NCERT Textbook: Circles | Mathematics (Maths) Class 9 PDF Download

 Page 1


116 MATHEMA TICS
CHAPTER 9
CIRCLES
9.1 Angle Subtended by a Chord at a Point
You have already studied about circles and its parts in Class VI.
Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR
(see Fig. 9.1). Then  ? PRQ is called the angle subtended by the line segment PQ at
the point R. What are angles POQ, PRQ and PSQ called in Fig. 9.2? ? POQ is the
angle subtended by the chord PQ at the centre O, ? PRQ and ? PSQ are respectively
the angles subtended by PQ at points R and S on the major and minor arcs PQ.
Fig. 9.1            Fig. 9.2
Let us examine the relationship between the size of the chord and the angle
subtended by it at the centre. You may see by drawing different chords of a circle
and  angles subtended by them at the centre that the longer is the chord, the bigger
will be the angle subtended by it at the centre. What will happen if you take two
equal chords of a circle? Will the angles subtended at the centre be the same
or not?
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116 MATHEMA TICS
CHAPTER 9
CIRCLES
9.1 Angle Subtended by a Chord at a Point
You have already studied about circles and its parts in Class VI.
Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR
(see Fig. 9.1). Then  ? PRQ is called the angle subtended by the line segment PQ at
the point R. What are angles POQ, PRQ and PSQ called in Fig. 9.2? ? POQ is the
angle subtended by the chord PQ at the centre O, ? PRQ and ? PSQ are respectively
the angles subtended by PQ at points R and S on the major and minor arcs PQ.
Fig. 9.1            Fig. 9.2
Let us examine the relationship between the size of the chord and the angle
subtended by it at the centre. You may see by drawing different chords of a circle
and  angles subtended by them at the centre that the longer is the chord, the bigger
will be the angle subtended by it at the centre. What will happen if you take two
equal chords of a circle? Will the angles subtended at the centre be the same
or not?
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CIRCLES 117
Draw two or more equal chords of a circle and
measure the angles subtended by them at the centre
(see Fig.9.3). You will find that the angles subtended
by them at the centre are equal. Let us give a proof
of this fact.
Theorem 9.1 : Equal chords of a circle subtend
equal angles at the centre.
Proof : You are given two equal chords AB and CD
of a circle with centre O (see Fig.9.4). You want to
prove that ? AOB = ? COD.
In triangles AOB and COD,
OA = OC (Radii of a circle)
OB = OD (Radii of a circle)
AB = CD (Given)
Therefore, ? AOB ? ? COD (SSS rule)
This gives ? AOB = ? COD
               (Corresponding parts of congruent triangles)
Remark : For convenience, the abbreviation CPCT will be used in place of
‘Corresponding parts of congruent triangles’, because we use this very frequently as
you will see.
Now if two chords of a circle subtend equal angles at the centre, what can you
say about the chords? Are they equal or not? Let us examine this by the following
activity:
Take a tracing paper and trace a circle on it. Cut
it along the circle to get a disc. At its centre O, draw
an angle AOB where A, B are points on the circle.
Make another angle POQ at the centre equal to
?AOB. Cut the disc along AB and PQ
(see Fig. 9.5). You will get two segments ACB and
PRQ of the circle. If you put one on the other, what
do you observe? They cover each other, i.e., they
are congruent. So AB  = PQ.
Fig. 9.3
Fig. 9.4
Fig. 9.5
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116 MATHEMA TICS
CHAPTER 9
CIRCLES
9.1 Angle Subtended by a Chord at a Point
You have already studied about circles and its parts in Class VI.
Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR
(see Fig. 9.1). Then  ? PRQ is called the angle subtended by the line segment PQ at
the point R. What are angles POQ, PRQ and PSQ called in Fig. 9.2? ? POQ is the
angle subtended by the chord PQ at the centre O, ? PRQ and ? PSQ are respectively
the angles subtended by PQ at points R and S on the major and minor arcs PQ.
