NCERT

10th Mathematics

Exercise 10.2

Topic: Circle

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In Q.1 to 3, choose the correct option and give justification.

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

### Solution:

1) QT is the tangent to the circle with center O at point T and QT = 24 cm.

2) OQ = 25 cm.

3) Let the radius OT be x cm.

4) In **∆** OTQ, by the theorem of Pythagoras, we get,

(OT)^{2} + (QT)^{2} = (OQ)^{2}

(OT)^{2} = (OQ)^{2} – (QT)^{2 }(OT)^{2} = (25)^{2} – (24)^{2}(OT)^{2} = (25 – 24) (25 + 24)

(OT)^{2} = 1(49)

(OT)^{2} = 49

OT = √49

OT = 7

5) Therefore, **answer is (A), OT = 7 cm**.

2. In the following fig., 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°

### Solution:

1) TQ and TP are the tangents to the circle with center O at points Q and P.

2) ∠ POQ = 110°.

3) In **□ **OPTQ,

∠ OQT = 90°

∠ OPT = 90°∠ PTQ + ∠ POQ = 180°

∠ PTQ + 110° = 180°

∠ PTQ = 180° – 110°

∠ PTQ = 70°

4) Therefore, **answer is (A), ****∠ PTQ = ****70°**.

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

(A) 50° (B) 60° (C) 70° (D) 80°

### Solution:

1) PA and PB are the tangents to the circle with center O at points A and B.

2) ∠ APB = 80°.

3) In ∆ POA and ∆ POB,

∠ PAO = ∠ PBO = 90°.

OA = OB radii of the same circle.

OP = OP common side of ∆ POA and ∆ POB.

∆ POA ≅ ∆ POB by Hypotenuse-side theorem -------- 1

4) From 1, using CACT, we have,

∠ OPA = ∠ OPB ------------- 2

∠ OPA + ∠ OPB = 80° ------------- 3 Given

5) From 2 and 3, we have,

∠ OPA + ∠ OPB = 80°

∠ OPA + ∠ OPBA = 80°

2 ∠ OPA = 80°

∠ OPA = 40° ------------- 4

6) In ∆ POA and From 4, we have,

∠ POA + ∠ OPA = 90°

∠ POA + 40° = 90°

∠ POA = 90° – 40°

∠ POA = 50°

7) Therefore, **answer is (A),**** ∠ POA = 50°.**

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

**parallel.**

1) AB and CD are the tangents to the circle with center O at points P and Q.

2) OQ ⊥ tangent CD and OP ⊥ tangent AB.

3) PQ is the diameter of a circle with center O.

4) So,

∠ OQC = 90°

∠ OPB = 90°

5) Here,line PQ is the transversal on the line AB and line CD, so ∠ OQC

and ∠ OPB are alternate interior angles.

6) As, alternate interior angles ∠ OQC = ∠ OPB, **line AB and line CD are**

**parallel. Hence proved.**

**5. Prove that the perpendicular at the point of contact to the tangent to a**

**circle passes ****through the centre. **

1) A tangent AB touching the circle at point P.

2) We know that the tangent of a circle is ⊥ to radius at point of contact,

therefore,

OP ⊥ tangent AB.

∠ OPA = 90° ----------------- equation 1

3) Now let us cosider that, QP ⊥ AB, so we have,

∠ QPA = 90° ----------------- equation 2

4) From equations 1 and 2, we can say that.

∠ OPA = ∠ QPA = 90°, is possible only if line QP passes through O.

5) So, **the perpendicular at the point of contact to the tangent to a**

**circle passes through the centre is proved.**

**6. The length of a tangent from point A at a distance of 5 cm from the centre**

**of ****the circle is 4 ****cm. Find the radius of the circle.**

### Solution:

1) QT is the tangent to the circle with center O at point T and QT = 4 cm.

2) OQ = 5 cm.

3) Let the radius OT be x cm.

