Circle
141
Introduction
In our day-to-day life, we come across many objects which are round in shape, such as dials
of many clocks, wheels of a vehicle, bangles, key rings, coins of denomination ` 1, ` 2, ` 5,
` 10, etc. In a clock, we observe that second's hand goes round the dial of the clock rapidly and
its tip moves in a round path. This path traced by the tip of the second's hand is called a circle.
In this chapter, we shall study about circles, some terms related with a circle, chord properties
of a circle, arc and chord properties of a circle.
14.1 Circle
s
Ra
←r
diu
→
A circle is the collection of all those points, say P, in a plane each of which is at a constant distance from
a fixed point in that plane.
In other words, a circle is the path of a point which moves in a plane so that it remains at
a constant distance from a fixed point in the plane.
The fixed point is called the centre and the constant distance is called the radius of the circle.
P
The radius of a circle is always positive.
The adjoining figure shows a circle with O as its centre and r as its Circle
radius.
O
Centre
Note that the centre of a circle does not lie on the circle.
Let O be the centre of a circle and r its radius. If P is a point on the
P
circle, then the line segment OP is a radius of the circle and its length is r. If Q
r
is another point on the circle then OQ is another radius of the circle. Note that
O
all radii (plural of radius) have one point in common, which is the centre of the
circle. Also OP = OQ = r. Thus:
All radii of a circle are equal.
Q
Note. The line segment joining the centre and any point on the circle is also called a radius
of the circle. Thus, ′radius′ is used in two senses — in the sense of a line segment
and also in the sense of its length.
14.1.1 Some terms related with a circle
1. Circle-interior and exterior
A circle is a closed curve. It divides the plane region into three parts. They are:
(i)Circle — The collection of all points P of the plane such that OP = r form a circle with
centre O and radius r (r > 0).
(ii)Interior of a circle — The collection of all points of the plane which lie
inside the circle i.e. the collection of all points P of the plane such that
OP < r form the interior of the circle.
The collection of all points of the plane which either lie on the circle or
are inside the circle form the circular region.
(iii)Exterior of a circle — The collection of all points of the plane
which lie outside the circle i.e. the collection of all points of the
plane such that OP > r form the exterior of the circle.
P
O ←r→
Interior
P
←r→
O
Exterior
2. Chord of a circle
A line segment joining any two points of a circle is called a chord of the
circle.
In the adjoining figure, PQ is a chord of the circle with centre O. The
distance PQ is called the length of the chord.
O
P
3. Diameter of a circle
A chord of a circle which passes through its centre is called a diameter of
the circle.
A diameter of a circle is the longest chord of the circle and all diameters
have equal length.
P
Q
rd
Cho
eter
Diam
O
Q
In the adjoining figure, PQ is a diameter of the circle with centre O.
Note that OP and OQ are both radii of the circle, so OP = OQ = r.
It follows that PQ = 2r = 2 × radius. Thus:
Length of a diameter = 2 × radius.
Note that there are infinitely many lines passing through the point O (centre of circle), so
a circle has infinitely many diameters.
4. Secant of a circle
O
A line which meets a circle in two points is called a secant of the
circle. In the adjoining figure, line PQ is a secant of the circle with
centre O.
nt
P
5. Arc of a circle
Seca
Q
Arc
A (continuous) part of a circle is called an arc of the circle.
Q
P
”.
The arc of a circle is denoted by the symbol ‘’

In the adjoining figure, PQ denotes the arc PQ of the circle with
centre O.
Arc of a circle is divided into following categories:
O
Minor arc
(i)Minor and major arc — An arc less than one-half of the whole
arc of a circle is called a minor arc of the circle and an arc
greater than one-half of the whole arc is called a major arc of
 is the major
 is the minor arc and PRQ
the circle. Here, PQ
arc.
Q
P
R
Major arc
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Understanding ICSE mathematics – Ix
Semicircle
(ii)Semicircle — When PQ is a diameter, then both arcs are equal
and each is called a semicircle. Thus, one-half of the whole arc
of a circle is called a semicircle.
P
Q
O
Circumference
(iii)Circumference — The whole arc of a circle is called the
circumference of the circle. The length of the circumference of
a circle is the length of the whole arc. However, in general, the
term ‘circumference of a circle’ refers to its length.