Fig. 9.1            Fig. 9.2
Let us examine the relationship between the size of the chord and the angle
subtended by it at the centre. You may see by drawing different chords of a circle
and  angles subtended by them at the centre that the longer is the chord, the bigger
will be the angle subtended by it at the centre. What will happen if you take two
equal chords of a circle? Will the angles subtended at the centre be the same
or not?
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CIRCLES 117
Draw two or more equal chords of a circle and
measure the angles subtended by them at the centre
(see Fig.9.3). You will find that the angles subtended
by them at the centre are equal. Let us give a proof
of this fact.
Theorem 9.1 : Equal chords of a circle subtend
equal angles at the centre.
Proof : You are given two equal chords AB and CD
of a circle with centre O (see Fig.9.4). You want to
prove that ? AOB = ? COD.
In triangles AOB and COD,
OA = OC (Radii of a circle)
OB = OD (Radii of a circle)
AB = CD (Given)
Therefore, ? AOB ? ? COD (SSS rule)
This gives ? AOB = ? COD
               (Corresponding parts of congruent triangles)
Remark : For convenience, the abbreviation CPCT will be used in place of
‘Corresponding parts of congruent triangles’, because we use this very frequently as
you will see.
Now if two chords of a circle subtend equal angles at the centre, what can you
say about the chords? Are they equal or not? Let us examine this by the following
activity:
Take a tracing paper and trace a circle on it. Cut
it along the circle to get a disc. At its centre O, draw
an angle AOB where A, B are points on the circle.
Make another angle POQ at the centre equal to
?AOB. Cut the disc along AB and PQ
(see Fig. 9.5). You will get two segments ACB and
PRQ of the circle. If you put one on the other, what
do you observe? They cover each other, i.e., they
are congruent. So AB  = PQ.
Fig. 9.3
Fig. 9.4
Fig. 9.5
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118 MATHEMA TICS
Though you have seen it for this particular case, try it out for other equal angles
too. The chords will all turn out to be equal because of the following theorem:
Theorem 9.2 : If the angles subtended by the chords of a circle at the centre are
equal, then the chords are equal.
The above theorem is the converse of the Theorem 9.1. Note that in Fig. 9.4, if
you take ? AOB = ? COD, then
           ? AOB ? ? COD (Why?)
Can you now see that AB = CD?
EXERCISE 9.1
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.
2. Prove that if chords of congruent circles subtend equal angles at their centres, then
the chords are equal.
9.2 Perpendicular from the Centre to a Chord
Activity : Draw a circle on a tracing paper. Let O
be its centre. Draw a chord AB. Fold the paper along
a line through O so that a portion of the chord falls on
the other. Let the crease cut AB at the point M. Then,
? OMA = ? OMB = 90° or OM is perpendicular to
AB. Does the point B coincide with A (see Fig.9.6)?
Yes it will. So MA = MB.
Give a proof yourself by joining OA and OB and proving the right triangles OMA
and OMB to be congruent. This example is a particular instance of the following
result:
Theorem 9.3 : The perpendicular from the centre of a circle to a chord bisects
the chord.
What is the converse of this theorem? To write this, first let us be clear what is
assumed in Theorem 9.3 and what is proved. Given that the perpendicular from the
centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus in the
converse, what the hypothesis is ‘if a line from the centre bisects a chord of a
circle’ and what is to be proved is ‘the line is perpendicular to the chord’. So the
converse is:
Fig. 9.6
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116 MATHEMA TICS
CHAPTER 9
CIRCLES
9.1 Angle Subtended by a Chord at a Point
You have already studied about circles and its parts in Class VI.
Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR
(see Fig. 9.1). Then  ? PRQ is called the angle subtended by the line segment PQ at
the point R. What are angles POQ, PRQ and PSQ called in Fig. 9.2? ? POQ is the
angle subtended by the chord PQ at the centre O, ? PRQ and ? PSQ are respectively
the angles subtended by PQ at points R and S on the major and minor arcs PQ.
