# NCERT Solutions, Chapter 10: Circles, Class 10 (Mathematics) Notes - Class 10

## Class 10: NCERT Solutions, Chapter 10: Circles, Class 10 (Mathematics) Notes - Class 10

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Mathematics
(Chapter – 10) (Circles)
(Class – X)

1
Exercise 10.2
Question 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 the circle is

(A) 7 cm   (B) 12 cm   (C) 15 cm   (D) 24.5 cm

Let O be the centre of the circle.
Given that,
OQ = 25cm and PQ = 24 cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, OP ? PQ
Applying Pythagoras theorem in ?OPQ, we obtain
OP
2
+ PQ
2
= OQ
2

OP
2
+ 24
2
= 25
2

OP
2
= 625 - 576
OP
2
= 49
OP = 7
Therefore, the radius of the circle is 7 cm.
Hence, alternative (A) is correct
Page 2

Mathematics
(Chapter – 10) (Circles)
(Class – X)

1
Exercise 10.2
Question 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 the circle is

(A) 7 cm   (B) 12 cm   (C) 15 cm   (D) 24.5 cm

Let O be the centre of the circle.
Given that,
OQ = 25cm and PQ = 24 cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, OP ? PQ
Applying Pythagoras theorem in ?OPQ, we obtain
OP
2
+ PQ
2
= OQ
2

OP
2
+ 24
2
= 25
2

OP
2
= 625 - 576
OP
2
= 49
OP = 7
Therefore, the radius of the circle is 7 cm.
Hence, alternative (A) is correct
Mathematics
(Chapter – 10) (Circles)
(Class – X)

2
Question 2:
In the given figure, if TP and TQ are the two tangents to a circle with centre O so that
?POQ = 110 , then ?PTQ is equal to

(A) 60   (B) 70    (C) 80  (D) 90

It is given that TP and TQ are tangents.
Therefore, radius drawn to these tangents will be perpendicular to the tangents.
Thus, OP ? TP and OQ ? TQ
?OPT = 90º
?OQT = 90º
Sum of all interior angles = 360
?OPT + ?POQ + ?OQT + ?PTQ = 360
? 90 + 110º + 90 + PTQ = 360
? PTQ = 70
Hence, alternative (B) is correct

Page 3

Mathematics
(Chapter – 10) (Circles)
(Class – X)

1
Exercise 10.2
Question 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 the circle is

(A) 7 cm   (B) 12 cm   (C) 15 cm   (D) 24.5 cm

Let O be the centre of the circle.
Given that,
OQ = 25cm and PQ = 24 cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, OP ? PQ
Applying Pythagoras theorem in ?OPQ, we obtain
OP
2
+ PQ
2
= OQ
2

OP
2
+ 24
2
= 25
2

OP
2
= 625 - 576
OP
2
= 49
OP = 7
Therefore, the radius of the circle is 7 cm.
Hence, alternative (A) is correct
Mathematics
(Chapter – 10) (Circles)
(Class – X)

2
Question 2:
In the given figure, if TP and TQ are the two tangents to a circle with centre O so that
?POQ = 110 , then ?PTQ is equal to

(A) 60   (B) 70    (C) 80  (D) 90

It is given that TP and TQ are tangents.
Therefore, radius drawn to these tangents will be perpendicular to the tangents.
Thus, OP ? TP and OQ ? TQ
?OPT = 90º
?OQT = 90º
Sum of all interior angles = 360
?OPT + ?POQ + ?OQT + ?PTQ = 360
? 90 + 110º + 90 + PTQ = 360
? PTQ = 70
Hence, alternative (B) is correct

Mathematics
(Chapter – 10) (Circles)
(Class – X)

3
Question 3:
If tangents PA and PB from a point P to a circle with centre O are inclined to each other
an angle of 80 , then ?POA is equal to
(A) 50   (B) 60    (C) 70   (D) 80

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ? PA and OB ? PB
?OBP = 90º and ?OAP = 90º
In AOBP,
Sum of all interior angles = 360
?OAP + ?APB + ?PBO + ?BOA = 360
90º  + 80º +90º + ?BOA = 360
?BOA = 100
In ?OPB and ?OPA,
AP = BP (Tangents from a point)
OA = OB (Radii of the circle)
OP = OP (Common side)
Therefore, ?OPB ? ?OPA (SSS congruence criterion)

And thus, ?POB = ?POA

Hence, alternative (A) is correct.