O
6. Sector of a circle
The part of the plane region enclosed by an arc of a circle and its two
bounding radii is called a sector of the circle.
Major
sector
O
The part containing the minor arc is called minor sector and the part
containing the major arc is called major sector.
Minor
sector
P
Q
7. Segment of a circle
A chord of a circle divides its circular region into two parts. Each part of
the circular region is called a segment of the circle.
The part of the circular region containing the minor arc is called a
minor segment and the part containing the major arc is called a major
segment.
or
Min ent
m
g
e
s
Q
O
P
r
Majo t
n
e
segm
8. Angle subtended by an arc
P
The angle subtended by the two bounding radii of an arc of a circle at the
centre of the circle is called the angle subtended by the arc.
In the adjoining figure, ∠POQ is the angle subtended by the minor arc
PQ of a circle with centre O and the reflex ∠POQ is the angle subtended by
the major arc PQ at O. If R is any point on the major arc, then ∠PRQ is called
the angle subtended by the minor arc PQ at the point R.
Q
O
R
9. Angle subtended by a chord
Let PQ be a chord of a circle with centre O and R, S be points on the minor
and major arcs of the circle (as shown in the adjoining figure).
Join OP and OQ, then ∠POQ is the angle subtended by the chord PQ
at the centre O.
Join PR, QR, PS and QS, then ∠PRQ and ∠PSQ are the angles subtended
by the chord PQ at the points R and S respectively on the minor and major
arcs.
R
Q
P
O
S
10. Concentric circles
Two or more circles are called concentric circles if and only if they have
same centre but different radii.
O
circle
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11. Equal (or congruent) circles
Two or more circles are called equal (or congruent) circles if
and only if they have same radius.
In the adjoining figures, two circles with centres A and B
have equal radius (= r), so, these are equal circles.
r
r
A
B
(a)
(b)
14.2 Chord Properties of Circles
Theorem 14.1
The straight line drawn from the centre of a circle to bisect a chord,
which is not a diameter, is perpendicular to the chord.
Given. A chord AB of a circle with centre O, and OM bisects the
chord AB.
To prove. OM ⊥ AB.
Construction. Join OA and OB.
Proof.
Statements
O
•
A
M
B
Reasons
In ∆s OAM and OBM
1. OA = OB
1. Radii of same circle.
2. AM = MB
2. M is mid-point of AB.
3. OM = OM
3. Common.
4. ∆ OAM ≅ ∆ OBM
4. S.S.S. axiom of congruency.
5. ∠ΑMO = ∠OMB
5. ‘c.p.c.t.’.
6. ∠ΑMO + ∠OMB = 180°
6. AMB is a st. line.
7. ∠ΑMO = 90°
Hence OM ⊥ AB.
7. From 5 and 6.
Q.E.D.
Theorem 14.2 (Converse of theorem 14.1)
The perpendicular to a chord from the centre of the circle bisects the
chord.
Given. A chord AB of a circle with centre O, and OM is perpendicular
to the chord AB.
To prove. AM = MB.
Construction. Join OA and OB.
Proof.
Statements
In ∆s OAM and OBM
1. OA = OB
2. ∠AMO = ∠OMB
3. OM = OM
4. ∆ OAM ≅ ∆ OBM
5. AM = MB
Q.E.D.
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Reasons
1. Radii of same circle.
2. Each = 90°, since OM ⊥ AB.
3. Common.
4. R.H.S. axiom of congruency.
5. ‘c.p.c.t.’.
Understanding ICSE mathematics – Ix
O
•
A
M
B
C
Theorem 14.3
Equal chords of a circle are equidistant from the centre.
Given. AB and CD are chords of a circle with centre O, and AB = CD.
To prove. AB and CD are equidistant from O i.e. if OM ⊥ AB and
ON ⊥ CD, then OM = ON.
Construction. Join OA and OC.
Proof.
Statements
N
•O
A
D
M
B
Reasons
1. AM = 1 AB
1. Perpendicular from centre bisects the chord
(Theorem 14.2).
2. CN = 1 CD
2. Same as above.
3. AM = CN
In ∆s OAM and OCN
3. AB = CD (given).
4. AM = CN
4. From 3.
5. ∠ AMO = ∠ CNO
5. Each = 90°, OM ⊥ AB and ON ⊥ CD.
6. OA = OC
6. Radii of same circle.
7. ∆ OAM ≅ ∆ OCN
7. R.H.S. axiom of congruency.
8. OM = ON
8. ‘c.p.c.t.’.