Fig. 9.1            Fig. 9.2
Let us examine the relationship between the size of the chord and the angle
subtended by it at the centre. You may see by drawing different chords of a circle
and  angles subtended by them at the centre that the longer is the chord, the bigger
will be the angle subtended by it at the centre. What will happen if you take two
equal chords of a circle? Will the angles subtended at the centre be the same
or not?
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CIRCLES 117
Draw two or more equal chords of a circle and
measure the angles subtended by them at the centre
(see Fig.9.3). You will find that the angles subtended
by them at the centre are equal. Let us give a proof
of this fact.
Theorem 9.1 : Equal chords of a circle subtend
equal angles at the centre.
Proof : You are given two equal chords AB and CD
of a circle with centre O (see Fig.9.4). You want to
prove that ? AOB = ? COD.
In triangles AOB and COD,
OA = OC (Radii of a circle)
OB = OD (Radii of a circle)
AB = CD (Given)
Therefore, ? AOB ? ? COD (SSS rule)
This gives ? AOB = ? COD
               (Corresponding parts of congruent triangles)
Remark : For convenience, the abbreviation CPCT will be used in place of
‘Corresponding parts of congruent triangles’, because we use this very frequently as
you will see.
Now if two chords of a circle subtend equal angles at the centre, what can you
say about the chords? Are they equal or not? Let us examine this by the following
activity:
Take a tracing paper and trace a circle on it. Cut
it along the circle to get a disc. At its centre O, draw
an angle AOB where A, B are points on the circle.
Make another angle POQ at the centre equal to
?AOB. Cut the disc along AB and PQ
(see Fig. 9.5). You will get two segments ACB and
PRQ of the circle. If you put one on the other, what
do you observe? They cover each other, i.e., they
are congruent. So AB  = PQ.
Fig. 9.3
Fig. 9.4
Fig. 9.5
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118 MATHEMA TICS
Though you have seen it for this particular case, try it out for other equal angles
too. The chords will all turn out to be equal because of the following theorem:
Theorem 9.2 : If the angles subtended by the chords of a circle at the centre are
equal, then the chords are equal.
The above theorem is the converse of the Theorem 9.1. Note that in Fig. 9.4, if
you take ? AOB = ? COD, then
           ? AOB ? ? COD (Why?)
Can you now see that AB = CD?
EXERCISE 9.1
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.
2. Prove that if chords of congruent circles subtend equal angles at their centres, then
the chords are equal.
9.2 Perpendicular from the Centre to a Chord
Activity : Draw a circle on a tracing paper. Let O
be its centre. Draw a chord AB. Fold the paper along
a line through O so that a portion of the chord falls on
the other. Let the crease cut AB at the point M. Then,
? OMA = ? OMB = 90° or OM is perpendicular to
AB. Does the point B coincide with A (see Fig.9.6)?
Yes it will. So MA = MB.
Give a proof yourself by joining OA and OB and proving the right triangles OMA
and OMB to be congruent. This example is a particular instance of the following
result:
Theorem 9.3 : The perpendicular from the centre of a circle to a chord bisects
the chord.
What is the converse of this theorem? To write this, first let us be clear what is
assumed in Theorem 9.3 and what is proved. Given that the perpendicular from the
centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus in the
converse, what the hypothesis is ‘if a line from the centre bisects a chord of a
circle’ and what is to be proved is ‘the line is perpendicular to the chord’. So the
converse is:
Fig. 9.6
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CIRCLES 119
Theorem 9.4 : The line drawn through the centre of a circle to bisect a chord is
perpendicular to the chord.
Is this true? Try it for few cases and see. You will
see that it is true for these cases. See if it is true, in
general, by doing the following exercise. W e will write
the stages and you give the reasons.
Let AB be a chord of a circle with centre O and
O is joined to the mid-point M of AB. You have to
prove that OM ? AB. Join OA and OB
(see Fig. 9.7). In triangles OAM and OBM,
OA = OB (Why ?)
AM = BM (Why ?)
OM = OM (Common)
Therefore, ?OAM ? ?OBM (How ?)
This gives ?OMA = ?OMB = 90°    (Why ?)