4) In **∆** OTQ, by the theorem of Pythagoras, we get,(OT)^{2} + (QT)^{2} = (OQ)^{2}

(OT)^{2} = (OQ)^{2} – (QT)^{2 }

(OT)^{2} = (5)^{2} – (4)^{2}

(OT)^{2} = (5 – 4) (5 + 4)

5) **Radius of a circle OT = 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:

1) The chord AB of larger circle touches the smaller circle at P.

2) OP = 3 cm and OB = 5 cm.

3) In **∆** OPB, by the theorem of Pythagoras, we get,(OP)^{2} + (PB)^{2} = (OB)^{2}

(PB)^{2} = (OB)^{2} – (PB)^{2 }

(PB)^{2} = (5)^{2} – (3)^{2}

(PB)^{2} = (5 – 3) (5 + 3)

4)** Length of the chord of the larger circle AB = 2 x PB = 2 x 4 = 8 cm.**

**8. A quadrilateral ABCD is drawn to circumscribe a circle (see following fig.).**

**Prove that ****AB + CD = AD + BC.**

1) AB, BC, CD, and DA are tangents to the circle touching at points P, Q, R,

and S respectively.

2) As, the lengths of tangents drawn from an external point to a circle are

equal, we have,

AP = AS ------------- equation 1

BP = BQ ------------- equation 2

CR = CQ ------------- equation 3

DR = DS ------------- equation 4

3) Adding equations 1, 2, 3, and 4, we get,

AP + BP + CR + DR = AS + BQ + CQ + DS ------------- equation 5

4) Rearranging the terms of equation 5, we get,

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

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

5) So, **AB + CD = AD + BC is proved**.

**9. In the following fig., 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:

1) XY and X’Y’ are two parallel tangents to a circle with centre O, touching at

points P and Q respectively.

2) AB is another tangent at C to the circle and intersects XY at A and X'Y' at B.

a) Tangents are ⊥ to the radii of a circle.

∠ OPA = ∠ OCA = 90° --------------- equation 1

b) As, the lengths of tangents drawn from an external point to a circle are

equal, we have,

AP = AC --------------- equation 2

c) Common side

AO = AO --------------- equation 3

4) From equation 1, 2, 3, and hypotenuse-side theorem we have,

∆ AOP ≅ ∆ AOC

5) So, using CPCT we have,

∠ POA = ∠ COA --------------- equation 4

∠ POC = ∠ POA + ∠ COA --------------- equation 5

6) From equation 4, and 5, we have

∠ POC = ∠ POA + ∠ COA

∠ POC = ∠ COA + ∠ COA

∠ POC = 2 ∠ COA --------------- equation 6

7) In the same way, we can prove that,

∠ QOC = 2 ∠ COB --------------- equation 7

8) Adding equations 6, and 7, we get,

2 ∠ COA + 2 ∠ COB = ∠ POC + ∠ QOC

2 ∠ COA + 2 ∠ COB = 180°

2 (∠ AOB) = 180°

**∠ AOB = 90****°, hence proved**.

**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:

1) TQ and TP are the tangents to the circle with center O at points Q and P.

2) ∠ POQ = 110°.

3) In **□** OPTQ, as tangents are ⊥ to the radii of a circle,

∠ OQT = 90° ------------- equation 1

∠ OPT = 90° ------------- equation 2

4) In **□** OPTQ and from equations 1, and 2,

∠ PTQ + ∠ POQ + ∠ OQT + ∠ OPT = 360°

∠ PTQ + ∠ POQ + 90° + 90° = 360°

∠ PTQ + ∠ POQ + 180° = 360°∠ PTQ + ∠ POQ = 180°

5) **∠ PTQ and ****∠ POQ are supplimentary.**

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

1) AB, BC, CD, and DA are tangents to the circle touching at points P, Q, R,

and S respectively.