2
2
Q.E.D.
Theorem 14.4 (Converse of theorem 14.3)
Chords of a circle that are equidistant from the centre of the
circle are equal.
Given. AB and CD are chords of a circle with centre O ;
OM ⊥ AB, ON ⊥ CD and OM = ON.
To prove. AB = CD.
Construction. Join OA and OC.
Proof.
Statements
C
N
•O
A
D
M
B
Reasons
In ∆s OAM and OCN
1. OM = ON
1. Given.
2. OA = OC
2. Radii of same circle.
3. ∠AMO = ∠CNO
3. Each = 90°, OM ⊥ AB and ON ⊥ CD.
4. ∆ OAM ≅ ∆ OCN
4. R.H.S. axiom of congruency.
5. AM = CN
5. ‘c.p.c.t.’.
6. AM = 1 AB
6. Perpendicular from centre bisects the chord.
7. CN = 1 CD
7. Same as above.
8. 1 AB = 1 CD
8. AM = CN, from 5.
2
2
2
2
⇒ AB = CD. Q.E.D.
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Theorem 14.5
There is one and only one circle passing through three given noncollinear points.
Given. Three non-collinear points A, B and C.
To prove. One and only one circle can be drawn passing through
the points A, B and C.
Construction. Join AB and BC.
Draw perpendicular bisectors of line segments AB and BC, say PQ
and RS respectively. Let these bisectors meet at O. Join OA, OB and OC.
Proof.
Statements
P
C
S
O
B
R
A
Q
Reasons
1. OA = OB
1.O lies on the perpendicular bisector PQ of AB
and every point on the perpendicular bisector
of a line segment is equidistant from its end
points.
2. OB = OC
2.O lies on the perpendicular bisector RS of BC
and every point on the perpendicular bisector
of a line segment is equidistant from its end
points.
3. OA = OB = OC
3. From 1 and 2.
4.If a circle is drawn with O as centre 4.From 3, O is equidistant from points A, B and
and radius = OA, it passes through
C.
the points B and C.
5.There is only one circle passing 5.Since two lines can intersect at only one point,
through the points A, B and C.
the perpendicular bisectors of AB and BC
meet at only one point O.
Hence, one and only one circle can be drawn passing through three given non-collinear
points.
Corollary 1. In the plane of a circle, the perpendicular bisector of a chord of a circle passes through
its centre.
Corollary 2. The perpendicular bisectors of two (non-parallel) chords of a circle intersect at the
centre of the circle.
Corollary 3. As there is one and only one circle passing through three non-collinear points, two
different circles can meet atmost in two different points.
Illustrative Examples
Example 1. A chord of length 16 cm is drawn in a circle of diameter 20 cm. Calculate its distance
from the centre of the circle.
Solution. Let AB be a chord of a circle with centre O such that AB = 16 cm and radius
of the circle = OA = 1 . diameter = 1 . 20 cm = 10 cm.
2
2
From O, draw OM ⊥ AB.
Since OM ⊥ AB, AB is bisected at M (Theorem 14.2)
∴AM = 1 AB = 1 . 16 cm = 8 cm.
2
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2
Understanding ICSE mathematics – Ix
O
A
M
B
From right angled ∆ OAM, by Pythagoras theorem, we get
OA2 = OM2 + AM2
⇒
102 = OM2 + 82
⇒
OM2 = 102 – 82 = 100 – 64 = 36
⇒
OM = 6 cm.
Hence, the given chord is at a distance of 6 cm from the centre of the circle.
Example 2. The centre of a circle of radius 13 units is the point (3, 6). P(7, 9) is a point inside the
circle. APB is a chord of the circle such that AP = PB. Calculate the length of AB.
Solution. Given the centre of the circle is C(3, 6) and P(7, 9) is a point on the chord AB
of the circle such that AP = PB i.e. P is mid-point of the chord.
Join CP.
Since CP bisects the chord AB, CP ⊥ AB (Theorem 14.1).
C(3, 6)
Radius of the circle = CA = 13 units (given).