9.3 Equal Chords and their Distances from the Centre
Let AB be a line and P be a point. Since there are
infinite numbers of points on a line, if you join these
points to P , you will get infinitely many line segments
PL
1
, PL
2
, PM, PL
3
, PL
4
, etc. Which of these is the
distance of AB from P? You may think a while and
get the answer. Out of these line segments, the
perpendicular from P to AB, namely PM in Fig. 9.8,
will be the least. In Mathematics, we define this least
length PM to be the distance of AB from P. So you
may say that:
The length of the perpendicular from a point to a line is the distance of the
line from the point.
Note that if the point lies on the line, the distance of the line from the point is zero.
A circle can have infinitely many chords. Y ou may observe by drawing chords of
a circle that longer chord is nearer to the centre than the smaller chord. You may
observe it by drawing several chords of a circle of different lengths and measuring
their distances from the centre.  What is the distance of the diameter, which is the
Fig. 9.7
Fig. 9.8
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116 MATHEMA TICS
CHAPTER 9
CIRCLES
9.1 Angle Subtended by a Chord at a Point
You have already studied about circles and its parts in Class VI.
Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR
(see Fig. 9.1). Then  ? PRQ is called the angle subtended by the line segment PQ at
the point R. What are angles POQ, PRQ and PSQ called in Fig. 9.2? ? POQ is the
angle subtended by the chord PQ at the centre O, ? PRQ and ? PSQ are respectively
the angles subtended by PQ at points R and S on the major and minor arcs PQ.
Fig. 9.1            Fig. 9.2
Let us examine the relationship between the size of the chord and the angle
subtended by it at the centre. You may see by drawing different chords of a circle
and  angles subtended by them at the centre that the longer is the chord, the bigger
will be the angle subtended by it at the centre. What will happen if you take two
equal chords of a circle? Will the angles subtended at the centre be the same
or not?
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CIRCLES 117
Draw two or more equal chords of a circle and
measure the angles subtended by them at the centre
(see Fig.9.3). You will find that the angles subtended
by them at the centre are equal. Let us give a proof
of this fact.
Theorem 9.1 : Equal chords of a circle subtend
equal angles at the centre.
Proof : You are given two equal chords AB and CD
of a circle with centre O (see Fig.9.4). You want to
prove that ? AOB = ? COD.
In triangles AOB and COD,
OA = OC (Radii of a circle)
OB = OD (Radii of a circle)
AB = CD (Given)
Therefore, ? AOB ? ? COD (SSS rule)
This gives ? AOB = ? COD
               (Corresponding parts of congruent triangles)
Remark : For convenience, the abbreviation CPCT will be used in place of
‘Corresponding parts of congruent triangles’, because we use this very frequently as
you will see.
Now if two chords of a circle subtend equal angles at the centre, what can you
say about the chords? Are they equal or not? Let us examine this by the following
activity:
Take a tracing paper and trace a circle on it. Cut
it along the circle to get a disc. At its centre O, draw
an angle AOB where A, B are points on the circle.
Make another angle POQ at the centre equal to
?AOB. Cut the disc along AB and PQ
(see Fig. 9.5). You will get two segments ACB and
PRQ of the circle. If you put one on the other, what
do you observe? They cover each other, i.e., they
are congruent. So AB  = PQ.
Fig. 9.3
Fig. 9.4
Fig. 9.5
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118 MATHEMA TICS
Though you have seen it for this particular case, try it out for other equal angles
too. The chords will all turn out to be equal because of the following theorem:
Theorem 9.2 : If the angles subtended by the chords of a circle at the centre are
equal, then the chords are equal.
The above theorem is the converse of the Theorem 9.1. Note that in Fig. 9.4, if
you take ? AOB = ? COD, then
           ? AOB ? ? COD (Why?)
Can you now see that AB = CD?
EXERCISE 9.1
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.
2. Prove that if chords of congruent circles subtend equal angles at their centres, then
the chords are equal.
