Chap 10 : Circles www.rava.org.

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CHAPTER 10
File Revision Date : 20 September 2019
Objective Questions
CLASS : 9 th
SUB : Maths
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Circles

1. OBJECTIVE QUESTIONS (a) 18 cm (b) 16 cm

(c) 12 cm (d) 10 cm
Ans : (d) 10 cm
1. In a cyclic quadrilateral, the difference between two
opposite angles is 58c, the measures of opposite angles AB = 16 cm
are EC = FD
(a) 158c, 22c (b) 129c, 51c Subtracting DE from both sides, we get CD = EF
(c) 109c, 71c (d) 119c, 61c = 12 cm
Ans : (d) 119c, 61c TAEG , TAFG , EG = GF = 6 cm

If +A - +C = 58c ...(1) In TAEG , AG = 1 AB = 8 cm

2
+A + +C = 180c ...(2) and AE 2 = AG 2 + EG 2
Adding (1) & (2), we get 2+A = 238c
r2 = 82 + 62
+A = 119c
= 64 + 36 & r = 10 cm
Subtracting (1) from (2), we get 2+C = 122c
+C = 61c 5. Which of the following statements is true for the
+A = 119c and +C = 61c longest chord of a circle?
(a) It is equal to radius
2. Which of the following statements is true for a regular (b) It is two times of radius
pentagon? (c) It is never equal to diameter
(a) All vertices are con-cyclic. (d) It is two times of diameter
(b) All vertices are not con-cyclic. Ans : (b) It is two times of radius
(c) Only four vertices are con-cyclic
(d) Cannot say anything about regular pentagon
6. The region between a chord and either of the arcs is
Ans : (a) All vertices are con-cyclic. called
(a) an arc (b) a sector
3. In a cyclic quadrilateral ABCD , if two sides are (c) a segment (d) a semicircle
parallel, which of the following statements is definitely Ans : (c) a segment
false?
(a) Remaining two sides are equal
7. When two chords of a circle bisect each other, then
(b) Diagonals are not equal
which of the following statements is true?
(c) Diagonals intersect at the centre of circle (a) Both chords are perpendicular to each other
(d) Both (a) and (c) (b) Both chords are parallel to each other
Ans : (b) Diagonals are not equal (c) Both chords are unequal
(d) Both are diameters of the circle.
4. A crescent is formed of two circular arcs ACB , ADB Ans : (d) Both are diameters of the circle.
of equal radius with respective, centres E and F as
shown in the given figure.
8. Equal chords of a circle subtend equal angles at
(a) centre (b) circumference
(c) Both (a) and (b) (d) None of these
Ans : (c) Both (a) and (b)

9. The line joining the centre of a circle to the midpoint

of a chord is always
The perpendicular bisector of AB cuts the crescent (a) parallel to the chord
at C and D , where CD = 12 cm, AB = 16 cm. The
(b) perpendicular to the chord
radius of arc ACB is

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(c) equal to the chord 15. ABCD is a cyclic quadrilateral with centre O in
(d) tangent to the chord the given figure. Chord AB is produced to E where
+CBE = 130c, the value of x is equal to
Ans : (b) perpendicular to the chord

10. There is one and only one circle passing through three
given .......... points
(a) collinear (b) non-collinear
(c) far-off (d) nearest (a) 130c (b) 260c
Ans : (b) non-collinear (c) 140c (d) 280c
Ans : (a) 130c
11. Diagonals of a cyclic quadrilateral are the diameters
+ADC = 180c - +CBA
of that circle, then quadrilateral is a
(a) parallelogram (b) square = +CBE = 130c
(c) rectangle (d) trapezium x = +ADC = 130c
Ans : (c) rectangle Download All PDF files from www.rava.org.in.

12. When two circles intersect at points A and B with 16. In the given figure, O is the centre of a circle. AB and
AC and AD being the diameters of the first and CD are its two chords. If OM = AB , ON = CD and
second circle then the points B, C and D are OM = ON , then
(a) concurrent (b) circumcentre
(c) orthocenter (d) collinear
Ans : (d) collinear
B, C and D are collinear.

