
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at angle of 80, then POA is equal to :
(a) 50 (b) 60 (c) 70 (d) 80
Sol.OAP = 90 [Angle between tangent andradius through the point of contact]
1OPA BPA
2 [The centre lies on the bisector
of the angle between the two tangents]1
(80) 402
InOPA,OAP +OPA +POA = 180
[Sum of the three angles of a triangle is 180]
90 + 40 +POA = 180
130 + POA = 180
POA = 180 130
POA = 50
Hence, the correct choice is (a).
Q.4. Prove that the tangents drawn at the
ends of a diameter of a circle are parallel.
[2011 (T-II)]
Sol. PQ and RS are tangents to the circle C(O, r) atpoints A and B respectively, AB being diameter.
Since tangent is perpendicular to the radius at the
point of contact.
ABPQ and ABRS
PAB = 90 andABS = 90
PAB =ABS
PQ || RS [ PAB andABS are
alternate angles]Proved.
Q.5. Prove that the perpendicular at the
point of contact to the tangent to a circle passes
through the centre.
Sol. We know that the tangent at any point of acircle is perpendicular to the radius through the point of
contact and the radius essentially passes through the
centre of the circle, therefore the perpendicular at the
point of contact to the tangent to a circle passes through
the centre.
Q.6. The length of a tangent from a point A
at distance 5 cm from the centre of the circle is
4 cm. Find the radius of the circle. [Imp.]
Sol. We knowthat the tangent at any point of a circleis perpendicular to the radius through the point of
contact.
OPA = 90
In right angled
OPA, OA2
= OP2
+ AP2
[By Pythagoras Theorem]
(5)2 = OP2 + (4)2 25 = OP2 + 16
OP2 = 25 16 = 9 OP 9 3 cm
Hence, the radius of the circle is 3 cm.
Q.7.Two concentric circles are of radii 5 cm
and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.
[Imp.]
Sol. Let O be the common centre of the twoconcentric circles.
Let AB be a chord of the larger circle which
touches the smaller circle at P.
Join OP and OA.
Then, OPA = 90
[ Th e t a n g e n t a t a n y p o i n t o f a c i rc l e i s
perpendicular to the radius through the point of contact]
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OBP = 90 ...(ii)
OAPB is a quadrilateral.
APB + AOB +OAP +OBP = 360
Putting the value ofOAP andOBP, we have
APB +AOB + 90 + 90 = 360
APB +AOB = 180
APB andAOB are supplementary.
Q.11.Prove that the parallelogram circums-
cribing a circle is a rhombus.
[2009, 2011 (T-II)]
Sol. Given : ABCD is a parallelogram circums-cribing circle.
To Prove : ABCD is a rhombus.Proof : Since the tangent segments from an
external point to a circle are equal.
AP = AS ...(i)
BP = BQ ...(ii)
CR = CQ ...(iii)
and DR = DS ...(iv)
Adding equations (i), (ii), (iii) and (iv), we get
(AP + BP) + (CR + DR)
= (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = AD + AD
2 AB = 2AD
AB = AD
But AB = CD and AD = BC
[Opposite sides of parallelogram] AB = BC = CD = AD
Hence, parallelogram ABCD is a rhombus.
Q.12. A t ri an gl e ABC i s d rawn t o
circumscribe a circle of radius 4 cm such that
the segments BD and DC into which BC is divided
by the point of contact D are of length 8 cm and
6cm respectively. (see figure). Find the sides AB
and AC. [HOTS]
Sol. Let the sides BC, CA and AB of the ABCtouch the circle at D, F and E respectively.
Join OE and OF. Also join OA, OB and OC.
Now, BD = BE = 8 cm and CF = CD = 6 cm
[ Tangent segments from an external point to a
circle are equal]
Also, let AE = AF =x cm. Then,
a= 14, b= x + 6, c= x + 8
Semi-perimeter ofABC2
a b c
14 ( 6) ( 8)( 14) cm
2
x xx
Area ofABC ( 14)( 14 14)x x
( 14 8)( 14 6) x x x x
( 14)( )(6)(8) x x
Now, area ofABC
= Area ofOBC + Area ofOCA
+ Area ofOAB
( 14)( )(6)(8)x x
(6 8)4 ( 6)4 ( 8)4
2 2 2
x x
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( 14)( )(6)(8) 28 2 12 2 16x x x x
( 14)( )(6)(8) 4 56x x x ( 14)( )(6)(8) 4( 14)x x x Squaring both sides, we have
(x+ 14)(x)(6)(8) = 16(x+ 14)2
3x= x + 14 3xx = 14
2x= 14 14
72
x
AB = x + 8 = 7 + 8 = 15 cm
and AC =x + 6 = 7 + 6 = 13 cm
Q.13. P ro ve th at o pp o site s ide s o f a
quadrilateral circumscribing a circle subtend
supplementary angles at the centre of the circle.[2011 (T-II)]
Sol. Given : ABCD is a quadrilateral circums-cribing a circle whose centre is O.