Page 4

Mathematics
(Chapter – 10) (Circles)
(Class – X)

1
Exercise 10.2
Question 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 the circle is

(A) 7 cm   (B) 12 cm   (C) 15 cm   (D) 24.5 cm

Let O be the centre of the circle.
Given that,
OQ = 25cm and PQ = 24 cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, OP ? PQ
Applying Pythagoras theorem in ?OPQ, we obtain
OP
2
+ PQ
2
= OQ
2

OP
2
+ 24
2
= 25
2

OP
2
= 625 - 576
OP
2
= 49
OP = 7
Therefore, the radius of the circle is 7 cm.
Hence, alternative (A) is correct
Mathematics
(Chapter – 10) (Circles)
(Class – X)

2
Question 2:
In the given figure, if TP and TQ are the two tangents to a circle with centre O so that
?POQ = 110 , then ?PTQ is equal to

(A) 60   (B) 70    (C) 80  (D) 90

It is given that TP and TQ are tangents.
Therefore, radius drawn to these tangents will be perpendicular to the tangents.
Thus, OP ? TP and OQ ? TQ
?OPT = 90º
?OQT = 90º
Sum of all interior angles = 360
?OPT + ?POQ + ?OQT + ?PTQ = 360
? 90 + 110º + 90 + PTQ = 360
? PTQ = 70
Hence, alternative (B) is correct

Mathematics
(Chapter – 10) (Circles)
(Class – X)

3
Question 3:
If tangents PA and PB from a point P to a circle with centre O are inclined to each other
an angle of 80 , then ?POA is equal to
(A) 50   (B) 60    (C) 70   (D) 80

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ? PA and OB ? PB
?OBP = 90º and ?OAP = 90º
In AOBP,
Sum of all interior angles = 360
?OAP + ?APB + ?PBO + ?BOA = 360
90º  + 80º +90º + ?BOA = 360
?BOA = 100
In ?OPB and ?OPA,
AP = BP (Tangents from a point)
OA = OB (Radii of the circle)
OP = OP (Common side)
Therefore, ?OPB ? ?OPA (SSS congruence criterion)

And thus, ?POB = ?POA

Hence, alternative (A) is correct.

Mathematics
(Chapter – 10) (Circles)
(Class – X)
4
Question 4:
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Page 5

Mathematics
(Chapter – 10) (Circles)
(Class – X)

1
Exercise 10.2
Question 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 the circle is

(A) 7 cm   (B) 12 cm   (C) 15 cm   (D) 24.5 cm

Let O be the centre of the circle.
Given that,
OQ = 25cm and PQ = 24 cm
As the radius is perpendicular to the tangent at the point of contact,
Therefore, OP ? PQ
Applying Pythagoras theorem in ?OPQ, we obtain
OP
2
+ PQ
2
= OQ
2

OP
2
+ 24
2
= 25
2

OP
2
= 625 - 576
OP
2
= 49
OP = 7
Therefore, the radius of the circle is 7 cm.
Hence, alternative (A) is correct
Mathematics
(Chapter – 10) (Circles)
(Class – X)

2
Question 2:
In the given figure, if TP and TQ are the two tangents to a circle with centre O so that
?POQ = 110 , then ?PTQ is equal to

(A) 60   (B) 70    (C) 80  (D) 90

It is given that TP and TQ are tangents.
Therefore, radius drawn to these tangents will be perpendicular to the tangents.
Thus, OP ? TP and OQ ? TQ
?OPT = 90º
?OQT = 90º
Sum of all interior angles = 360
?OPT + ?POQ + ?OQT + ?PTQ = 360
? 90 + 110º + 90 + PTQ = 360
? PTQ = 70
Hence, alternative (B) is correct

Mathematics
(Chapter – 10) (Circles)
(Class – X)

3
Question 3:
If tangents PA and PB from a point P to a circle with centre O are inclined to each other
an angle of 80 , then ?POA is equal to
(A) 50   (B) 60    (C) 70   (D) 80

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ? PA and OB ? PB
?OBP = 90º and ?OAP = 90º
In AOBP,
Sum of all interior angles = 360
?OAP + ?APB + ?PBO + ?BOA = 360
90º  + 80º +90º + ?BOA = 360
?BOA = 100
In ?OPB and ?OPA,
AP = BP (Tangents from a point)
OA = OB (Radii of the circle)
OP = OP (Common side)
Therefore, ?OPB ? ?OPA (SSS congruence criterion)

And thus, ?POB = ?POA

Hence, alternative (A) is correct.

Mathematics
(Chapter – 10) (Circles)
(Class – X)
4
Question 4:
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Mathematics
(Chapter – 10) (Circles)
(Class – X)
5
Question 5:
Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the centre.
Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.
We have to prove that the line perpendicular to AB at P passes through centre O. We
shall prove this by contradiction method.
Let us assume that the perpendicular to AB at P does not pass through centre O. Let it
pass through another point O’. Join OP and O’P.
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