13. If PQ is a chord of a circle with radius r units and R

is a point on the circle such that +PRQ = 90c, then
the length of PQ is
(a) r units (b) 2r units (a) AB ! CD (b) AB < CD
(c) r units (d) 4r units (c) AB > CD (d) AB = CD
2
Ans : (b) 2r units Ans : (d) AB = CD
Since PQ is a chord of a circle and R is a point on the Chords of a circle which are equidistant from the
circle such that +PRQ = 90c, therefore, the arc PRQ centre of the circle are equal.
is a semicircle. PQ is a diameter.
Hence, Length of PQ = 2 # radius = 2r units 17. In the given figure, O is the centre of circle.
+OPQ = 27c and +ORQ = 21c. The values of
14. If an equilateral triangle PQR is inscribed in a circle +POR and +PQR respectively are
with centre O , then +QOR is equal to
(a) 60c (b) 30c
(c) 120c (d) 90c
Ans : (c) 120c
As PQR is an equilateral triangle inscribed in a circle,
+QPR = 60c (a) 84c, 42c (b) 96c, 48c
Hence, +QOR = 2 # +QPR (c) 54c, 42c (d) 108c, 54c

= 2 # 60c = 120c Ans : (b) 96c, 48c

Draw a line passing through Q and O .
a = 27c 6OP = OQ@
b = 21c 6OR = OQ@

+PQR = a + b

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= 27c + 21c = 48c 21. In the given figure, E is any point in the interior
+POR = 2 # +PQR = 2 # 48c of the circle with centre O . Chord AB = AC . If
+OBE = 20c, the value of x is
= 96c (a) 40c (b) 45c
! (c) 50c (d) 70c
18. In the given figure, chord RS = chord NS . How RS
! Ans : (d) 70c
is related with NS ?
Since, AB = AC
Hence, +AOB = +AOC
[Equal chords subtend equal angles at the centre]
AO = BC 6+BOA + +COA = 180c@
Now, in TOBE
! !
(a) RS is smaller than NS (b) Both are equal 20c + x + +BOE = 180c
! !
(c) RS is greater than NS (d) None of these
Ans : (b) Both are equal
When chords are equal, their arcs are also equal.

19. In the given figure, O is the centre of the circle. For 20c + x + 90c = 180c
what values of x and y , chord BC will pass through
the centre of circle where points A, B and C are on x = 70c
the circle?
22. In the given figure, PQRS is a cyclic quadrilateral in
which PS = RS and +PQS = 60c. The value of x is

(a) x = 90c, y = 60c (b) x = 75c, y = 30c

(c) x = 65c, y = 90c (d) x = 90c, y = 65c
(a) 30c (b) 60c
Ans : (d) x = 90c, y = 65c
(c) 75c (d) 80c
When chord BC passes through centre,
Ans : (b) 60c
then, x = 90c.
+RPS = +RQS = x
Now, x + y + 25c = 180c
(Angles in the same segment)
90c + y + 25c = 180c
+PRS = +PQS = 60c
y = 65c
+RPS = +PRS = 60c 6PS = RS @
20. In the given pentagon ABCDE , AB = BC = CD Hence, x = +RPS = +RQS
= DE = AE . The value of x is
= +PRS = 60c

23. In the given figure, chords AB = AC = 6 cm. The

length of BC , if radius is 5 cm, is

(a) 36c (b) 54c

(c) 72c (d) 108c
Ans : (b) 54c
Since, equal chords subtend equal angles at the centre. (a) 9.6 cm (b) 4.8 cm
(c) 19.2 cm (d) 8.0 cm
+AOE = 360c = 72c
5
Ans : (a) 9.6 cm
Now, OE = OA
Let OD = x & AD = 5 - x
+OEA = +OAE = x
In TOCD , OC 2 = OD2 + CD2
In TOAE , x + x + +AOE = 180c
52 = x2 + CD2
2x + 72c = 180c
CD2 = 25 - x2 ...(1)
x = 108c = 54c
2
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In TACD , AC2 = AD2 + CD2 value of x is