To Prove :(i)AOB +COD = 180
(ii)BOC +AOD = 180
Construction :Join OP, OQ, OR and OS.
Proof : Since tangent segments from an external
point to a circle are equal, therefore
AP = AS, BP = BQ, CQ = CR, DR = DS
InOPB andOBQ,
OP = OQ [Radii of the same circle]
OB = OB [Common]
BP = BQ [Proved above]
OBP OBQ [SSS congruence criterion]
1 =2 [CPCT]
Similarly, 3 =4,5 =6 and7 = 8
So, 1 +2 +3 +4 +5 +6 +7 +
8 = 360
1 +1 +4 +4 +5 +5 +8 +
8 = 360
2(1 +4 +5 +8) = 360
1 +4 +5 +8 = 180
(1 +8) + (4 +5) = 180
AOB +COD = 180
Similarly, we can prove that
BOC +AOD = 180. Proved.
OTHER IMPORTANT QUESTIONS
Q.1. The common point of the tangent and
the circle is called the :
(a) point of contact
(b) point of intersection
(c) point of concurrence
(d) none of the above
Sol.(a) The common point of the tangent and the
circle is called the point of contact.
Q.2. How many tangents of a circle passingthrough a point lying inside the circle?
(a) zero (b) one
(c) two (d) infinite
Sol. (a) There is no tangent to a circle passing
through a point lying inside the circle.
Q.3. How many tangents can we draw to a
circle from a point lying outside the circle?
(a) 0 (b) 1 (c) 2 (d) infinite
Sol. (a) There are exactly 2 tangents to a circle
through a point lying outside the circle.
Q.4. The tangents drawn at the ends of a
diameter of a circle are :
(a) parallel (b) perpendicular
(c) inclined at an angle of 60
(d) none of the above
Sol. (a) The tangents drawn at the ends of a
diameter of a circle are parallel.
Q.5. In the given figure, PT is a tangent to
the circle with centre O. If OT = 6 cm and
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OP = 10 cm, then the length of tangent PT is :
(a) 8 cm (b) 12 cm
(c) 10 cm (d) 16 cm
Sol. (a) By Pythagoras Theorem in OPT,
PT2 = OP2 OT2
PT2 = (10)2 (6)2 = 64 PT = 8 cm
Q.6. In a circle of radius 7 cm, tangent PTis drawn from a point P such that PT = 24 cm.
If O is the centre of the circle, then length of OP
is:
(a) 30 cm (b) 28 cm
(c) 25 cm (d) 31 cm
Sol.(c) OP2 = PT2 + OT2 = (24)2 + (7)2
= 576 + 49 = 625
OP 625 25 cm
Q.7.A point P is 26 cm away from the centre
of a circle and the length of tangent drawn from
P to the circle is 24 cm. The radius of the circleis :
(a) 8 cm (b) 10 cm
(c) 12 cm (d) 14 cm
Sol. (b) Let O be the centre of the circle and PT be
the tangent. Then, radius of the circle is OT.
OT2 = OP2 PT2 = (26)2 (24)2
= 676 576 = 100
OT2 = 100 OT 100 10 cm
Q.8. In the given figure, O is the centre of
two concentric circles of radii 3 cm and 5 cm.
AB is a chord of outer circle which touches theinner circle. The length of chord AB is :
(a) 4 cm (b) 8 cm
(c) 34 cm (d) 7 cm
Sol.(b) Draw OP AB. Join OA.Then, OP = 3 cm and OA = 5 cm.
OA2 = OP2 + AP2
AP2 = OA2 OP2 = 52 32
AP2 = 16 AP = 16 4 cm
AB = 2 AP = (2 4) cm = 8 cm
Q.9. In the figure, ABC is circumscribed
touching the circle at P, Q, R. If AP = 4 cm, BP
=6 cm, AC =12 cm and BC = x cm. Then, x = ?
(a) 10 cm (b) 6 cm
(c) 14 cm (d) 18 cm
Sol. (c) Since, the lengths of tangents drawn froma n e xt e rn al p oi nt t o a c ir c le a re e q ua l, t he n
AR = AP = 4 cm
CR = (AC AR) = (12 4) cm = 8 cm = CQ
And, BQ = BP = 6 cm
BC = BQ + CQ = (6 + 8) cm = 14 cm.
Q.10. In the given figure, quad. ABCD is
circumscribed, touching the circle at P, Q, R and
S. If AP =5 cm, BC =7 cm and CS = 3 cm, then
length AB = ?
(a) 9 cm (b) 10 cm
(c) 12 cm (d) 8 cm
Sol. (a) Since the lengths of tangents drawn froman external point to a circle are equal, then
AQ = AP = 5 cm
CR = CS = 3 cm, BR = (BC CR)
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= (7 3) cm = 4 cm
BQ = BR = 4 cm
AB = AQ + BQ = (5 + 4) cm = 9 cm.
