Class 10 - Maths - Circles

Exercise 10.1

Question 1:

How many tangents can a circle have?

Answer:

A circle can have infinite number of tangents because a circle can have infinite number of points on it and at every point,

a tangent can be drawn.

Question 2:

Fill in the blanks:

(i) A tangent to a circle intersects it in _______ point (s).

(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 one point (s).

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

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

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

Question 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 OQ = 12 cm.

Length PQ is :

(A) 12 cm                            (B) 13 cm                        (C) 8.5 cm                       (D) √119 cm.

Answer:

In ΔOPQ, P is right angle.                          [Since radius is Ʇ to tangent]

Class_10_Maths_Circles_Tangets_Drawn_To_Circle8

Now from Pythagoras Theorem,

      OQ2 = PQ2 + OP2  

=> 122 = PQ2 + 52

=> 144 = PQ2 + 25

=> PQ2 = 144 - 25

=> PQ2 = 119

=> PQ = √119

Hence, the option D is the right answer.

Question 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:

Consider a circle with centre O. Let AB is the given line.

     Class_10_Maths_Circles_Line                                        

             Class_10_Maths_Circles_Circle                                                      

Now, draw a perpendicular from O to line AB, which intersect AB at P.

                                     Class_10_Maths_Circles_Drawing_Perpendicular_To_The_Circle                

Now take two points on the line PO, one at circle X and another Y inside the circle. Draw lines

parallel to AB and passing through X and Y. So, CD and EF are the required lines.

Class_10_Maths_Circles_Drawing_Perpendicular_To_The_Circle1

 

                                                                  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:

Let O be the centre of the circle.

Given that, OQ = 25 cm and PQ = 24 cm

As the radius is perpendicular to the tangent at the point of contact, Therefore, OP ⊥ PQ

Class_10_Maths_Circles_Tangets_Drawn_To_Circle7

Applying Pythagoras theorem in ∆OPQ, we obtain,

      OP2 + PQ2 = OQ2

=> OP2 + 242 = 252

=> OP2 + 576 = 625

=> OP2 = 625 − 576

=> OP2 = 49

=> OP = √49

=> OP = 7

Therefore, the radius of the circle is 7 cm.

Hence, option (A) is correct.

 

Question 2:

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°,

then ∠PTQ is equal to 

 (A) 600                       (B) 700                     (C) 800                        (D) 900

Class_10_Maths_Circles_Tangets_Drawn_To_Circle6

Answer:

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 = 900

∠OQT = 900

In quadrilateral POQT,

Sum of all interior angles = 3600

∠OPT + ∠POQ +∠OQT + ∠PTQ = 3600

=> 900 + 1100 + 900 + ∠PTQ = 3600

=> 2900 + ∠PTQ = 700

=> ∠PTQ = 3600 – 2900

=> ∠PTQ = 700

Hence, option (B) is correct.

 

Question 3:

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠ POQ = 110°, then ∠ PTQ is equal to

(A) 500                        (B) 600                              (C) 700                           (D) 800

Answer 3:

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Class_10_Maths_Circles_Tangets_Drawn_To_Circle5

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 900 and ∠OAP = 900

In AOBP,

Sum of all interior angles = 3600

=> ∠OAP + ∠APB +∠PBO + ∠BOA = 3600

=> 900 + 800 +900 + ∠BOA = 3600

=> 2600 + ∠BOA = 3600

=> ∠BOA = 3600 - 2600

=> ∠BOA = 1000

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

∠POA = ∠AOB/2 = 1000/2 = 500

Hence, option (A) is correct.

Question 4:

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

Answer:

Let AB be a diameter of the circle. Two tangents PQ and RS

are drawn at points A and B respectively.

Class_10_Maths_Circles_Tangets_Drawn_To_Circle4

Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA Ʇ RS and OB Ʇ PQ.

Now, ∠OAR = 90, ∠OAS = 90, ∠OBP = 90, ∠OBQ = 90

It can be observed that

∠OAR = ∠OBQ                (Alternate interior angle)

∠OAS = ∠OBP                 (Alternate interior angle) 

Since alternate interior angles are equal, lines PQ and RS will be parallel.

Question 5:

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

Answer:

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

Class_10_Maths_Circles_Tangets_Drawn_To_Circle3

                                                       

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 passes

through another point O’. Join OP and O’P.