6 = ^5 - x h + CD
2 2 2

CD2 = 11 + 10x - x2 ...(2)

From (1) and (2), we get
11 + 10x - x2 = 25 - x2
10x = 14 & x = 1.4 cm
Hence, CD2 = 25 - ^1.4h2 = 23.04
CD = 4.8 cm
(a) 140c (b) 70c
Hence, BC = 2 # CD = 2 # 4.8 cm (c) 290c (d) 210c
= 9.6 cm Ans : (a) 140c
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as, OA = OC = radius of circle
24. In the given figure, AB is diameter, +AOC = 40c.
The value of x is Hence, +OAC = +OCA = 40c
Similarly, in TOBC
+OBC = +OCB = 30c
We know that angle subtended at the arc is half of the
angle subtended at the centre.
Hence, +ACB = 1 +AOB
2
(a) 50c (b) 60c
(c) 70c (d) 80c +AOB = 2 (+ACB) = 2 (70) = 140c
Ans : (c) 70c
27. In the given figure, ABCD is a cyclic quadrilateral.
+BCA = 90c [ Since AB is diameter] BA is produced to E and DC | | AB . If y : x is equal
to 4:5, then value of z is
Also, +ABC = 1 # +AOC = 20c
2
[Angle subtended by an arc at centre is double the
angle subtended by it at remaining part of the circle.]
In TABC , 20c + x + 90c = 180c & x = 70c
(a) 15c (b) 20c
25. In the given figure, AEDF is a cyclic quadrilateral.
(c) 25c (d) 30c
The values of x and y respectively are
Ans : (c) 25c

DC | | AB y + x = 180c [Co-interior angles]

But y :x = 4 :5
y = 180c = 80c
4+5#4
(a) 79c, 47c (b) 89c, 37c and x = 180c # 5 = 100c
4+5
(c) 89c, 47c (d) 79c, 37c
Also, +EAD = y
Ans : (b) 89c, 37c
[Exterior angle of cyclic quadrilateral ABCD ]
In TABE , +EAF + +FAD = y & z + 55c = 80c
35c + 54c + +AEB = 180c z = 80c - 55c = 25c
+AEB = 91c
+AFD + +AED = 180c 28. In the given figure, +AOB = 90c and +ABC = 30c.
Then, +CAO is
[Opposite angles of cyclic quadrilateral]
x + +AEB = 180c
x + 91c = 180c & x = 89c
In TACF , 54c + y + x = 180c
54c + y + 89c = 180c
y = 180c - 143c = 37c (a) 30c (b) 45c
(c) 60c (d) 90c
26. In the given figure, O is the centre of the circle. The Ans : (c) 60c

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+AOB = 2+ACB +A = 60c

+ACB = 1 +AOB = b 1 # 90cl = 45c Hence, +A = +B = +O = 60c
2 2 TOAB is an equilateral triangle.
Now, +COA = 2+CBA = ^2 # 30ch = 60c
+COD = 180c - +COA 31. In the given figure, BC passes through the centre of
a circle where points A, B and C are con-cylic and
= ^180c - 60ch = 120c +B is 44c more than +C . The values of x and y
+CAO = 1 +COD = b 1 # 120cl respectively are
2 2
= 60c

29. In the given figure, PQRS is a cyclic quadrilateral. If

+SPR = 25c and +PRS = 60c, then the value of x is

(a) x = 4 ; y = 3 (b) x = 3 ; y = 5
(c) x = 7 ; y = 2 (d) x = 5 ; y = 2
Ans : (d) x = 5 ; y = 2