Q.11. I n t he g iv en f i gu re , A BC D i s
circumscribed touching the circle at P, Q, R and
S. If AP = 6 cm, BP =5 cm, CQ = 3 cm and DR
= 4 cm, then perimeter of quad. ABCD is :
(a) 18 cm (b) 27 cm
(c) 36 cm (d) 22 cm
Sol. (c) Since the length of tangents from ane xt er na l p oi nt t o a c ir cl e a re e qu al , t he n
AP = AS = 6 cm, BP = BQ = 5 cm,
CR = CQ = 3 cm, DS = DR = 4 cm
Perimeter of quad. ABCD
= (AP + BP) + (BQ + CQ) + (CR + DR)
+ (AS + DS) = 36 cm.
Q.12. In the given figure, quad. ABCD is
circumscribed, touching the circle at P, Q, R and
S such thatDAB =90. If CS = 27 cm and CB
= 38 cm and the radius of the circle is 10 cm,
then AB = ?
(a) 17 cm (b) 28 cm(c) 19 cm (d) 21 cmSol.(d) Since the lengths of tangents drawn from
an external point to the circle are equal, thenCR = CS = 27 cm.So, BR = (BC CR) = (38 27) cm = 11 cm BQ = BR = 11 cm.Join OQ. Then, PAQO is a rectangle. AQ = PQ = 10 cm.Hence, AB = (AQ + BQ) = (10 + 11) cm = 21 cm.
Q.13. In the given figure, ABC is right-
angled at B such that BC = 6 cm and AB = 8 cm.
A circle with centre O has been inscribed inside
th e tr ia n gle . O P A B, O Q B C a nd
OR C. If OP = OQ = OR = x cm, then x = ?
[2011 (T-II)]
(a) 2 cm (b) 3 cm
(c) 2.5 cm (d) 4 cm
Sol.(a) AC2 = (AB2 + BC2) = 82 + 62
= 64 + 36 = 100
AC = 100 = 10 cmNow, CR = CQ = BC BQ = (6 x) cm
and AR = AP = (AB BP) = (8 x) cm
AC = CR + AR = (6 x) + (8 x)
= (14 2x)
14 2x= 10 x= 2 cm
Hence, radius of the circle = 2 cm
Q.14. Quadrilateral ABCD circumscribes a
circle as shown in figure. The side of the
quadrilateral which is equal to AP + BR is :
[2011 (T-II)]
(a) AD (b) AC
(c) AB (d) BC
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Sol.(c) Since the lengths of tangents drawn from
an external point to the circle are equal, then
AP = AQ and BR = QB.
AP + BR = AQ + QB = AB [AQ + QB = AB]
Q.15. In figure, if OC = 9 cm and OB = 15
cm, then BC + BD is equal to : [2011 (T-II)]
(a) 18 cm (b) 12 cm
(c) 24 cm (d) 36 cm
Sol.(c) InOCB, by Pythagoras theorem
CB = 2 2(15) (9) 225 81 144 = =
CB = 12 cm
BD = BC = 12 cm [length of tangents
drawn from an external point to the circle are equal]
Now, BC + BD = 12 cm + 12 cm
= 24 cm
Q.16. A tangent PA is drawn from an
external point P to a circle of radius 3 2 cmsuch that the distance of the point P from O is
6cm as shown in figure. The value ofAPO is:
[2011 (T-II)]
(a) 30 (b) 45
(c) 60 (d) 75
Sol.(b) LetAPO =.
In rightOAP, we have
sin=OA
OP
3 2 1sin
6 2 = =
45 =
Q.17. The figure, shows two concentric
circles with centre O. AB and APQ are tangents
to the inner circle from point A lying on the outer
circle. If AB = 7.5 cm, then AQ is equal to :
[2011 (T-II)]
(a) 18 cm (b) 15 cm
(c) 12 cm (d) 10 cm
Sol.(b) AP = AB = 7.5 cm [since the length of
tangents from an external
point to a circle are equal]
Now, AQ = 2AP = 2 7.5 cm = 15 cm
[since perpendicular from the centre bisects the chord]
Q.18. In figure, APB is a tangent to a circle
with centre O, at point P. If QPB = 50, then
the measure of POQ is : [2011 (T-II)]
(a) 120 (b) 100 (c) 140 (d) 150
Sol.(b) OPB = 90 [the tangent at any point
of a circle is perpendicular to the radius through the
point of contact]
OPQ = 90 QPB = 90 50 = 40
Now, since OP = OQ (radii)
OQP =OPQ = 40
40 + 40 +POQ = 180
POQ = 180 80 = 100
Q.19. In the given figure, the length of PR
is: [2011 (T-II)]
(a) 20 cm (b) 26 cm
(c) 24 cm (d) 28 cm
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Sol.(b) In rightOSR, OR2 = 52 + 122 = 169
OR = 169 OR = 13 cmIn rightPQO,
PO2 = 42 + 32 = 25PO= 25
PO= 5 cm
Now, OO= (3 + 5) cm = 8 cm
PR = PO + OO+ OR = 5 cm + 8 cm + 13 cm
= 26 cm
Q.20. CP and CQ are tangents to a circle
