Lesson: Circle
Exercise 10.1
Question: 1
Fill in the blanks:
(i) The centre of a circle lies in the____________ of
the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle
is greater than its radius lies in the__________ of the
circle. (exterior/ interior)
(iii) The longest chord of a circle is a _____________
of the circle.
(iv) An arc is a ___________ when its ends are the
ends of a diameter.
(v) Segment of a circle is the region between an arc
and a_____________ of the circle.
(vi) A circle divides the plane, on which it lies, in
_____________ parts.
Solution
(i) The centre of a circle lies in the interior of the
circle.
(ii) A point, whose distance from the centre of a circle
is greater than its radius lies in the exterior of the
circle.
(iii) The longest chord of a circle is a diameter of the
circle.
(iv) An arc is a semicircle when its ends are the ends
of a diameter.
(v) Segment of a circle is the region between an arc
and a chord of the circle.
(vi) A circle divides the plane, on which it lies,
in three parts.
Question: 2
(i) Line segment joining the centre to any point on the
circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each of
them is a major arc.
(iv) A chord of a circle, which is twice as long as its
radius, is a diameter of the circle.
(v) Sector is the region between the chord and its
corresponding arc.
(vi) A circle is a plane figure.
Solution
(i) True.
All line segments from the centre of the circle to the
point on the circle are of equal length and are called
(ii) False.
There can be infinite numbers of equal chords.
(iii) False.
Let us consider a circle with three equal arcs PQ, QR
and PR as shown in the figure below.
Now, for arc PQ, the major arc is PRQ and the same is
true for the other two arcs.
Hence, all the three arcs are minor arcs.
(iv) True.
A chord which is twice as long as a radius must pass
through the centre of the circle. Hence, it is called the
diameter of the circle.
(v) False.
Sector is the region between the two radii of the circle
and the arc.
(vi) True.
A circle is a two-dimensional figure and can be drawn
on the plane.
Exercise 10.2
Question: 1
Recall that two circles are congruent if they have the
same radii. Prove that equal chords of congruent
circles subtend equal angles at their centres.
Solution
Given two congruent circles.
Let AB and CD be two equal chords of two congruent
circles having centres O and O respectively.
AB CD
(Equal chords)
Now,
In ΔAOB and ΔCOD,
OA OC
OB OD
AB CD
(Given)
Δ ΔAOB COD
(SSS congruence condition)
Thus,
AOB COD
by CPCT.
Hence, equal chords of congruent circles subtend
equal angles at their centres.
Question: 2
Prove that if chords of congruent circles subtend equal
angles at their centres, then the chords are equal.
Solution
Given two congruent circles with equal radius and
centres O and
O
.
Given,
AOB COD
(Equal angles)
Now, in ΔAOB and
ΔCOD
,
OA OC
AOB COD
(Given)
OB OD
Δ ΔAOB COD
(SAS congruence condition)
Thus,
AB CD
by CPCT.
Hence, if chords of congruent circles subtend equal
angles at their centres, then the chords are equal.
Exercise 10.3
Question: 1
Draw different pairs of circles. How many points does
each pair have in common? What is the maximum
number of common points?
Solution
No point in common.
No point in common.
One point P is common.
One point P is common.
Two points, P and Q, are common.
Question: 2
Suppose you are given a circle. Give a construction to
find its centre.
Solution
Construction of a circle to find its centre:
Step I: Draw a circle.
Step II: Draw two chords AB and CD as shown in the
figure.
Step III: Draw perpendicular bisector of the chords AB
and CD.
Step IV: Let these two perpendicular bisectors meet at
a point O’. Hence, O is the centre of the given circle.
Question: 3
If two circles intersect at two points, prove that their
centres lie on the perpendicular bisector of the
common chord.
Solution
Let two circles with centres as O and
O
intersect
each other at points P and Q.
We need to prove that
is the perpendicular
bisector of PQ.
