Chapter 10: Circles — Class 10 Mathematics NCERT Solutions | CBSE

 Chapter 10: Circles – NCERT Class 10 Mathematics Solutions | CBSE

 Chapter 10: Circles 

 EXERCISE 10.1

1. How many tangents can a circle have ?

Answer: A circle have infinitely many tangents .

2. Fill in the blanks :

(i) A tangent to a circle intersects it in  points .

(ii) A line intersecting a circle in two points is called a  .

(iii)  A circle can have   parallel tangents at the most .

(iv) The common point of a tangent to a circle and the circle is called  .

Answer: (i) A tangent to a circle intersects it in  points .

(ii) A line intersecting a circle in two points is called a  .

(iii)  A circle can have   parallel tangents at the most .

(iv) The common point of a tangent to a circle and the circle is called  .

3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that  . Length PQ is :

(A) 12 cm          (B) 13 cm    

(C) 8.5 cm         (D)  cm

Answer:  (D)   cm .

[  In figure :

          

       In  , we have

           cm .  ]

4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle .

Answer:  Step of construction:

             

(i) We draw a straight line EF (the given line).

(ii) We draw a circle below the line.

(iii) We draw a line CD parallel to EF so that it just touches the circle at one point. This is the tangent.

(iv) We draw another line AB parallel to EF , passing through the circle so that it cuts the circle at two points. This is the secant.

EXERCISE 10.2

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

 Answer: (A)  7 cm 

[  Here , OQ = 25 cm and  QT = 24 cm

            

In , we have

  cm  ]

2. In fig. 10.11,if TP and TQ are the two tangents to a circle with centre O so that  , then   is equal to :

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

Answer : (B)  70° 

[  Here,  and

In quadrilateral OPTQ , We have,

   ]

3. If the tangents PA and PB from a point P to a circle with centre O are inclined to each other  at an angle of 80°, then  is equal to

(A)  50°     (B) 60°       (C)  70°         (D) 80°

Answer : (A) 50°

[ In given figure: 

                 

 Since , ∆OPA≅∆OPB  and ∠OPA=∠OPB 

In  , we have   

  ]

4.  Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution:    let AB and CD are two tangent touch at X and Y of the diameter XY of the circle with centre O .

To prove :  .

          

Proof : Since, the tangent to a circle is perpendicular to the radius through the point of contact .

  and  

     

  

So,  and XY is a transversal .

     Proved .

5.  Prove that the perpendicular at the point of contact to the tangent  to a circle passes  through  the centre .

Solution:  Let, AXB and CYD are two parallel tangents at the point X and Y of the circle with centre O .

To Prove : XY is a line passes through the centre of the circle .

Construction : join OX and OY . we draw AB∥OZ  .

                          

Proof :   Since, the tangent to a circle is perpendicular to the radius through the point of contact .

 and 

    and OX is a transversal .

    

  

 

   and OY is a transversal .

  

So, XOY is straight line .

Therefore, XY  is a line passes through the centre of the circle . Proved

6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm .Find  the radius of the circle .

Solution:  In given figure:

                

Here, OP = 5 cm  and AP = 4 cm

 In ∆OAP , We have

    

 cm  

Therefore, the radius of the circle is 3 cm .

7.   Two concentric circles are of radii 5 cm and 3 cm .Find the length of the chord  of the larger circle which touches the smaller circle.

Solution:  In given figure: 

                 

Here, OM = 3 cm and OA = 5 cm .

In ∆OMA, we have   

   

Since,                                                                                                                                                                                             

   

Therefore, the length of the chord is 8 cm .

8.  A quadrilateral  ABCD is drawn to circumscribe  a circle a circle (see Fig. 10.12). Prove that

Solution:   Let,   be a quadrilateral and its sides are AB, BC , CD and DA in which the point P, Q , R and  S are touch of a circle with centre O respectively . 

       To Prove  :    .

  Proof :  Since the length of the two tangents from an external point to a circle are equal .

    So ,    (From A)…….(i)  

         (From B)…….(ii)

        (From C)…….(iii) 

          (From D )…….(iv) 

 

       Proved.

9.   In Fig. 10.13, XY and    are two parallel tangents to a circle with centre O and another tangent AB with  point of contact C intersecting XY at A and  X’Y’ at B .Prove that   .

            

Solution:  Given , XY and   are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting  at A and  at B .

To Prove  

 Construction :  Join   and .

         

Proof :  Since the lengths of the two tangents from an external point to a circle are equal and also the subtend equal angles at the centre .

    Therefore , OA is a bisector of  

 

     And is a bisect of  

 Since ,  and AB is a transversal 

   

  [From  and  ]  

        ∆AOB   we have,

     

   [ From  ]

   Proved.

10.  Prove that the angle between the two tangents drawn from an external point  to a circle is supplementary to the nagle subtended by the line –segment joining the points of contact at the centre.

Solution: Given, AP and BP be two tangents drawn from an external point P to a circle with centre O .

To prove : 

                

Proof : Since, the tangent to a circle is perpendicular to the radius through the point of contact . 

Therefore ,  and  .

So, and .

 OAPB is cyclic quadrilateral , we have  

 

 

   proved .

11.  Prove that the parallelogram circumscribing  a circle is a rhombus.

Solution:   Let,   be a quadrilateral and its sides are AB, BC , CD and DA in which the point P, Q , R and  S are touch of a circle with centre O respectively . 

       To Prove  :    .

                  

  Proof :  Since the length of the two tangents from an external point to a circle are equal .

    So ,    (From A)…….(i)  

         (From B)…….(ii)

        (From C)…….(iii) 

          (From D )…….(iv) 

 

      

    Since,   be a parallelogram . So,  and  

    So ,  

 

 

     

Therefore,   is a rhombus. 

12.  A triangle ABC is drawn circumscribe a circle of radius 4cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.  

        

Solution:  Since, the lengths of the two tangents from an external point to a circle are equal.

                

  cm  ;   cm    

Let,   cm 

  Therefore,    ,   ,  

     

      ;     ;   

      

   So,   

        

  

A/Q,    

 

 

   (impossible)  or   

  Thus,   cm  and  cm 

13.  Prove that the opposite sides of a quadrilateral circumscribing  a circle subtended supplementary  angles at the centre of the circle.

Solution: Given , ABCD is a quadrilateral circumscribing a circle with centre O and touches the quadrilateral at P , Q , R and S respectively .

 To Prove :   (i)       (ii)   

 Construction : Join OP , OQ , OR and OS respectively .  

          

Proof : Since the lengths of the two tangents from an external point to a circle are equal .

    i.e.,    ,   ,   and  .

    In  and  , we have

               (Given)

       

             [ Common side]

        [ R.H.S rule]

        [ C.P.C.T]   

    Similarly , ∠BOP=∠BOQ  ,   ,  

Let,   , ,   ,   ,   ,  ,   and  

We know that the sum of the all angles of subtended at a point is  360° .

        

   

 Similarly ,    Proved

---

Posted 5 years ago