+B - +C = 44c ...(1)
and +C + +B = 90c ...(2)
[BC is diameter of circle;
+A = 90c]
(a) 105c (b) 95c
From (1) and (2), we get
(c) 115c (d) 85c
+B = 67c and +C = 23c
Ans : (d) 85c
& 10x + 17 = 67 and 15y - 7 = 23
InTPRS , +PSR + 25c + 60c = 180c & x = 5 and y = 2
+PSR = 95c
Now, +PQR + +PSR = 180c 32. In the given figure, O is the centre of the circle,
(sum of opp. angles of a cyclic quad. is 180c) +CBE = 25c and +DEA = 60c. The measure of
+ADB is
x + 95c = 180c
x = 85c

30. In the given figure, chord AB subtends +AOB equal

to 60c at the centre of the circle. If OA = 5 cm , then
length of AB (in cm) 0 is

(a) 90c (b) 85c

(c) 95c (d) 120c
Ans : (c) 95c
(a) 5 cm (b) 5 3 cm
2 2 +DEA = +BEC
(c) 5 cm (d) 5 3 cm (Vertices. opp. angles)
4
Ans : (c) 5 cm 60c = +BEC
Now, in TBEC
OA = OB +E + +B + +C = 180c
(radius of circle) 60 + 25 + +C = 180c
5 = OB +C = 95c
Thus, +A = +B (since OA = OB ) Also, +C = +D
Thus, in TOAB
+D = 95c
+A + +B + +O = 180c
2+A + 60c = 180c
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33. In the given figure, ABCD is a cyclic quadrilateral, O (a) 6 cm (b) 8 cm

is the centre of the circle and a : b = 2 : 5 . The value (c) 14 cm (d) 16 cm
of x is
Ans : (c) 14 cm
E is mid point of CD and F is the mid point of AB
Now, in TOAF , AF 2 = OA2 - OF 2
2
b 48
2l
= 262 - OF 2

OF 2 = 676 - 576
(a) 20c (b) 25c OF = 10 cm
(c) 30c (d) 35c
In TOCE , OE 2 = OC 2 - CE 2
Ans : (c) 30c 2
OE 2 = 262 - b 20 l
a = 2 & 5a - 2b = 0 2
...(1)
b 5 OE 2 = 676 - 100
and x + b = 180c OE = 24 cm
a + b = 180c a
dx = n Hence, EF = OE - OF = ^24 - 10h cm
2 2
= 14 cm
a + 2b = 360c ...(2)
From (i) and (ii), we get 36. In the given figure, MNQS is a cyclic quadrilateral in
a = 60c, b = 150c. which +QNR = 61c and x : y is 2 :1. The values of x
and y respectively are
Also, x = a = 30c
2

34. In the given figure, chords AB and CD of a circle

when produced meet at P . If +APD = 35c and
+BCD = 25c, then +ADC is equal to

(a) 18 1c, 37 3c (b) 38 2c, 19 1c

4 4 3 3
(c) 21 1c, 33 2c (d) 19 1c, 38 1c
3 3 4 4
Ans : (b) 38 , 19 1c
2c
3 3
In 3 QNR , +SQN = y + 61c,
+MNP = +QNR = 61c
(a) 60c (b) 70c
(c) 50c (d) 120c +SMN = x + +MNP

Ans : (a) 60c = x + 61c

Now, +SQN + +SMN = 180c
+BAD = +BCD
(angles in the same segment of a circle)
or ^y + 61ch + ^x + 61ch = 180c
x + y = 58c but x = 2y
+PAD = +BAD = 25c
(since, +BCD = 25c, given) Thus, y = b 58 lc = 19 1c
3 3
Now, +ADC = +PAD + +APD
and x = b 116 lc = 38 2c
(ext. angle of a T = sum of two 3 3
internal opposite angles)
2. FILL IN THE BLANK
+ADC = 25c + 35c = 60c
DIRECTION : Complete the following statements with an
35. In the given figure, OE = CD , OF = AB , AB | | CD , appropriate word/term to be filled in the blank space(s).
AB = 48 cm, CD = 20 cm, radius OA = 26 cm. The
length of EF is
1. The longest chord of a circle is a .......... of the circle.
Ans : diameter

2. Segment of a circle is the region between an arc and

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the related .......... of the circle. from the corresponding centres are equal.
Ans : chord Ans : True