Now, in
ΔPOO
and
Δ 'QOO
,
OP OQ
' 'OO OO
(Common)
' 'O P O Q
Δ ' Δ 'POO QOO
(SSS congruence condition)
Hence,
' 'POO QOO
--- (i)
Again, in
ΔPOR
and
ΔQOR
,
OP OQ
POR QOR
(From equation (i))
OR OR
(Common)
Δ ΔPOR QOR
(SAS congruence condition)
Hence,
PRO QRO
and PR = RQ (By CPCT)
also,
180PRO QRO
180
90
2
PRO QRO
Therefore,
OO' is the perpendicular bisector of PQ.
Exercise 10.4
Question: 1
Two circles of radii 5 cm and 3 cm intersect at two
points and the distance between their centres is 4 cm.
Find the length of the common chord.
Solution
Let two circles with centres as O and S intersect at
points P and Q as shown in the figure.
Given,
5 cm, 3 cmOP PS
and
4 cmOS
.
also,
2PQ PR
Let
RS
be a.
Therefore, if two circles intersect at two points, then
their centres lie on the perpendicular bisector of the
common chord.
90ORP PRS
Now, in
ΔPOR
,
2 2 2
OP OR PR
(By Pythagoras theorem)
2
2 2
2 2
2 2
5 4
25 16 8
9 8 i
a PR
a a PR
PR a a
In
ΔPRS
,
2 2 2
2 2 2
2 2
3
9 ... ii
PS PR RS
PR a
PR a
Equating (i) and (ii),
2 2
9 8 9
8 0
0
a a a
a
a
Putting the value of a in (ii), we get,
2 2
9 0
3 cm
PR
PR
Length of the chord
2 2 3 6 cmPQ PR
Question: 2
If two equal chords of a circle intersect within the
circle, prove that the segments of one chord are equal
to the corresponding segments of the other chord.
Solution
Let AB and CD be the chords of a circle with centre O,
intersecting at point E as shown in the figure.
Also,
AB CD
(Given)
Now, draw a perpendicular line from the centre of the
circle to chords AB and CD, such that,
OM AB and ON CD.
Also, join O to point E.
Now,
OM bisects
( )AB OM AB
ON bisects
( )CD ON CD
And
AB CD
, therefore,
... iAM ND
and
... iiMB CN
In
ΔOME
and
ΔONE
,
OME ONE
(Perpendiculars)
OE OE
(Common)
OM ON
(
AB CD
and therefore equidistant from
the centre)
Δ ΔOME ONE
by RHS congruence condition.
ME EN
by CPCT --- (iii)
From (i) and (iii), we get,
AM ME ND EN
AE ED
From (ii) and (iii), we get,
MB ME CN EN
EB CE
Hence, it is proved.
Question: 3
If two equal chords of a circle intersect within the
circle, prove that the line joining the point of
intersection to the centre makes equal angles with the
chords.
Solution
Let AB and CD be the chords of a circle, with centre O,
intersecting at point E as shown in the figure.
Also,
AB CD
(Given)
Now, draw a perpendicular line from the centre of the
circle to the chords AB and CD, such that,
OM AB
and
ON CD
.
Also, join O to point E. PQ is the diameter.
Now, in
ΔOEM
and
ΔOEN
,
OM ON
( Equal chords are equidistant from the
centre)
OE OE
(Common)
OME ONE
(Perpendicular)
Δ ΔOEM OEN
by RHS congruence condition.
Therefore,
MEO NEO
(By CPCT)
BEQ CEQ
Question: 4
If a line intersects two concentric circles (circles with
the same centre) with centre O at A, B, C and D, prove
that
AB CD
(See Fig. 10.25).
Solution
Draw a perpendicular OM to line AD from point O’.
OM
bisects
as
.
... iAM MD
Also,
bisects
BC
as
OM BC
.
... iiBM MC
From (i) and (ii),
AM BM MD MC
AB CD
Hence, it is proved.
Question: 5
Three girls Reshma, Salma and Mandip are playing a
game by standing on a circle of radius 5 m drawn in a
park. Reshma throws a ball to Salma, Salma to
Mandip, Mandip to Reshma. If the distance between
Reshma and Salma and between Salma and Mandip is
6 m each, what is the distance between Reshma and
Mandip?
Solution
Let Reshma, Salma and Mandip stand at positions A,
B and C on a circle with centre at O.
6 mAB
and
6 mBC
(Given)
5 mOA
(Given)
Draw a perpendicular BM to AC as shown in the
figure.
