NCERT Solutions, Chapter 10: Circles, Class 10 (Mathematics) Class 10 Notes | EduRev

Class 10 : NCERT Solutions, Chapter 10: Circles, Class 10 (Mathematics) Class 10 Notes | EduRev

 Page 1


Mathematics 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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  
 
Answer 1:  
 
 
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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  
 
Answer 1:  
 
 
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
Answer 2:  
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  
In quadrilateral POQT,  
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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  
 
Answer 1:  
 
 
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
Answer 2:  
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  
In quadrilateral POQT,  
Sum of all interior angles = 360  
?OPT + ?POQ + ?OQT + ?PTQ = 360  
? 90 + 110º + 90 + PTQ = 360  
? PTQ = 70  
Hence, alternative (B) is correct  
 
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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  
 
Answer 3:  
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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  
 
Answer 1:  
 
 
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
Answer 2:  
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  
In quadrilateral POQT,  
Sum of all interior angles = 360  
?OPT + ?POQ + ?OQT + ?PTQ = 360  
? 90 + 110º + 90 + PTQ = 360  
? PTQ = 70  
Hence, alternative (B) is correct  
 
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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  
 
Answer 3:  
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
www.tiwariacademy.com  
4 
Question 4:  
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.  
Answer 4:  
Page 5


Mathematics 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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  
 
Answer 1:  
 
 
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
Answer 2:  
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  
In quadrilateral POQT,  
Sum of all interior angles = 360  
?OPT + ?POQ + ?OQT + ?PTQ = 360  
? 90 + 110º + 90 + PTQ = 360  
? PTQ = 70  
Hence, alternative (B) is correct  
 
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
 
www.tiwariacademy.com  
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  
 
Answer 3:  
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 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
www.tiwariacademy.com  
4 
Question 4:  
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.  
Answer 4:  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 10) (Circles)  
(Class – X) 
www.tiwariacademy.com  
5 
Question 5:  
Prove that the perpendicular at the point of contact to the tangent to a circle passes 
through the centre.   
Answer 5:  
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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