with centre O. ARB is another tangent touching
the circle at R. If CP = 11cm, BC = 7 cm, then
the length BR is : [2011 (T-II)]
(a) 11 cm (b) 7 cm
(c) 3 cm (d) 4 cm
Sol.(d) CP = CQ [length of tangents from
an exterior point to a circle are equal]
11 cm = QB + 7 cm
QB = 11 cm 7 cm = 4 cm BR = QB = 4 cm [length of tangents from
an exterior point to a circle are equal]
Q.21. In the given figure, O is the centre of
the circle. If PA and PB are tangents from an
external point P to the circle, then AQB is
equal to : [2011 (T-II)]
(a) 100 (b) 80
(c) 70 (d) 50
Sol.(d)PAO =PBO = 90
[Angle between tangent and
radius through the point of contact]
AOB = 360 (80 + 90 + 90)
= 360 260 = 100
Now,AQB =1
2AOB
[The angle subtended by an arc at the centre is
double the angle subtended by it at any point on the
remaining part o f the circle]
1100 50
2= =
Q.22. In the given figure, if AOB = 125,
then COD is equal to : [2011 (T-II)]
(a) 62.5 (b) 45
(c) 125 (d) 55
Sol. (d) If a c i rc l e t o u c h e s t h e si d e s o f a
quadrilateral, then the angles subtended at the centre by
a pair of opposite sides are supplementary. AOB +COD = 180
COD = 180 125 = 55
Q.23.In the given figure, AT is the tangent to the
circle with centre O such that OT = 4 cm and
OTA =30. Then AT is equal to : [2011 (T-II)]
(a) 4 cm (b) 2 cm
(c) 2 3 cm (d) 8 cm
Sol.(c) Let AT = x cm
cos 30 = 3
4 2 4 =
x x
x= 2 3 cm
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Q.24. In the given figure, CP and CQ are
tangents from an external point C to a circle with
centre O. AB is another tangent which touches
the circle at R. If CP = 11 cm and BR = 4 cm,
find the length of BC. [2010]
Sol.We have, CP = 11 cm, BR = 4 cm
Since, the tangent from an external point are equal,
we have CQ = CP = 11 cm and BQ = BR = 4 cm
BC = CQ BQ = (11 4) cm = 7 cm
Q.25. In the given figure, there are two
concentric circles with centre O and of radii 5
cm and3 cm. From an external point P, tangents
PA and PB are drawn to these circles. If AP =
12 cm, find the length of BP. [2010]
Sol.Join OA, OB and OP.
Then OA = 5 cm, OB = 3 cm and PA = 12 cm
InOPA, OP2 = OA2 + AP2
= 52 + 122
= 25 + 144 = 169
OP = 169 = 13 cm
Now, inOBP, OP2 = OB2 + BP2
BP2 = OP2 OB2 = (13)2 (3)2
= 160
= BP = 160 cm = 4 10 cm
Q.26. In the given figure, if ATO = 40,
findAOB. [2008]
Sol. InOAT andOBT,
4 =1 = 90 [Tangent isto the
radius through the point of contact]
OT = OT [Common]
OA = OB [Radius]
OAT OBT [RHS criterion] 3 =2 ....(i) [CPCT]
InOAT,3 +4 +5 = 180
[Angle sum property of a]
3+ 90 + 40 = 180
3 + 130 = 180 3 = 50 AOB =2 +3 =3 +3 [From (i)]
= 50 + 50 = 100.
Q.27. From a point P, the length of the
tangent to a circle is 15 cm and distance of P
from the centre of the circle is 17 cm. Then what
is the radius of the circle? [2008C]
Sol.OAP = 90
[Radius isto the tangent through the point of
contact]
In rightOAP, OA2 + AP2 = OP2
[Pythagoras Theorem]
r2 + (15)2 = (17)2 r2 + 225 = 289
r2 = 289 225 = 64r= 8 cm.
Q.28. The two tangents from an external
point P to a circle with centre O are PA and PB.
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If PB = 70, what is the value of OB?
[2008C]
Sol. 1 =2 = 90[Tangent isto the radius through the point of
contact]
In quad. OAPB,
AOB +1 +APB +2 = 360
[Angle sum property of a quad.] AOB + 90 + 70 + 90 = 360
AOB + 250 = 360
AOB = 360 250 = 110.
Q.29. In the figure, ABC is circumscribing
a circle. Find the length of BC. [2009]
Sol. We know that the lengths of tangents drawn
from an external point to a circle are equal.
So, AR = AQ = 4 cm ...(i)
BR = BP = 3 cm ...(ii)
CP = CQ = (AC AQ) = (11 4) cm
= 7 cm
So, BC = BP + CP = (3 + 7) cm = 10 cm.
