Class 10 maths chapter 10 exercise 10.2

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2

NCERT Solutions for Maths Chapter 10, Exercise 10.2 involve complete  answers for each question in the exercise 10.2. The solutions provide students a strategic methods  to prepare for their exam. Class 10 Maths Chapter 10 Circles Exercise 10.2 questions and answers helps students to perform better in exam and it will  clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems.NCERT Solutions for Chapter 10 Circles Exercise 10.2 prepared by www.mathematicsandinformationtechnology.com team in very delicate, easy and creative way. 

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:

• Distance of Q from the centre, OQ = 25 cm
• Length of the tangent to a circle, PQ = 24 cm
• Radius, OP = ?

We know that, Radius is perpendicular to the tangent at the point of contact

Hence, OP ⊥ PQ

Therefore, OPQ forms a Right Angled Triangle

Applying Pythagoras theorem for ΔOPQ,

OP² + PQ² = OQ²

By substituting the values in the above Equation,

OP² + 24² = 25²

OP² = 625 – 576 (By Transposing)

OP² = 49

OP = 7 (By Taking Square Root)

Therefore, the radius of the circle is 7 cm.

Hence, alternative (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° (B) 70° (C) 80° (D) 90°

Given:

•Tangents: TP and TQ

We know that, Radius is perpendicular to the tangent at the point of contact

Thus, OP ⊥ TP and OQ ⊥ TQ

• Since the Tangents are Perpendicular to Radius

• ∠OPT = 90º
• ∠OQT = 90º

Now, POQT forms a Quadrilateral

We know that, Sum of all interior angles of a Quadrilateral = 360°

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360°

⇒ 90° + 110º + 90° + PTQ = 360° (By Substituting)

⇒ ∠PTQ = 70°

Hence, alternative (B) is correct.

Question 3. 

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

(A) 50° (B) 60° (C) 70° (D) 80°

Answer:

Given:

• Tangents are PA and PB

We know that, Radius is perpendicular to the tangent at the point of contact

Thus, OA ⊥ PA and OB ⊥ PB

• Since the Tangents are Perpendicular to Radius

• ∠OBP = 90º
• ∠OAP = 90º

Now, AOBP forms a Quadrilateral

We know that, Sum of all interior angles of a Quadrilateral = 360°

∠OAP + ∠APB +∠PBO + ∠BOA = 360°

90° + 80° +90º + ∠BOA = 360° (By Substituting)

∠BOA = ∠AOB = 100°

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)

A ↔ B, P ↔ P, O ↔ O

And thus, ∠POB = ∠POA

∠POA = ¹/₂ (∠AOB)=^100°/₂ = 50°

Hence, alternative (A) is correct.

Question 4:

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

Solution 4:

From the figure,

Given

• Let PQ be a diameter of the circle.

• Two tangents AB and CD are drawn at points P and Q respectively.

To Prove:

Tangents drawn at the ends of a diameter of a circle are parallel.

Proof:

We know that, Radius is perpendicular to the tangent at the point of contact

Thus, OP ⊥ AB and OQ ⊥ CD

Since the Tangents are Perpendicular to Radius

o ∠OQC = 90º

o ∠OQD = 90º

o ∠OPA = 90º

o ∠OPB = 90º

From Observation,

o ∠OPC = ∠OQB (Alternate interior angles)

o ∠OPD = ∠OQA (Alternate interior angles)

If the Alternate interior angles are equal then lines AB and CD should be parallel.

We know that AB & CD are the tangents to the circle.

Hence, it is proved that Tangents drawn at the ends of a diameter of a circle are parallel.

Question 5:

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

Solution 5:

From the figure,

Given:

Let ‘O’ be the centre of the circle

 Let AB be a tangent which touches the circle at P.

To Prove:

Line perpendicular to AB at P passes through centre O.

Proof:

Consider the figure below,

Let us assume that the perpendicular to AB at P does not pass through centre O.

Let It pass through another point Q. Join OP and QP.

We know that, Radius is perpendicular to the tangent at the point of contact

Hence, AB ⊥ PQ

∴ ∠QPB = 90° … (1)

We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

∴ ∠OPB = 90° … (2)

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

∴ ∠QPB = ∠OPB … (3)

From the figure, it can be observed that,

∴ ∠QPB < ∠OPB … (4)

Therefore, in reality ∠QPB ≠ ∠OPB

∠QPB = ∠OPB only if QP = OP which is possible in a scenario when the line QP coincides with OP.

Hence it is proved that the perpendicular to AB through P passes through centre 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.

Solution 6:

From the Figure:

Given:

o Let point ‘O’ be the centre of a circle

o AB is a tangent drawn on this circle from point A, AB = 4 cm

o Distance of A from the centre, OA = 5 cm

o Radius, OB = ?

In ΔABO,

We know that, OB ⊥ AB (Radius ⊥ tangent at the point of contact)

OAB forms a Right Angled Triangle.

