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]

Now from Pythagoras Theorem,

OQ^{2} = PQ^{2} + OP^{2 }

=> 12^{2} = PQ^{2} + 5^{2}

=> 144 = PQ^{2} + 25

=> PQ^{2} = 144 - 25

=> PQ^{2} = 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.

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

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.

**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

Applying Pythagoras theorem in ∆OPQ, we obtain,

OP^{2} + PQ^{2} = OQ^{2 }

=> OP^{2} + 24^{2} = 25^{2}

=> OP^{2} + 576 = 625

=> OP^{2} = 625 − 576

=> OP^{2} = 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) 60^{0} (B) 70^{0} (C) 80^{0} (D) 90^{0}

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 = 90^{0}

∠OQT = 90^{0}

In quadrilateral POQT,

Sum of all interior angles = 360^{0}

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360^{0}

=> 90^{0} + 110^{0} + 90^{0} + ∠PTQ = 360^{0}

=> 290^{0} + ∠PTQ = 70^{0}

=> ∠PTQ = 360^{0} – 290^{0}

=> ∠PTQ = 70^{0}

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) 50^{0} (B) 60^{0} (C) 70^{0} (D) 80^{0}

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 = 90^{0} and ∠OAP = 90^{0}

In AOBP,

Sum of all interior angles = 360^{0 }

=> ∠OAP + ∠APB +∠PBO + ∠BOA = 360^{0}

=> 90^{0} + 80^{0} +90^{0} + ∠BOA = 360^{0}

=> 260^{0} + ∠BOA = 360^{0}

=> ∠BOA = 360^{0} - 260^{0}

=> ∠BOA = 100^{0}

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 = 100^{0}/2 = 50^{0}

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.

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.

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.

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

∠O’PB = 90^{0} …………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 = 90^{0} …………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:

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

AB^{2} + BO^{2} = OB^{2}

=> 4^{2} + BO^{2} = 5^{2}

=> 16 + BO^{2} = 25

=> BO^{2} = 25 - 16

=> BO^{2} = 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.

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

OA^{2} + AP^{2} = OP^{2}

=> 3^{2} + AP^{2} = 5^{2}

=> 9 + AP^{2} = 25

=> AP^{2} = 25 - 9

=> AP^{2} = 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

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 = 90^{0}

Answer:

Let us join point O to C.

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 = 180^{0}

From equation 1 and 2, we get

=> 2∠COA + 2∠COB = 180^{0}

=> ∠COA + ∠COB = 180^{0}/2

=> ∠COA + ∠COB = 90^{0}

=> ∠AOB = 90^{0}

**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.

It can be observed that

OA (radius) ⊥ PA (tangent)

Therefore, ∠OAP = 90^{0}

Similarly, OB (radius) ⊥ PB (tangent)

∠OBP = 90^{0}

In quadrilateral OAPB,

Sum of all interior angles = 360^{0}

∠OAP + ∠APB + ∠PBO + ∠BOA = 360^{0}

=> 90^{0} + ∠APB + 90^{0} + ∠BOA = 360^{0}

=> 180^{0} + ∠APB + ∠BOA = 360^{0}

=> ∠APB + ∠BOA = 360^{0 }- 180^{0}

=> ∠APB + ∠BOA = 180^{0}

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.

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.

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.

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 + x^{2})]

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 + x^{2})] = 28 + 12 + 2x + 16 + 2x

=> 4√[3(14x + x^{2})] = 56 + 4x

=> 4√[3(14x + x^{2})] = 4(14 + x)

=> √[3(14x + x^{2})] = 14 + x

Squaring on both sides, we get

=> 3(14x + x^{2}) = (14 + x)^{2}

=> 42x + 3x^{2} = 196 + x^{2} + 28x

=> 42x + 3x^{2} - 196 - x^{2} - 28x = 0

=> 2x^{2} + 14x – 196 = 0

=> x^{2} + 7x – 98 = 0

=> x^{2} + 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.

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 = 360^{0}

(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360^{0}

2∠1 + 2∠2 + 2∠5 + 2∠6 = 360^{0}

2(∠1 + ∠2) + 2(∠5 + ∠6) = 360^{0}

(∠1 + ∠2) + (∠5 + ∠6) = 180^{0}

∠AOB + ∠COD = 180^{0}

Similarly, we can prove that ∠BOC + ∠DOA = 180^{0}

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

at the centre of the circle.

.