Q.30. In the figure, CP and CQ are tangents
to a circle with centre O. ARB is another tangent
touching the circle at R. If CP = 11 cm and
BC = 7 cm, then find the length of BR. [2009]
Sol. We know that the lengths of tangents drawn
from an external point to a circle are equal.
So, CP = CQ ... (i)
AP = AR ... (ii)
BQ = BR ... (iii)
Now, CP = CQ
11= BC + BQ
11= 7 + BR [From (iii)]
BR= 11 7 = 4 cm.
Q.31. Two tangents PA and PB are drawn to
a circle with centre O from an external point P.
Prove thatAPB = 2OAB.
[2009, 2011 (T-II)]
Sol. OAP = 90 [Radius through the point
of contact is perpendicular to the tangent]
PAB = 90 OAB
InAPB, PA = PB [Tangents from an
external point to a circle are equal]
PBA =PAB = 90 OAB ...(i)
Also, APB = 180 (PBA +PAB)
= 180 2 (90 OAB) APB = 180 180 + 2OAB
APB = 2OAB Proved.
Q.32. ABC is an isosceles triangle, in which
AB = AC, circumscribed about a circle. Show
that BC is bisected at the point of contact.
[2008, 2011 (T-II)]
Sol. Given :An isosceles triangle ABC, in which
AB = AC, circumscribed about a circle.
To prove :BF = FC
Prove :We know that the tangents drawn from an
external point to a circle are equal.
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So, AD = AE (i)
CE = CF (ii)
BD = BF (iii)
Also, AB = AC [Given]
AD+ BD = AE + CE
DB = CE (iv) [From (i)]
From (ii), (iii) and (iv), we have BF = FC
Hence, BC is bisected at the point of contact.
Proved.
Q.33. In the figure a circle is inscribed in a
quadrilateral ABCD in which B = 90. If AD =
23 cm, AB = 29 cm and DS = 5 cm, find the
radius (r) of the circle. [2008]
Sol. We know that the tangents drawn from an
external point to a circle are equal.
AR = AQ (i)
DR = DS (ii)
BQ = BP (iii)AD = AR + DR
AD = AR + DS [From (ii)]
23 = AR + 5
AR = 23 5 = 18 cm
AB = AQ + BQ
AB = AR + BQ [From (i)]
29 = 18 + BQ
BQ = 29 18 = 11 cm (iv)
In quadrilateral OPBQ,
B = 90
OQB = OPB = 90
[Tangent is perpendicular to the radius at the point
of contact]
BQ = BP [From (iii)]
OQ = OP =r [Given]
OPBQ is a square.
OP = OQ = PB = BQ
r = 11 cm [From (iv)]
Q.34. In the figure, OP is equal to diameter
of the circle. Prove that ABP is an equilateral
triangle. [2008, 2011 (T-II)]
Sol. Suppose OP meets the circle at Q. Join QA
and AB.
We have, OP = diameter
OQ + PQ = diameter
PQ = diameter radius.
PQ = radius
So, OQ = PQ = radius
Thus, OP is the hypotenuse of right triangle OAP
and Q is the mid-point of OP.
OA = AQ = OQ
[ Mid-point of hypotenuse of a right triangle is
equidistant from the vertices]
OAQ is an equilateral triangle.
AOQ = 60
So, APO = 30
APB = 2APO = 60
Also, PA = PB PAB = PBA
But, APB = 60
Therefore, PAB = PBA= 60
Hence, APB is equilateral. Proved.
Q.35. In the figure, ADC = 90, BC =
38 cm, CD = 28 cm and BP = 25 cm. Find the
radius of the circle. [2006, 2011 (T-II)]
P
P
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Sol. Since tangent to a circle is perpendicular to
the radius through the point of contact.
OSD = ORD = 90, OR = OS
DROS is a square.
Als,o BP = BQ
[Tangents from an external point are equal]
BQ = 25 cm [BP = 25 cm]
BC CQ = 25
38 CQ = 25 [BC = 38 cm]
CQ = 38 25 = 13 cm
CR = CQ = 13 [CQ = 13 cm]
CD DR = 13 [CR = CD DR]
28 DR = 13 [CD = 28 cm]
DR = 28 13 = 15 cm
Since DROS is a square, so OR = DR = 15 cm.
Hence, radius of the circle = 15 cm.
Q.36. PQR is a right-angled triangle with
QR =12 cm and PQ = 5 cm. A circle with centre
O is inscribed in PQR. Find the radius of the
circle.
Sol.
Since OA PQ, OB QR and OA = OB
AOBQ is a square.
In right-angled triangle PQR,
PR2 = PQ2 + QR2
[Pythagoras theorem]
PR2 = 52 + 122
PR2 = 25 + 144
PR2 = 169
PR = 13 cm
PA = PC [Tangents drawn
from an external point are equal]
PQ AQ = PR RC
5 r= 13 RC
RC = 13 5 +r
RC = 8 +r
RB = RC [Tangents drawn
from an external point are equal]
QR BQ = RC 12 r= 8 + r
2r= 4 r= 2 cm
Hence, radius of the circle is 2 cm.