CP =
=
(7 − 3)2 + (9 − 6)2 42
+
32
=
| Distance formula
A
B
P(7, 9)
16 + 9 = 5 units
From right angled ∆ CAP, by Pythagoras theorem, we get
AC2 =AP2 + CP2
⇒
132 =AP2 + 52
⇒AP2 = 132 – 52 = 169 – 25 = 144
⇒AP = 12 units.
∴ The length of chord AB = 2 × AP = (2 × 12) units = 24 units.
Example 3. In the figure given below, the diameter CD of a circle with centre O is perpendicular to
the chord AB. If AB = 8 cm and CM = 2 cm, find the radius of the circle.
Solution. Since CD ⊥ AB i.e. OM ⊥ AB, AB is bisected at M (Theorem 14.2),
D
∴ AM = 1 AB = 1 . 8 cm = 4 cm.
2
2
Let radius of circle = r cm, then
OM = OC – MC = (r – 2).
From right angled ∆ OAM, by Pythagoras theorem, we get
OA2 = AM2 + OM2
⇒
r2 = 42 + (r – 2)2
⇒
r2 = 16 + r2 – 4r + 4
⇒
4r = 20 ⇒ r = 5.
∴ Radius of the circle = 5 cm.
•O
A
B
M
C
Example 4. Two chords AB, CD of lengths 24 cm, 10 cm respectively of a circle are parallel. If the
chords lie on the same side of centre and the distance between them is 7 cm, find the length of a diameter
of the circle.
Solution. Let O be the centre of the circle.
Draw OM ⊥ AB and ON ⊥ CD.
∴AM = 1 AB = 1 . 24 cm = 12 cm and
2
1
CN = CD =
2
2
1
. 10 cm = 5 cm.
2
•O
M
A
N
C
Since AB CD, points O, M and N are collinear.
MN = distance between AB and CD = 7 cm (given).
circle
B
D
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Let OM = x cm, then ON = OM + MN = (x + 7) cm.
Let the radius of the circle be r cm.
From right angled ∆s OAM and OCN (by Pythagoras theorem), we get
OA2 = AM2 + OM2 and OC2 = CN2 + ON2
⇒
r2 = (12)2 + x2
and r2 = 52 + (x + 7)2
From (i) and (ii), we get
52 + (x + 7)2 = (12)2 + x2
⇒
25 + x2 + 14x + 49 = 144 + x2
⇒
14x = 70 ⇒ x = 5.
∴ From (i), r2 = (12)2 + 52 = 144 + 25 = 169 ⇒ r = 13.
∴ The length of a diameter of the circle = 2r = 26 cm.
…(i)
…(ii)
Example 5. ABC is an isosceles triangle inscribed in a circle. If AB = AC = 25 cm and
BC = 14 cm, find the radius of the circle.
Solution. Let O be the centre of the circle.
From A, draw AM ⊥ BC.
Since ∆ ABC is isosceles with AB = AC, M is the mid-point of BC i.e. AM is perpendicular
bisector of BC, therefore, O lies on AM.
BM = 1 BC = 1 . 14 cm = 7 cm.
2
2
From right angled ∆ ABM, by Pythagoras theorem, we get
AM2 = AB2 – BM2 = (25)2 – 72 = 625 – 49 = 576
⇒AM = 24 cm.
Let r cm be the radius of the circle, then
OM = AM – AO = (24 – r) cm.
From right angled ∆ OBM, by Pythagoras theorem, we get
OB2 = BM2 + OM2
⇒
r2 = 72 + (24 – r)2
⇒
r2 = 49 + 576 + r2 – 48r
A
•O
B
M
C
625
1
= 13 .
48
48
1
Hence, the radius of the circle = 13 cm.
48
⇒
48r = 625
⇒
r=
Example 6. Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on
the bisector of ∠BAC.
Given. AB and AC are chords of a circle with centre O,
and AB = AC.
O
•
To prove. O lies on the bisector of ∠BAC i.e.
∠BAO = ∠OAC.
B
C
Construction. Join OB and OC.
A
Proof.
Statements
Reasons
In ∆s OBA and OCA
2184
1. OB = OC
1. Radii of same circle.
2. AB = AC
2. Given.
Understanding ICSE mathematics – Ix
3. OA = OA
3. Common.
4. ∆ OBA ≅ ∆ OCA
4. S.S.S. axiom of congruency.
5. ∠BAO = ∠OAC
Q.E.D.