ABC is an isosceles triangle, then
AB BC
and M is
the mid-point of AC and BM is a perpendicular
bisector of AC. Hence, BM passes through the centre
of the circle.
Now, let
AM y
and
OM x
then
5BM x
.
2 2 2
OA OM AM
(By Pythagoras theorem)
2 2 2
5 ... ix y
Also,
2 2 2
AB BM AM
(By Pythagoras theorem)
2
2 2
6 5 ... iix y
Subtracting (i) from (ii), we get
2
2
36 25 5
11 25 10
10 14
7
5
x x
x
x
x
Substituting the value of x in (i), we get
2
2
2
2
49
25
25
49
25
25
625 49
25
576
25
24
5
y
y
y
y
y
Therefore,
24 48
2 2 2 m m 9.6 m
5 5
AC AM y
Distance between Reshma and Mandip is 9.6 m.
Question: 6
A circular park of radius 20 m is situated in a colony.
Three boys Ankur, Syed and David are sitting at equal
distance on its boundary each carrying a toy telephone
in his hand and talking to each other. Find the length
of the string of each phone.
Solution
Let Ankur, Syed and David be sitting at positions A, B,
and C on the boundary of a circular park with centre at
O and radius as 20 m.
All three boys are sitting at equal distances from each
other. Thus, ABC is an equilateral triangle.
median of ΔABC and it passes through the centre O.
Again, O is the centroid of ΔABC. OA is the radius of
the circle.
2
3
Let the side of a triangle be a metres.
2
a
BD
Applying Pythagoras theorem in ΔABD,
 
2 2 2
2 2 2
2
2 2
2
2
2
3
3 / 2, 2/ 3
20 2/ 3 3 / 2
20 3
4
a
a
m a
a m
Length of the string is 20√3 m.
Exercise 10.5
Question: 1
In Fig. 10.36, A, B and C are three points on a circle
with centre O such that
30BOC
and
60AOB
. If D is a point on the circle other than
the arc ABC, find
.
Solution
Given,
30BOC
and
60AOB
60 30
90
AOC AOB BOC
AOC
AOC
The angle subtended by an arc at the centre is
double the angle subtended by it at any point on the
remaining part of the circle,
1 1
90 45
2 2
Question: 2
A chord of a circle is equal to the radius of the circle.
Find the angle subtended by the chord at a point on the
minor arc and also at a point on the major arc.
Solution
Let OA and OB be the radius of the circle with centre
as O, and AB as the chord of a circle.
OA OB AB
Therefore,
ΔOAB
is an equilateral triangle.
60AOB
Also,
1 1
60 30
2 2
ACB AOB
(Opposite angles of cyclic
Hence, angle subtended by the chord at a point on the
minor arc and at a point on the major arc are 150° and
30° respectively.
Question: 3
In Fig. 10.37,
100PQR
, where P, Q and R are
points on a circle with centre O. Find
OPR
.
Solution
Let us take any point S on the major arc of chord PR
as shown in the figure.
Join PS and RS.
180PQR PSR
(Sum of opposite angles of a
180 100 80PSR
(Given,
100PQR
) …. (i)
Now, the angle subtended by an arc at the centre is
double the angle subtended by it at any point on the
remaining part of the circle.
Here, arc PR is subtending POR at the centre and
PSR at point S on the remaining part of the circle.
2 2 80 160POR PSR
(Using equation
(i)) (ii)
In POR,
OP OR
ORP OPR
(Angles opposite to equal sides
in a triangle) (iii)
Also,
180POR OPR ORP
(Angle sum
property of a triangle)
160 180OPR OPR
(Using equations (ii)
and (iii))
10OPR
Question: 4
In Fig. 10.38,
69 , 31 ,ABC ACB
find
BDC
.
Solution
In
ΔABC
and
,
BAC BDC
(Angles in the segment of the circle
are equal.)
Now, in
ΔABC
,
180BAC ABC ACB
(Sum of the angles in
a triangle)
69 31 180
180 100
80
BAC
BAC
BAC
Therefore,
80BDC
.
Question: 5
In Fig. 10.39, A, B, C and D are four points on a circle.