Q.37. A circle touches the side BC ofABC,
at P and touch AB and AC produced at Q and R
respectively. Prove that AQ =1
2Perimeter of
ABC. [2006, 2011 (T-II)]
Sol.Since tangents from an exterior point to acircle are equal in length.
BP = BQ ...(i)
CP = CR ...(ii)
and AQ = AR ...(iii)
Now, perimeter ofABC
= AB + BC + AC= AB + (BP + PC) + AC
= (AB + BQ) + (AC + CR)
[From (i) and (ii)]
= AQ + AR = 2AQ [AQ = AR]
So, AQ =1
2(Perimeter ofABC).
Q.38. In the figure, O is the centre of the
circle, PA and PB are tangent segments. Show
that the quadrilateral AOBP is cyclic.
[2011 (T-II)]Sol.OA AP and OB BP
[Tangent to a circle is perpendicular to the radius at
the point of contact.]
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OAP = 90 and OBP = 90
OAP + OBP = 90 + 90 = 180 ...(i)
OAPB is cyclic because in a quadrilateral if a
pair of opposite angles are supple mentary, then the
quadrilateral is cyclic. Proved.
Q.39. In the figure, circles C(O, r) and
C(O,2
r)touch internally at a point A and AB is
a chord of the circle C(O, r), intersect ing
C(O,2
r) at C. Prove that AC = CB. [HOTS]
Sol.Join OC, OA.
OCA is the angle in a semicircle.
OCA = 90 ... (i)
OC ABIn circle C(O,r), AB is chord and OC AB
[From (i)]
AC = CB
[Perpendicular from centre of circle to a chord
bisects the chord] Proved.
Q.40. If two tangents are drawn to a circle
from an external point, then (i) they subtends
equal angles at the centre, (ii) they are equally
inclined to the segment joining the centre to that
point. Prove. [HOTS]
Sol.
Given : PT and PT are two tangents drawn to a
circle C (O,r) from an external point P.
To prove : (i) POT = POTand
(ii) OPT = OPT
Proof :In POT and POT
PT = PT
[Tangents drawn from an external point]
OT = OT [Radii of the same circle]
PTO =PTO [Each = 90]
POT POT [SAS congruency]
POT = POT
and OPT = OPT [CPCT]
Q.41. In the figure below, the incircle of
ABC touches the sides BC, CA and AB at D, E
and F respectively. Show that :AF + BD + CE = AE + BF + CD
=1
2(Perimeter of ABC)
[Imp.]
Sol. We know that the tangents from an exterior
point to a circle are equal in length.
AF = AE ...(i)
BD = BF ...(ii)
CE = CD ...(iii)
Adding (i), (ii) and (iii), we get
AF + BD + CE = AE + BF + CD
Perimeter ofABC = AB + BC + AC
= (AF + FB) + (BD + CD) + (AE + EC)= (AF + AE) + (BF + BD) + (CD + CE)
= (2AF + 2BD + 2CE)
= 2(AF + BD + CE) [From (i), (ii) and (iii)]
AF + BD + CE =1
2 (Perimeter ofABC)
Hence, AF + BD + CE = AE + BF + CD
=1
2 (Perimeter ofABC) Proved.
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Q.42. In the figure, if AB = AC, prove that
BE = EC. [2011 (T-II)]
Sol. We know that the tangents from an exterior
point to a circle are equal in length.
AD = AF ...(i)
BD = BE ...(ii)
CE = CF ...(iii)
Now, AB = AC
AB AD = AC AD
[Subtracting AD from both sides]
AB AD = AC AF [Using (i)]
BD = CF
BE = CE Proved.
[Using (ii), (iii)]
Q.43. PQ is a chord of length 8 cm of acircle of radius 5 cm. The tangents at P and Q
intersect at a point T. Find the length TP.
[HOTS]
Sol.Let TR = y.
Since OT is perpendicular bisector of PQ.
PR = QR = 4 cm
In right triangle ORP, we have
OP2 = OR2 + PR2
OR2 = OP2 PR2 = 52 42 = 9
OR = 3 cm
In right triangles PRT and OPT, we have
TP2
= TR2
+ PR2
...(i)and, OT2 = TP2 + OP2
OT2 = (TR2 + PR2) + OP2 [From (i)]
(y+ 3)2 =y2 + 16 + 25
6y = 32
y =16
3
TR =16
3
From (i), we have
TP2 =
2216 256 4004 16
3 9 9
TP = 20
3cm
Q.44. Let A be one point of intersection of
two intersecting circles with centres O and Q.
The tangents at A to the two circles meet the
circles again at B and C, respectively. Let the
po int P be loc at ed so tha t AO PQ is a
parallelogram. Prove that P is the circumcentreof the triangle ABC. [HOTS]
Sol. If P is the circumcentre ofABC, then OP and
PQ are perpendicular bisectors of AB and AC
respectively. i.e., we have to show that OP and PQ are
perpendicular bisectors of AB and AC respectively.