3. The chords of a circle which are .......... from the centre 5. A diameter of a circle divides the circular region into
are equal. two parts. Each part is called a semi-circular region.
Ans : equidistant Ans : True

4. A radius of a circle is a line segment with one end 6. Equal chords subtend equal angles at the centre.
point at the .......... and the other end on the .......... Ans : True
Ans : centre, circle
7. Every circle has a unique centre and it lies inside the
5. The .......... bisectors of two chords to a circle intersect circle.
at the centre. Ans : True
Ans : perpendicular
8. A chord of a circle which is twice as long as its radius
6. A diameter of a circle is a chord that passes through is a diameter of the circle.
the .......... of the circle. Ans : True
Ans : centre
9. Line segment joining the centre to any point on the
7. Angles in the same segment of a circle are .......... . circle is a radius of the circle.
Ans : equal Ans : True

8. The centre of a circle lies in the .......... of the circle. 10. A circle has only a finite number of equal chords.
Ans : interior Ans : False

9. A point, whose distance from the centre of a circle is 11. A circle is a plane figure.
greater than its radius, lies on the .......... of the circle. Ans : True
Ans : exterior
12. If a circle is divided into three equal arcs, each is a
10. An arc is a .......... when its ends are the ends of a major arc.
diameter. Ans : False
Ans : semicircle

4. MATCHING QUESTIONS
11. The sum of either pair of opposite angles of a cyclic
quadrilateral is .......... .
DIRECTION : Each question contains statements given in
Ans : 180c two columns which have to be matched. Statements (P, Q, R, S,
T) in Column-I have to be matched with statements (1, 2, 3, 4,
3. TRUE/FALSE 5) in Column-II.

DIRECTION : Read the following statements and write your 1. Math the following
answer as true or false.
Column-I Column-II
(P) In the given figure, (1) 120c
1. Sector is the region between the chord and its
+ADC =
corresponding arc.
Ans : False

2. A cyclic trapezium is always isoscele.

Ans : True

3. Diameter is the longest chord of the circle.

Ans : True

4. Chords of congruent circles which are equidistant

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(Q) Distance of a chord AB (2) 75c

of a circle from the centre
is 12 cm and length of
the chord is 10 cm. The
diameter of the circle is
.......... cm.
(R) In the figure given below, (3) 45c
O is the centre of the (R) +BOC = 70c
circle. If AB = BC and (Since, equal chords of a circle subtend equal angles
+AOB = 70c then +OBC at the centre).
is equal to +OBC + +OCB = 180c - 70c = 110c
Hence, +OBC = 55c
(Since OB = OC & +OBC = +OCB )
(S) Since angles in the same segment of a circle are
equal
Hence, +ADB = +ACB = 75c
(T) Since PQR is an equilateral triangle inscribed in
a circle,
Hence, +QPR = 60c
(S) In the given figure, (4) 55c
the points A, B, C Hence, +QOR = 2 # +QPR = 120c
and D lie on a circle.
If +ACB = 75c, then
+ADB is equal to