AC and BD intersect at a point E such that
130BEC
and
20 .ECD
Find
BAC
.
Fig. 10.39
Solution
In
ΔABC
and
BAC CDE
(Angles in the segment of the circle
are equal.)
In
ΔCDE
,
CEB CDE DCE
(Exterior angles of a triangle
is the sum of the interior opposite angles.)
130 20
110
CDE
CDE
Therefore,
110BAC
.
Question: 6
ABCD is a cyclic quadrilateral whose diagonals
intersect at a point E. If
70 ,DBC BAC  
=
30
,
find
BCD
. Further, if
AB BC
, find
ECD
.
Solution
In
and
ΔDBC
,
(Angles in same segment of a circle
are equal.)
(Opposite angles of a cyclic
In
100 180
80
Δ
BCD
BCD
ABC
AB BC
(Given)
BCA CAB
(Angles opposite to equal sides of a
triangle)
30BCA
also,
80BCD

80
30 80
50
50
BCA ACD
ACD
ACD
ECD
Question: 7
If diagonals of a cyclic quadrilateral are diameters of
the circle through the vertices of the quadrilateral,
prove that it is a rectangle.
Solution
Let ABCD be a cyclic quadrilateral as shown in figure
below and its diagonals AC and BD be the diameters
of the circle through the vertices of the quadrilateral.
90ABC BCD CDA DAB
(Angles in
the semi-circle)
Therefore, ABCD is a rectangle as each internal angle
is 90°.
Question: 8
If the non-parallel sides of a trapezium are equal,
prove that it is cyclic.
Solution
Let AD and BC be the non-parallel sides of trapezium
ABCD, such that,
Let us draw DM and CN perpendicular to AB.
Now, in
ΔDAM
and
Δ ,CBN
(Given)
AMD BNC
(Right angles)
DM CN
(Distance between the parallel lines)
Δ ΔDAM CBN
by RHS congruence condition.
Now,
A B
by CPCT
also,
180B C
(Sum of the co-interior angles)
180A C
Therefore, ABCD is a cyclic quadrilateral as sum of
the pair of opposite angles is 180°.
Question: 9
Two circles intersect at two points B and C. Through
B, two line segments ABD and PBQ are drawn to
intersect the circles at A, D and P, Q respectively (see
Fig. 10.40). Prove that
ACP QCD
.
Solution
Let us join A and P, and D and Q as shown in the
figure below.
Now, for chord AP,
PBA ACP
( Angles in the same segment) --- (i)
Similarly, for chord DQ,
DBQ QCD
(Angles in same segment) --- (ii)
ABD and PBQ are line segments intersecting at B.
PBA DBQ
(Vertically opposite angles) --- (iii)
By the equations (i), (ii) and (iii),
ACP QCD
Hence, it is proved.
Question: 10
If circles are drawn taking two sides of a triangle as
diameters, prove that the point of intersection of these
circles lies on the third side.
Solution
Let the triangle be ABC. Now, let us draw two circles
on the sides AB and AC of the
ΔABC
as diameters
such that the circles intersect at point D.
Join AD. Now, to prove that D lies on BC, we need to
prove that BDC is a straight line.
(Angle in the semi-circle)
Now,
BDC
is straight line.
Therefore, D lies on the side BC.
Question: 11
ABC and ADC are two right triangles with common
hypotenuse AC. Prove that
.
Solution
Let
ΔABC
and
be the two right angled
triangles such that hypotenuse AC is common.
(Given)
Since,
ABC
and
are 90°, these angles are
in the semi circle. Hence, both the triangles lie on the
semi circle and AC is the diameter of the circle.
Points A, B, C and D are concyclic.
Thus, CD is the chord.
(Angles in the same segment of
the circle)
Hence, it is proved.
Question: 12
Prove that a cyclic parallelogram is a rectangle.
Solution
Let ABCD be a cyclic parallelogram as shown in the
figure.
1 2 180
(Opposite angles of a cyclic
parallelogram)
Also, opposite angles of a cyclic parallelogram are
equal.
Therefore,
1 2
1 1 180
2 1 90
One of the interior angle of the parallelogram is right
angled. Therefore, ABCD is a rectangle.