We have, OA AC
[ Tangent to a circle isto the radius through
the point of contact]
PQ AC
[ OAQP is a parallelogram OA || PQ]
PQ is the perpendicular bisector of AC
[ Q is the centre of the circle]
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Similarly,
BA AQ
BA OPAQPO is a parallelogram
OP || AQ
OP is the perpendicular bisector of AB.
Thus, P is the point of intersection of perpendicular
bisectors PQ and PO of sides AC and AB respectively.
Hence, P is the circumcentre ofABC.Proved.
Q.45. If AB, AC, PQ are tangents in figure
below and AB = 5 cm, find the perimeter of
APQ. [2011 (T-II)]
Sol.We have, PB = PX, QC = QX and AB = AC
Perimeter ofAPQ
= AP + PQ + AQ
= AP + (PX + XQ) + AQ
= (AP + PX) + (AQ + XQ)
= (AP + PB) + (AQ + QC)
= AB + AC = (5 + 5) cm = 10 cm
Q.46. In figure below, PQ is tangent at a
point R of the circle with centre O.
IfTRQ = 30, findPRS. [Imp.]
Sol.We have,TRQ = 30
Since ST is a diameter and angle in a semicircle
is a right angle.
SRT = 90
Now,TRQ +SRT +PRS = 180
[Straight angle]
30 + 90 +PRS = 180
PRS = 180 120 = 60.
Q.47.In figure below, a circle touches all the
four sides of a quadrilateral ABCD with AB = 6
cm, BC = 7 cm and CD = 4 cm. Find AD.
[2011 (T-II)]
Sol. Since tangents drawn from an exterior point to
the circle are equal.
AP = AS ...(1)
BP = BQ ...(2)
CR = CQ ...(3)
and DR = DS ...(4)
Adding (1), (2), (3) and (4), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR)
= (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
6 + 4 = AD + 7 AD = 3 cm.
Q.48. In the figure, two circles touch each
other externally at C. Prove that the common
tangent at C bisects the other two common
tangents. [2007, 2011 (T-II)]
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Sol. Given : Two circles with centres at A and B
touch each other externally at C. RS and PQ are two
common tangents to the circles.
To prove : (i) PE = EQ
(ii) RF = FS.
Proof :We know that the tangents drawn from an
exterior point are equal.
PE = EC (1)
Similarly, EQ = EC (2)
PE = EQ [From (1) and (2)]
EC bisects PQ at E.
Similarly, we can prove that F is the bisector of RS.
So, EF bisects PQ and RS. Proved.
Q.49. Prove that the lengths of tangents
drawn from an external point to a circle are
equal. Using the above, prove the following :
q ua dr il at er al A BC D i s d ra wn t o
circumscribe a circle (see fig.). Prove that
AB + CD = AD + BC. [2008]
Sol.Part I
Given : PT and PT are two tangents drawn from an
external point P to a circle C (O,r).
To prove :PT = PT
Construction :Join TO, TO and PO.
Proof :InPOT andPOT
PO =PO (Common)
OT = OT (Radii of the same circle)
PTO =PTO = 90 [OT PT and OT PT']
POT POT (RHS congruence)
PT = PT (CPCT)Proved.
Part II :See Q.8(Textbooks Exercise10.2)
Q.50. Prove that the lengths of the tangents
drawn from an external point to a circle are
equal.
Using the above, do the following :
In the figure, TP and TQ are tangents from T
to the circle with centre O and R is any point on
the circle. If AB is a tangent to the circle at R,
prove that
TA + AR = TB + BR. [2008, 2011 (T-II)]
Sol. Part I : Same as Q.49.
Part II
From part I, we have
TP = TQ (i)
AP = AR (ii)
BQ = BR (iii)
From (i), TA + AP = TB + BQ
TA + AR = TB + BR [From (ii) and (iii)]
Proved.
Q.51. Prove that the tangent at any point of
a circle is perpendicular to the radius through
the point of contact.
Using the above, do the following :
In the figure, O is the centre of the two
concentric circles. AB is a chord of the larger
circle touching the smaller circle at C. Prove
that AC = BC. [2009, 2011 (T-II)]
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1. A point P is 13 cm from the centre of a
circle. Radius of the circle is 5 cm. Then,
the length of the tangent drawn from P to the
circle is :
(a) 10 cm (b) 11 cm(c) 12 cm (d) 13 cm
2. The length of the tangent drawn from a
point, whose distance from the centre of a circle
is 20 cm and radius of the circle is 16 cm is :
(a) 12 cm (b) 144 cm
(c) 169 cm (d) 25 cm
3. A tangent PQ at a point P of a circle of
radius 15 cm meets a line through the centre O
at a point Q so that OQ = 25 cm. Length of PQ
is:
(a) 5 cm (b) 25 cm
(c) 16 cm (d) 20 cm4. In the figure, if AD, AE and BC are
tangents to the circle at D, E and F respectively,
then :
(a) AD = AB + BC + CA
(b) 2AD = AB + BC + CA
(c)AD
2 = AB + BC + CA
(d) AD
4= AB + BC + CA
5. Two circles touch each other externally at
C and AB is a common tangent to the circles.