9.2 Perpendicular from the Centre to a Chord
Activity : Draw a circle on a tracing paper. Let O
be its centre. Draw a chord AB. Fold the paper along
a line through O so that a portion of the chord falls on
the other. Let the crease cut AB at the point M. Then,
? OMA = ? OMB = 90° or OM is perpendicular to
AB. Does the point B coincide with A (see Fig.9.6)?
Yes it will. So MA = MB.
Give a proof yourself by joining OA and OB and proving the right triangles OMA
and OMB to be congruent. This example is a particular instance of the following
result:
Theorem 9.3 : The perpendicular from the centre of a circle to a chord bisects
the chord.
What is the converse of this theorem? To write this, first let us be clear what is
assumed in Theorem 9.3 and what is proved. Given that the perpendicular from the
centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus in the
converse, what the hypothesis is ‘if a line from the centre bisects a chord of a
circle’ and what is to be proved is ‘the line is perpendicular to the chord’. So the
converse is:
Fig. 9.6
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CIRCLES 119
Theorem 9.4 : The line drawn through the centre of a circle to bisect a chord is
perpendicular to the chord.
Is this true? Try it for few cases and see. You will
see that it is true for these cases. See if it is true, in
general, by doing the following exercise. W e will write
the stages and you give the reasons.
Let AB be a chord of a circle with centre O and
O is joined to the mid-point M of AB. You have to
prove that OM ? AB. Join OA and OB
(see Fig. 9.7). In triangles OAM and OBM,
OA = OB (Why ?)
AM = BM (Why ?)
OM = OM (Common)
Therefore, ?OAM ? ?OBM (How ?)
This gives ?OMA = ?OMB = 90°    (Why ?)
9.3 Equal Chords and their Distances from the Centre
Let AB be a line and P be a point. Since there are
infinite numbers of points on a line, if you join these
points to P , you will get infinitely many line segments
PL
1
, PL
2
, PM, PL
3
, PL
4
, etc. Which of these is the
distance of AB from P? You may think a while and
get the answer. Out of these line segments, the
perpendicular from P to AB, namely PM in Fig. 9.8,
will be the least. In Mathematics, we define this least
length PM to be the distance of AB from P. So you
may say that:
The length of the perpendicular from a point to a line is the distance of the
line from the point.
Note that if the point lies on the line, the distance of the line from the point is zero.
A circle can have infinitely many chords. Y ou may observe by drawing chords of
a circle that longer chord is nearer to the centre than the smaller chord. You may
observe it by drawing several chords of a circle of different lengths and measuring
their distances from the centre.  What is the distance of the diameter, which is the
Fig. 9.7
Fig. 9.8
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120 MATHEMA TICS
longest chord from the centre? Since the centre lies on it, the distance is zero. Do you
think that there is some relationship between the length of chords and their distances
from the centre? Let us see if this is so.
Fig. 9.9
Activity : Draw a circle of any radius on a tracing paper. Draw two equal chords
AB and CD of it and also the perpendiculars OM and ON on them from the centre
O. Fold the figure so that D falls on B and C falls on A [see Fig.9.9 (i)]. You may
observe that O lies on the crease and N falls on M. Therefore, OM = ON. Repeat
the activity by drawing congruent circles with centres O and O' and taking equal
chords AB and CD one on each. Draw perpendiculars OM and O'N on them [see
Fig. 9.9(ii)]. Cut one circular disc and put it on the other so that AB coincides with
CD. Then you will find that O coincides with O' and M coincides with N. In this
way you verified the following:
Theorem 9.5 : Equal chords of a circle (or of congruent circles) are equidistant
from the centre (or centres).
Next, it will be seen whether the converse of this theorem is true or not. For
this, draw a circle with centre O. From the centre O, draw two line segments OL
and OM of equal length and lying inside the circle [see Fig. 9.10(i)]. Then draw
chords PQ and RS of the circle perpendicular to OL and OM respectively [see Fig
9.10(ii)]. Measure the lengths of PQ and RS. Are these different? No, both are
equal. Repeat the activity for more equal line segments and drawing the chords
perpendicular to them. This verifies the converse of the Theorem 9.5 which is stated
as follows:
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