5. ‘c.p.c.t.’.
Example 7. In a circle of radius 5 cm, AB and AC are two chords such that AB = AC = 6 cm.
Find the length of the chord BC.
Solution. Let O be the centre of the circle.
Let the bisector of ∠BAC meet the chord BC at M.
O
•
In ∆s BAM and CAM
1. ∠BAM = ∠MAC
(by const.)
C
B
M
2.AB = AC
(given)
A
3.AM = AM
(common)
∴ ∆ BAM ≅ ∆ CAM
(S.A.S. axiom of congruency)
∴ BM = MC and ∠ΒMA = ∠AMC
(‘c.p.c.t.’)
But ∠BMA + ∠AMC = 180°
( BMC is a st. line)
⇒
∠BMA = 90°.
Therefore, AM is perpendicular bisector of BC, and hence it passes through the centre
O of the circle.
LetBM = y cm and OM = x cm, then
MA = OA – OM = (5 – x) cm
[ radius = 5 cm]
2
2
2
From right angled ∆ OBM, y + x = 5 …(i)
2
2
2
From right angled ∆ BAM, y + (5 – x) = 6
i.e.
y2 + x2 – 10 x = 11
…(ii)
Subtracting (ii) from (i), we get
10 x = 25 – 11 ⇒ 10 x = 14 ⇒
Substituting this value of x in (i), we get
∴
x=
7
.
5
49
625 − 49
576
=
⇒ y2 =
⇒
25
25
25
24
48
Length of chord BC = 2BM = 2 . cm =
cm = 9∙6 cm.
5
5 y2 +
49
= 25
25
⇒
y2 = 25 –
y=
24
5
Example 8. Prove that the line joining mid-points of two parallel chords of a circle passes through
the centre of the circle.
Given. AB, CD are two chords of a circle with centre O. AB CD,
N
D
C
M and N are mid-points of AB and CD respectively.
To prove. MN passes through O.
E
F
•
O
Construction. Join OM, ON and through O draw a straight line
A
B
M
parallel to AB.
Proof.
Statements
Reasons
1. OM ⊥ AB
1. M is mid-point of AB, and the line drawn from the centre
of circle to bisect a chord is perpendicular to it.
2. ∠AMO = 90°
2. OM ⊥ AB.
circle
2185
3. ∠MOE = 90°
3. OE AB, alternate angles are equal.
4. ∠NOE = 90°
4. Similarly, N is mid-point of CD and same reasons as
above.
5. ∠NOE + ∠MOE = 180°
5. Adding 3 and 4.
6. MON is a straight line.
Hence, MN passes through O.
Q.E.D.
6. Sum of adjacent angles is 180°.
Example 9. If two equal chords of a circle intersect, prove that their
segments will be equal.
Given. AB and CD are chords of a circle with centre O. AB and CD
intersect at P and AB = CD.
To prove. (i) AP = PD
(ii) PB = CP.
Construction. Draw OM ⊥ AB, ON ⊥ CD. Join OP.
Proof.
Statements
D
O
•
B
Reasons
1. Perpendicular from centre bisects the chord.
2. CN = ND = 1 CD
2. Same as above.
3. AM = ND and MB = CN
In ∆s OMP and ONP
3. AB = CD (given).
2
P
M
1. AM = MB = 1 AB
2
N
A
4. OM = ON
4. Equal chords of a circle are equidistant from the centre.
(Theorem 14.3)
5. ∠OMP = ∠ONP
5. Each = 90°.
6. OP = OP
6. Common.
7. ∆ OMP ≅ ∆ ONP
7. R.H.S. axiom of congruency.
8. MP = PN
8. ‘c.p.c.t.’.
9. AM + MP = ND + PN
⇒ AP = PD.
9. From 3 and 8, adding.
10. MB – MP = CN – PN
⇒ PB = CP.
Hence (i) AP = PD and
(ii) PB = CP. Q.E.D.
10. From 3 and 8, subtracting.
Example 10. Of two unequal chords of a circle, prove that greater chord is
nearer to the centre of the circle.
Given. AB and CD are chords of a circle with centre O. AB > CD,
and OM ⊥ AB, ON ⊥ CD.
To prove. OM < ON.