ACB is :
(a) 60 (b) 45 (c) 90 (d) 180
6. PQ is a tangent drawn from a point P to
a circle with centre O and QOR is a diameter of
the circle such thatPOR = 120.OPQ is :
(a) 50 (b) 40 (c) 30 (d) 25
7. In the figure, a circle touches all the foursides of a quadrilateral ABCD, whose three sides
are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find
AD. [2011 (T-II)]
PRACTICE EXERCISE 10.1A
Sol.Part I :
Given : A circle (O,r) and a tangent XY at P.
To prove :OP XY
Construction :Take any point Q on XY. Join OQ
which intersects the circle at R.
Proof : OP = OR (Radii of the circle)
Now, OQ = OR + RQ
OQ > OR
OQ > OP (OP = OR)
OP < OQ
Thus, OP is shorter than any other point on XY.
Hence, OPXY
(Perpendicular distance is the shortest distance
from a point to a line) Proved.
Part II :
Join OC. Then,
OCB = 90 [From part I]
For outer circle, AB is a chord
AC = BC
[Perpendicular from the centre bisects the chord]
Proved.
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17.In the figure, O is the centre of the circle,
AB is a diameter and AD is a tangent. IfBOC
= 120, findADC.
18. In the figure, PA and PB are the tangentsegments to a circle with centre O. Show that A,
O, B, P are concyclic.
19. Prove that in two concentric circles, the
chord of the larger circle which touches the
smaller circle is bisected at the point of contact.20. Prove that the line segment joining the
point of contact of two parallel tangents to a
circle is a diameter of the circle. [2011 (T-II)]
21. Prove that there is one and only one
tangent at any point on the circumference of a
circle.
22. From a point P, two tangents PA and PB
are drawn to a circle C(O, r). If OP = 2r, show
thatAPB is equilateral. [Imp.]
23. In the given figure PO QO. The
tangents to the circle with centre O at P and Q
intersect at a point T. Prove that PQ and OT are
right bisectors of each other. [2011 (T-II)]
24. If from an external point B of a circle
with centre O, two tangents BC and BD are
drawn such that DBC = 120, prove thatBO = 2BC. [2011 (T-II)]
B. FORMATIVE ASSESSMENT
Activity
Objective : To show the following with the help of an activity.
(a) Lengths to tangents drawn from an external point are equal.
(b) Tangents are equally inclined to the segment joining the centre to that point.
Materials Required : White sheet of paper (or rice paper), Geometry box, Sketch pens, A pair
of scissors.
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Figure 2
Figure 1
Figure 3
Preparation for the Activity
1. Draw a circle of any radius with centre 'O' on a rice paper.2. Take a point, say P, outside of the circle.
3. From point P, draw a line segment touching the circle at A
( th e p oi nt o f c on ta ct ), w hi ch i s t he r eq ui re d t an gen t.
[See figure 1]
4. Following step 3, draw one more tangent PB, B being the
point of contact. [See figure 1]
5. Join OA, OP and OB. [See figure 1]
(a) Lengths of tangents drawn from an external point are equal.
Presentation : Fold the paper along OP.
Observation :
You will observe that point A coincides with point B and
line segment PA (the tangent from point P) concides with line
segment PB (another tangent from the same point on the circle).
[See figure 2]
Thus, length of PA = length of PB.
Hence, lengths of tangents drawn from an external point are equal.
The above property can be verified a number of times by taking external point at differentlocations.
Result : Lengths of tangents drawn from an external point are equal.
(b) Tangents are equally inclined to the line segment joining the centre with the external
point.
1. Cut out the two triangles, OPA andOPB, so formed figure 3.
2. Colour the two triangles with different colours.
3. Put one triangle on the other.
Observation :
You will observe that two triangles are congruent to each other
(i.e., one triangle exactly superimposes the other) with the
following (angle) correspondence.
OPA =OPB, (Proves the required result)
PAO =PBO,
AOP =BOP (See figure 4)
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Figure 4
Hence, tangents drawn from an external point, equally
inclined to the line segment joining the centre with that point.Result : Tangents drawn equally inclined to the line segment
joining the centre with the external point.
Note :The tangent at any point of a circle is perpendicular to
the radius through the point of contactOAP =OBP = 90.
ANSWERS
A. SUMMATIVE ASSESSMENTPractice Exercise 10.1A
1. (c) 2. (a) 3. (d) 4. (b) 5. (c) 6. (c)
7. 3 cm 8. 28 cm 9. 10.9 cm 10. 10 cm 11. 24 cm 12. 5 cm
13. 7 cm, 5 cm, 3 cm 14. 8 2cm 15. 140 16. 8 cm 17. 60
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