2. Match the following :

(T) If an equilateral triangle (5) 26c Column-I Column-II

PQR is inscribed in a (P) The radius of circle is 8 cm (1) 23 cm
circle with centre O , then and the length of one of its
+QOR is equal to chords is 12 cm. The distance
of the chord from the centre
is
(Q) Two parallel chords of (2) 5.196 cm
lengths 30 cm and 16 cm are
drawn on the opposite sides
of the centre of a circle of
radius 17 cm. The distance
between the chords is
Ans : P–3, Q–5, R–4, S–2, T–1 (R) The length of a chord which (3) 5.291 cm
is at a distance of 4 cm from
(P) +ADC = 1 +AOC = 1 # 90c = 45c the centre of the circle of
2 2
radius 6 cm is
(Since, angle subtended by an arc at the centre is
double the angle formed by it on the remaining part (S) An equilateral triangle of side (4) 8.94 cm
of the circle) 9 cm is inscribed in a circle.
(Q) The perpendicular from the centre to the chord The radius of the circle is
bisects the chord.
AL = 5 cm and OL = 12 cm P Q R S
2 2
AO = (12) + (5) (a) 3 1 4 2
= 169 = 13 cm (b) 3 4 1 2
Radius = 13 cm (c) 1 2 3 4
Diameter = 26 cm (d) 1 3 2 4
Ans : (a) P–3, Q–1, R–4, S–2
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(P) Let PQ be the chord of a circle with centre O and circle to a chord bisects the chord, we have
radius 8 cm such that PQ = 12 cm . From O , draw PQ = 2 # PL
OL = PQ . Join OP . Since the perpendicular from
the centre of a circle to chord bisects the chord. = 2 # 4.472 = 8.94cm
(S) Let 3 PQR be an equilateral triangle of side 9
PL = LQ = 1 PQ = 6cm cm. Let PS be one of its median. Then PS = QR
2
In right angled triangle OLP , we have and QS = 9 cm
2
OP2 = OL2 + PL2 PS = PQ2 - QS2
2
82 = OL2 + 62 = 92 - b 9 l cm = 9 3 cm
2 2
OL2 = 82 - 62 = 64 - 36 = 28
OL = 28 = 5.291cm
Hence, the distance of the chord from the centre
is 5.291 cm.

In an equilateral triangle, the centroid and circum

centre coincide and
PG : PS = 2 : 3
(Q) We know that the perpendicular from the centre
of a circle to a chord bisects the chord. Radius = PG = 2 PS = 2 # 9 3
3 3 2
PL = 1 PQ = 15cm = 3 3 = 5.196cm
2
RM = 1 RS = 8cm
2 3. Match the following:
OP = OR = 17cm
Column-I Column-II
(P) C is a point on the minor arc (1) 60c
AB of the circle with centre O
. If +ACB = x calculate x , if
ACBO is a parallelogram.
(Q) Chord ED is parallel to the (2) 120c
From the right angled 3 OLP , we have diameter AC of the circle. If
OL2 = OP2 - PL2 +CBE = 55c, then +DEC is
OL2 = 172 - 152 = 64
OL = 8cm
From the right angled 3 ORM , we have
OM2 = OR2 - RM2 = 172 - 82 = 225
OM = 15cm
Since OL = PQ , OM = RS and PQ | | RS , the points (R) In the given figure, O is (3) 35c
L, O, M are collinear. the centre of the circle. If
LM = LO + OM = 8 + 15 = 23cm +ACB = 60c, find +OAB .

(R) Let PQ be a chord of a circle with centre O and

radius 6cm . Draw OL = PQ . Join OP .
Then OL = 4cm and OP = 6cm .
From the right angled 3 OPL , we have
OP2 = OL2 + PL2
(S) In the given figure, O is the (4) 30c
PL2 = OP2 - OL2
centre of a circle, +AOB
PL2 = 62 - 42 = 40c and +BDC = 100c. Find
= 36 - 16 = 20 +OBC .
PL = 4.472cm

Since the perpendicular from the centre of the

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P Q R S +ACB = 1 # +AOB
2
(a) 2 4 3 1
= 1 # 40c = 20c
(b) 2 3 4 1 2
In 3 BDC, +DBC + +BDC + +DCB = 180c
(c) 1 2 3 4
+OBC + 100c + 20c = 180c
(d) 1 3 2 4
+OBC = 180c - 120c = 60c
Ans : (b) P–2, Q–3, R–4, S–1
(P) Clearly, major arc BA subtends x angle at a point
on the remaining part of the circle.

4. Match the following:

Column-I Column-II
(P) In the given figure, ABCD (1) 60c
Reflex, +AOB = 2x is a cyclic quadrilateral, O
360c- y = 2x is the centre of the circle.
If +BOD = 160c, find the
y = 360c - 2x measure of +BPD .
Since, ACBO is a parallelogram.
x = y & x = 360c - 2x
3x = 360c & x = 120c
(Q) Since, +CBE and +CAE are the angles in the
same segment of arc CDE .