             Class_10_Maths_Circles_Tangets_Drawn_To_Circle2                                                  

As perpendicular to AB at P passes through O’, therefore

∠O’PB = 900  …………1

O is the center of circle and P is the point of contact. We know that the line joining the center and the point of contact to the

tangent of the circle are perpendicular to each other.

So, ∠OPB = 900  …………2

Comparing equation 1 and 2, we get

∠O’PB = ∠OPB   ……….3

From the figure, it can be observed that

∠O’PB < ∠OPB  ………..4

Therefore, ∠O’PB = ∠OPB is not possible.

It is only possible when the line O’P coincides with OP.

Therefore, the perpendicular to AB through P passes through center O.

Question 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.

Answer:

Class_10_Maths_Circles_Tangets_Drawn_To_Circle1

Let us consider a circle centered at point O.

AB is a tangent drawn on this circle from point A.

Given that OA = 5 cm and AB = 4 cm

In ΔABO,

OB Ʇ AB     [radius is Ʇ to tangent at the point of contact]

Apply Pythagoras theorem in ΔABO, we get

      AB2 + BO2 = OB2

=> 42 + BO2 = 52

=> 16 + BO2 = 25

=> BO2 = 25 - 16

=> BO2 = 9

=> BO = √9

=> BO = 3

Hence, the radius of the circle is 3 cm.

Question 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.

Answer:

Let two concentric circles be centered at point O.

Let PQ be the chord of the larger circle which touches the smaller circle at point A.

Class_10_Maths_Circles_2_Concentric_Circle

Therefore, PQ is tangent to the smaller circle.

Now, OA Ʇ PQ     [Since OA is the radius of the circle]

Apply Pythagoras theorem in ΔOAP, we get

      OA2 + AP2 = OP2

=> 32 + AP2 = 52

=> 9 + AP2 = 25

=> AP2 = 25 - 9

=> AP2 = 16

=> AP = √16

=> AP = 4

In ΔOAP,

Since OA Ʇ PQ,

=> AP = AQ         [Perpendicular from the center of the circle bisects the chord]

So, PQ = 2 * AP = 2 * 4 = 8

Hence, the length of the chord of the larger circle is 8 cm.

 

 

Question 8:

A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that: AB + CD = AD + BC

Class_10_Maths_Circles_Quadrilateral_Circumscribing_In_A_Circle1

Answer:

It can be observed that

DR = DS (Tangents on the circle from point D) .............. (1)

CR = CQ (Tangents on the circle from point C) ............... (2)

BP = BQ (Tangents on the circle from point B) ............... (3)

AP = AS (Tangents on the circle from point A) ............... (4)

Adding all these equations, we get

       DR + CR + BP + AP = DS + CQ + BQ + AS

=> (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

=> CD + AB = AD + BC

=> AB + CD = AD + BC

 

 

Question 9:

In Fig. 10.13, XY and X′Y′ 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 ∠AOB = 900

Class_10_Maths_Circles_2_Parallel_Tangets_Drawn_To_Circle1

Answer:

Let us join point O to C.

Class_10_Maths_Circles_2_Parallel_Tangets_Drawn_To_Circle

In ΔOPA and ΔOCA,

OP = OC           [radii of the same circle]

AP = AC            [tangents from point A]  

AO = AO           [Common side]

ΔOPA ≅ ΔOCA       [SSS Congruence criterion]

∠POA = ∠COA  ………..1

Similarly,

ΔOQB ≅ ΔOCB      

∠QOB = ∠COB  ………..2

Since POQ is a diameter of the circle, it is a straight line.

So, ∠POA + ∠COA + ∠QOB + ∠COB = 1800

From equation 1 and 2, we get

=> 2∠COA + 2∠COB = 1800

=> ∠COA + ∠COB = 1800/2

=> ∠COA + ∠COB = 900

=> ∠AOB = 900

Question 10:

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the

angle subtended by the line-segment joining the points of contact at the centre.

Answer:

Let us consider a circle centered at point O. Let P be an external point from which two

tangents PA and PB are drawn to the circle which are touching the circle at point A and B

respectively and AB is the line segment, joining point of contacts A and B together such that it

subtends ∠AOB at center O of the circle.