Hence using, Pythagoras theorem in ΔABO,

AB² + OB² = OA²

42 + OB² = 5²  (By Substituting)

16 + OB²  = 25

OB²  = 9

Radius, OB = 3 (By Taking Square Roots)

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.

Solution 7:

From the Figure,

Given,

o Let ‘O’ be the centre of the two concentric circles

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

o PQ = ?

By Observation,

Line PQ is tangent to the smaller circle.

Hence, OA ⊥ PQ (Radius ⊥ tangent at the point of contact)

ΔOAP forms a Right Angled Triangle

By applying Pythagoras theorem in ΔOAP,

OA² + AP² = OP²

32 + AP²  = 52 (By Substituting)

9 + AP²  = 25

AP²  = 16

AP = 4 (By Taking Square Roots)

In ΔOPQ,

Since OA ⊥ PQ,

AP = AQ (Perpendicular from the center of the circle bisects the chord)

∴ PQ = 2 times AP = 2 × 4 = 8 (Substituting AP = 4 cm)

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

Question 8:

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

Solution 8:

From the Figure,

Given,

o DC , DA, BC, AB are sides of the Quadrilaterals which also form the tangents to the circle inscribed within Quadrilateral ABCD

To Prove:

AB + CD = AD + BC

Proof:

We know that length of tangents drawn from an external point of the circle are equal.

DR = DS ………..… (1)

CR = CQ ………..… (2)

BP = BQ ………….. (3)

AP = AS ………..… (4)

Adding (1), (2), (3), (4), we obtain

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

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) (By regrouping) ------------- (5)

From the figure,

o DR +CR = DC

o BP + AP = AB

o DS + AS = AD

o CQ +BQ = BC

Hence substituting the above values in Equation (5),

CD + AB = AD + BC

Hence it is proved.

Question 9:

In the given figure, 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°.

Solution 9:

From the Figure,

Given,

• Let ‘O’ be the centre of the circle

• XY and X’Y’ are two parallel tangents to circle

• AB is another tangent such that with point of contact C intersecting XY at A and X’Y’ at B.

To Prove:

∠AOB=90°.

Proof:

Join point O to C.

From the Figure above,

Consider ΔOPA and ΔOCA,

Here,

o OP = OC (Radii of the same circle)

o AP = AC (Tangents from external point A)

o AO = AO (Common side)

Therefore, ΔOPA ≅ ΔOCA (SSS congruence criterion)

Hence, P ↔ C, A ↔ A, O ↔ O

We can also say that,

∠POA = ∠COA … (i)

Similarly, ΔOQB ≅ ΔOCB

∠QOB = ∠COB … (ii)

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

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º--------------- (3)

Substituting Equation (i) and (ii) in the Equation (3),

2∠COA + 2 ∠COB = 180º

2(∠COA + ∠COB) = 180º

∠COA + ∠COB = 90º (By Transposing)

∠AOB = 90°

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.

Solution 10:

From the figure,

Given,

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

AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

To Prove:

∠APB is supplementary to ∠AOB.

Proof:

Join OP

Consider the ΔOAP & ΔOBP,

PA = PB( Tangents drawn from an external point are equal)

OA = OB ( Radii of the same circle)

OP = OP(Common Side)

Therefore, ΔOAP ≅ ΔOBP (SSS congruence criterion)

Hence,

∠OPA = ∠OPB

∠AOP = ∠BOP

Also,

∠APB = 2 ∠OPA -------------(1)

∠AOB= 2 ∠AOP--------------(2)

In the Right angled Triangle ΔOAP,

∠AOP +∠OPA = 90º

∠AOP = 90º - ∠OPA ----------------- (3)

Multiplying the Equation (3) by 2,

2∠AOP = 180º - 2∠OPA

By Substituting (1) and (2) in the equation above,

∠AOB = 180º - ∠APB

∠AOP + ∠OPA = 180º

Hence it is proved 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.

Solution 11:

From the figure,

Given,

ABCD is a parallelogram,

Hence

AB = CD …(1)

BC = AD …(2)

To Prove:

Parallelogram circumscribing a circle is a rhombus.

Proof:

From the figure,

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 the above equations, we obtain

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

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) (By Rearranging) ------ (3)

From the figure,

DR + CR = CD,

(BP + AP) = AB

(DS + AS) = AD

(CQ + BQ) = BC

Substituting the above values in (3),

CD + AB = AD + BC ------------- (4)

On putting the values of equations (1) and (2) in the equation (4), we obtain

2AB = 2BC

AB = BC ……………………… (5)

Comparing equations (1), (2), and (5), we get

AB = BC = CD = DA satisfies the property of Rhombus.

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 given figure). Find the sides AB and AC.

Solution 12:

From the figure,

Given,

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

• Length of the line segment AF be x.

• In ΔABC,

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

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

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

o AB = ?

o AC = ?

In ΔABE,

AB = AE + BE = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + AF = 6 + x

We know that, 2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

s = 14 + x

We also known that,

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

Either x+14 = 0 or x − 7 =0

Therefore, x = −14 and 7

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

Therefore, x = 7

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

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