(Q) In given figure, ABCD is a (2) 65c

cyclic quadrilateral whose side
AB is a diameter of the circle
through A, B, C, D . If +ADC
= 130c, find +BAC .
+CAE = +CBE
+CAE = 55c 6+CBE = 55c@
Since, AC is the diameter of the circle and the angle
in semi-circle is a right angle +AEC = 90c Now, in
3 ACE ,
+ACE + +AEC + +CAE = 180c
(R) In the given figure, BD = DC (3) 40c
+ACE = 35c and +CBD = 30c, find
But, +DEC and +ACE are alternate angles, because +BAC .
AC || DE .
+DEC = +ACE = 35c
(R) +AOB = 120c
(The angle subtended by an arc of a circle at the
centre is double the angle subtended by it at any point
on the remaining part of the circle)
(S) In the given figure, D is the (4) 100c
centre of the circle and arc
ABC subtends an angle of
130c at the centre. If AB is
extended to P , find +PBC .

OA = OB (Radius of the circle)

x + x + 120c = 180c
2x = 60c
x = 30c
(S) +AOB = 2 # +ACB

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P Q R S +PBC = 65c (If one side of a cyclic quadrilateral

is produced, then the exterior angle is equal to the
(a) 1 2 3 4 interior opposite angle)
(b) 4 1 3 2
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(c) 4 3 1 2
(d) 1 4 2 3 5. ASSERTION AND REASON
Ans : (c) P-4, Q-3, R-1, S-2
DIRECTION : In each of the following questions, a statement
(P) Consider the arc BCD of the circle. This arc of Assertion is given followed by a corresponding statement
makes +BOD = 160c at the centre of the circle of Reason just below it. Of the statements, mark the correct
and +BAD at a point A on the circumference. answer as
(a) Both assertion and reason are true and reason is
the correct explanation of assertion.
(b) Both assertion and reason are true but reason is
not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
+BAD = 1 +BOD = 80c
2
1. Assertion : The length of a chord which is at a distance
Now, ABPD is a cyclic quadrilateral.
of 5 cm from the centre of a circle of radius 10 cm is
+BAD + +BPD = 180c 17.32 cm.
80c + +BPD = 180c Reason : The perpendicular from the centre of a circle
to a chord bisects the chord.
+BPD = 100c
(Q) Since ABCD is a cyclic quadrilateral Ans : (a) Both assertion and reason are true and
reason is the correct explanation of assertion.

+ADC + +ABC = 180c Let, PQ be a chord of a circle with centre O and

130c + +ABC = 180c radius 10cm . Draw OR = PQ .
+ABC = 50c Now, OP = 10cm and OR = 5cm
In right triangle ORP , we get
Since +ACB = 90c
Now, in 3 ABC , we have OP2 = PR2 + OR2
+BAC + +ACB + +ABC = 180c PR2 = OP2 - OR2
+BAC + 90c + 50c = 180c PR2 = 102 - 52 = 75
+BAC = 40c PR = 75 = 8.66
Since, the perpendicular from the centre to a chord
(R) BD = DC bisects the chord.
= +BCD Therefore, PQ = 2 # PR
= +CBD = 30c = 2 # 8.66 = 17.32cm
So, +BDC = 180c - 30c - 30c
= 120c 2. Assertion : The circumference of a circle must be a
As ABCD is cyclic quadrilateral positive real number.
Reason : If r (> 0) is the radius of the circle, then its
+BAC = 180c - 120c = 60c
circumference 2pr is a positive real number.
(S) Let Q be a point on circumference.
Ans : (a) Both Assertion and Reason are correct and
Reason is the correct explanation of Assertion.