Class_10_Maths_Circles_Tangets_Drawn_To_Circle

It can be observed that

OA (radius) ⊥ PA (tangent)

Therefore, ∠OAP = 900

Similarly, OB (radius) ⊥ PB (tangent)

∠OBP = 900

In quadrilateral OAPB,

Sum of all interior angles = 3600

∠OAP + ∠APB + ∠PBO + ∠BOA = 3600

=> 900 + ∠APB + 900 + ∠BOA = 3600

=> 1800 + ∠APB + ∠BOA = 3600

=> ∠APB + ∠BOA = 3600 - 1800

=> ∠APB + ∠BOA = 1800

Hence, it can be observed that the angle between the two tangents drawn from an external

point to a circle is supplementary to the angle subtended by the line segment joining the

points of contact at the centre.

Question 11:

Prove that the parallelogram circumscribing a circle is a rhombus.

Class_10_Maths_Circles_Parallelogram_Circumscribing_In_A_Circle

Answer:

Since ABCD is a parallelogram,

AB = CD ..........................(1)

BC = AD ..........................(2)

It can be observed that

DR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BC

AB = BC ..............................(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

 

 

Question 12:

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and

DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm

respectively (see Fig. 10.14). Find the sides AB and AC.

Class_10_Maths_Circles_Triangle_Circumscribing_In_A_Circle1

Answer:

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively

and the length of the line segment AF be x.

Class_10_Maths_Circles_Triangle_Circumscribing_In_A_Circle

In ΔABC,

CF = CD = 6cm      (Tangents on the circle from point C)

BE = BD = 8cm     (Tangents on the circle from point B)

AE = AF = x           (Tangents on the circle from point A)

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

Now, 2s = AB + BC + CA

               = x + 8 + 14 + 6 + x

               = 28 + 2x

=> s = 14 + x

Area of ΔABC = √{s(s - a)(s - b)(s - c)}

                         = √[(14 + x){14 + x - 14}{(14 + x) – (6 + x)} {(14 + x) – (8 + x)}]

                         = √[(14 + x) * x * 8 * 6]

                         = 4√[3(14x + x2)]

Area of ΔOBC = 1/2 * OD * BC

                         = 1/2 * 4 * 14

                         = 28

Area of ΔOCA = 1/2 * OF * AC

                         = 1/2 * 4 * (6 + x)

                         = 2 * (6 + x)

                         = 12 + 2x   

Area of ΔOAB = 1/2 * OE * AB

                         = 1/2 * 4 * (8 + x)

                         = 2 * (8 + x)

                         = 16 + 2x   

Now, Area of ∆ABC = Area of ∆OBC + Area of ∆OCA + Area of ∆OAB

=> 4√[3(14x + x2)] = 28 + 12 + 2x + 16 + 2x

=> 4√[3(14x + x2)] = 56 + 4x

=> 4√[3(14x + x2)] = 4(14 + x)

=> √[3(14x + x2)] = 14 + x

Squaring on both sides, we get

=> 3(14x + x2) = (14 + x)2

=> 42x + 3x2 = 196 + x2 + 28x

=> 42x + 3x2 - 196 - x2 - 28x = 0

=> 2x2 + 14x – 196 = 0

=> x2 + 7x – 98 = 0

=> x2 + 14x - 7x - 98 = 0

=> x(x + 14) – 7(x + 14) = 0

=> (x - 7)(x + 14) = 0

=> x = 7, -14

Here, x = -14 is not possible as the length of the sides will be negative.

So, x = 7

Hence, AB = x + 8 = 7 + 8 = 15 cm

             CA = 6 + x = 6 + 7 = 13 cm

Question 13:

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

Answer:

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle

at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the center of the circle.

Class_10_Maths_Circles_Quadrilateral_Circumscribing_In_A_Circle

From ∆OAP and ∆OAS,

AP = AS         (Tangents from the same point)

OP = OS        (Radii of the same circle)

OA = OA       (Common side)

∆OAP ≅ ∆OAS        (SSS congruence criterion)

thus, ∠POA = ∠AOS

∠1 = ∠8

Similarly,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 3600

(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 3600

2∠1 + 2∠2 + 2∠5 + 2∠6 = 3600

2(∠1 + ∠2) + 2(∠5 + ∠6) = 3600

(∠1 + ∠2) + (∠5 + ∠6) = 1800

∠AOB + ∠COD = 1800

Similarly, we can prove that ∠BOC + ∠DOA = 1800

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles

at the centre of the circle.

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