Construction. Join OA and OC.
2186
Understanding ICSE mathematics – Ix
C
N
D
•O
A
M
B
Proof.
Statements
Reasons
1. AM = 1 AB
1. Perpendicular from centre bisects the chord.
2. CN = 1 CD
2. Same as above.
3. AM > CN
3. AB > CD (given).
4. OA2 = AM2 + OM2
4. In ∆ OAM, ∠AMO = 90°.
5. OC2 = CN2 + ON2
5. In ∆ OCN, ∠CNO = 90°.
6. AM2 + OM2 = CN2 + ON2
⇒ OM2 – ON2 = – (AM2 – CN2)
6. OA = OC, radii of same circle.
7. OM2 – ON2 < 0
⇒ OM2 < ON2
⇒ OM < ON.
Q.E.D.
7. From 3, AM > CN
⇒ AM2 > CN2
⇒ AM2 – CN2 is +ve.
2
2
Example 11. If two circles intersect in two points, then prove that the
line through their centres is perpendicular bisector of the common chord.
Given. Two circles with centres C, D, and intersecting at points
A, B so that AB is their common chord.
To prove. CD is perpendicular bisector of AB i.e. AM = MB and
∠CMA = 90°.
Construction. Join CA, CB, AD and BD.
Proof.
Statements
A
C•
M
•D
B
Reasons
In ∆s ACD and BCD
1. CA = CB
1. Radii of same circle.
2. AD = BD
2. Radii of same circle.
3. CD = CD
3. Common.
4. ∆ ACD ≅ ∆ BCD
4. S.S.S. axiom of congruency.
5. ∠ACM = ∠MCB
In ∆s ACM and BCM
5. ‘c.p.c.t.’.
6. ∠ACM = ∠MCB
6. From 5.
7. CA = CB
7. Radii of same circle.
8. CM = CM
8. Common.
9. ∆ ACM ≅ ∆ BCM
9. S.A.S. axiom of congruency.
10. AM = MB and
∠CMA = ∠CMB
10. ‘c.p.c.t.’.
11. ∠CMA + ∠CMB = 180°
11. AMB is a st. line
12. ∠CMA = 90°
Hence, CD is perpendicular
bisector of AB. Q.E.D.
12. From 10 and 11.
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Exercise 14.1
1. Calculate the length of a chord which is at a distance 12 cm from the centre of a circle of
radius 13 cm.
2. A chord of length 48 cm is drawn in a circle of radius 25 cm. Calculate its distance from
the centre of the circle.
3. A chord of length 8 cm is at a distance 3 cm from the centre of the circle. Calculate the
radius of the circle.
4. Calculate the length of a chord which is at a distance 6 cm from the centre of a circle of
diameter 20 cm.
5. A chord of length 16 cm is at a distance 6 cm from the centre of the circle. Find the length
of the chord of the same circle which is at a distance 8 cm from the centre.
6. In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm
respectively. Calculate the distance between the chords, if they are on
(i) the same side of the centre. (ii) the opposite sides of the centre.
7. (a)In the figure (i) given below, O is the centre of the circle. AB and CD are two chords
of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24
cm, OM = 5 cm, ON = 12 cm. Find the:
(i) radius of the circle
(ii) length of chord CD.
(b)In the figure (ii) given below, CD is a diameter which meets the chord AB in E such
that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.
D
D
N
C
O•
A
M
O•
B
B
4 cm
4 cm
E 3 cm
A
C
(i)
(ii)
8. In the adjoining figure, AB and CD are two parallel chords and O is
the centre. If the radius of the circle is 15 cm, find the distance MN
between the two chords of length 24 cm and 18 cm respectively.
M
A
O•
C
B
N
D
9. AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm respectively. If
the chords lie on the same side of the centre and the distance between them is 3 cm, find
the diameter of the circle.
10. ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12 5 cm and
BC = 24 cm, find the radius of the circle.
11. An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.
12. AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and
MB = 8 cm. Find the length of the shortest chord through M.
13.A rectangle with one side of length 4 cm is inscribed in a circle of diameter 5 cm. Find the
area of the rectangle.
14. The length of the common chord of two intersecting circles is 30 cm. If the radii of the
two circles are 25 cm and 17 cm, find the distance between their centres.
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Understanding ICSE mathematics – Ix