3. Assertion : The measure of +AOC = 60c

Join QA and QC .
Now, ABCQ is a cyclic quadrilateral
+AQC = 1 +ADC = 65c
2

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Reason : Angle subtended by an arc of a circle at the Reason is not the correct explanation of Assertion.
centre of the circle is double the angle subtended by
arc on the circumference.
7. Assertion : Two diameters of a circle intersect each
Ans : (d) Assertion is false but reason is true. other at right angles. Then the quadrilateral formed
Join BO . by joining their end-points is a square.
Reason : Equal chords subtend equal angles at the
centre.
Ans : (a) Both assertion and reason are true and
reason is the correct explanation of assertion.

In 3 AOB , we have
OA = OB [radius]
+OBA = +OAB
[Angle opposite to equal sides of a triangle are equal]
+OBA = 30c ...(1) Let AB and CD be two perpendicular diameters of a
Similarly, in 3 BOC , we get OB = OC circle with centre O .
+OCB = +OBC Now, +ABC = 90c
+OBC = 40c ...(2) [Angle in semicircle is a right angle]
+ABC = +OBA + +OBC Similarly +ACD = +ADC
= 30c + 40c = 70c = +BAD = 90c ...(1)
[Using (1) and (2)] In 3 AOB and 3 AOD , we have
Since angle subtended by an arc of a circle at the AO = AO (Common)
centre of the circle is double the angle subtended by
the arc on the circumference. +AOB = +AOD (Each 90c, given)
+AOC = 2 # +ABC BO = OD (Radii of circle)
= 2 # 70c = 140c 3 AOB b +AOD (By SAS congruence)
AB = AD (By C.P.C.T.)
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AD = DC , DC = BC; BC = AB
4. Assertion : Given a circle of radius r and with centre
O . A point P lies in a plane such that OP > r then Hence, AB = BC = CD = DA ...(2)
point P lies on the exterior of the circle. Also, it is given that diagonals of ABCD intersect at
Reason : The region between an arc and the two radii, 90c ...(3)
joining the centre of the end points of the arc, is called By (1), (2) and (3) ABCD is a square.
a sector.
Ans : (b) Both Assertion and Reason are correct, but 8. Assertion : The sum of either pair of opposite angles
Reason is not the correct explanation of Assertion. of a cyclic quadrilateral is 180c.
Reason : Two or more circles are called concentric
circles if and only if they have different centre and
5. Assertion : In a cyclic quadrilateral ABCD , radii.
+A - +C = 60c, then the smaller of two is 60c.
Ans : (c) Assertion is correct but Reason is incorrect.
Reason : Opposite angles of cyclic quadrilateral are
supplementary. Two or more circles are called concentric circles if and
Ans : (a) Both assertion and reason are true and only if they have same centre but different radii.
reason is the correct explanation of assertion.
9. Assertion : In an isosceles triangles ABC with
Since ABCD is a cyclic quadrilateral, so, its opposite
AB = AC , a circle is passing through B and C
angles are supplementary
intersects the sides AB and AC at D and E
+A + +C = 180c ...(1) respectively. Then DE | | BC .
Also, +A - +C = 60c ...(2) Reason : Exterior angle of a cyclic quadrilateral is
On solving (i) and (ii), we get equal to interior opposite angle of that quadrilateral.
+A = 120c, +C = 60c Ans : (a) Both assertion and reason are true and
reason is the correct explanation of assertion.
6. Assertion : If P and Q are any two points on a circle,
then the line segment PQ is called a chord of the
circle.
Reason : Equal chords of a circle subtend equal angles
at the centre.
Ans : (b) Both Assertion and Reason are correct, but
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Chap 10 : Circles www.rava.org.in

To prove DE || BC
i.e., +B +ADE .
In 3 ABC , we have
AB = AC
+B = +C ...(1)
In the cyclic quadrilateral CBDE ,
side BD is produced to A. We know that an exterior
angle of cyclic quadrilateral is equal to interior
opposite angle of cyclic quadrilateral.
+ADE = +C ...(2)
From (1) and (2), we get
+B = +ADE .
Hence, DE || BC

10. Assertion : A diameter of a circle is the longest chord

of the circle and all diameters have equal length.
Reason : Length of a diameter = radius.
Ans : (c) Assertion is correct but Reason is incorrect.

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