2) As, the lengths of tangents drawn from an external point to a circle are

equal, we have,

AP = AS ------------- equation 1

BP = BQ ------------- equation 2

CR = CQ ------------- equation 3

DR = DS ------------- equation 4

3) Adding equations 1, 2, 3, and 4, we get,

AP + BP + CR + DR = AS + BQ + CQ + DS ------------- equation 5

4) Rearranging the terms of equation 5, we get,

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

(AB) + (CD) = (AD) + (BC) ------------- equation 6

5) As, ABCD is parallelogram, we have,

CD = AB ------------- equation 7

BC = AD ------------- equation 8

6) From equations 6, 7, and 8, we have

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

(AB) + (AB) = (AD) + (AD)

2(AB) = 2(AD)

AB = AD ------------- equation 9

7) From equations 7, 8, and 9, we can say that **all the sides of the**

**parallelogram are equal, so ABCD is the rhombus. Hence proved.**

**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 the following fig.). Find the sides AB ****and AC.**

1) AB, BC, and CA are tangents to the circle touching at points E, D, F.

2) As, the lengths of tangents drawn from an external point to a circle are

equal, we have,

CD = CF = 6

BD = BE = 8

AE = AF = x

3) So, we have,

a = BC = 8 + 6 = 14 ------------- equation 1

b = CA = (x + 6) ------------- equation 2

c = AB = (x + 8) ------------- equation 3

4) So, we have,

s = (a + b + c)/2

s = (14 + x + 6 + x + 8)/2

s = (2x + 14 + 6 + 8)/2

s = (x + 7 + 3 + 4)

s = (x + 14)

5) Using Heron's formula, we have,

A**△**(ABC) = √[s(s - a)(s - b)(s - c)]

A**△**(ABC) = √[(x + 14)(x + 14 - 14)(x + 14 - x - 6)(x + 14 - x - 8)]

A**△**(ABC) = √[(x + 14)(x)(8)(6)]

A**△**(ABC) = √[48x(x + 14)] ------------- equation 4

6) We know that,

A**△**(ABC) = A**△**(AOB) + A**△**(BOC) + A**△**(COA)

A**△**(ABC) = [(1/2)x(4)x(c)] + [(1/2)x(14)x(a)] + [(1/2)x(4)x(b)]

A**△**(ABC) = [(1/2)x(4)x(x + 8)] + [(1/2)x(4)x(14)] + [(1/2)x(4)x(x + 6)]

A**△**(ABC) = [(2)x(x + 8)] + [(2)x(14)] + [(2)x(x + 6)]

A**△**(ABC) = 2[(x + 8) + (14) + (x + 6)]

A**△**(ABC) = 2[2x + 8 + 14 + 6]

A**△**(ABC) = 4[x + 4 + 7 + 3]

A**△**(ABC) = 4[x + 14] ------------- equation 5

7) From equation 1, and 2, we have,

√[48x(x + 14)] = 4[x + 14]

[48x(x + 14)] = 16[x + 14]^{2}

[3x(x + 14)] = [x + 14]^{2}

x = 7 ------------- equation 6

8) Put x = 7 from equation 6 in euations 2 and 3, we have,

b = CA = 13 ------------- equation 7

c = AB = 15 ------------- equation 8

**CA = 13 cm and AB = 15 cm.**

**13. Prove that opposite sides of a quadrilateral ****circumscribing a circle**

**subtend supplementary ****angles at the centre of the circle. **

1) AB, BC, CD, and DA are tangents to the circle touching at points P, Q, R,

and S respectively.

2) ∆ AOP and ∆ AOS, we have,

AP = AS ------------- tangent segments are same,

OA = OA ------------- common side of two triangles,

OP = OS ------------- radii of the same circle.

3) Using sss theorem, we have,

∆ AOP ≅ ∆ AOS ------------- by CPCT are equal, we have,

∠ a = ∠ h ------------- equation 1

4) Simillarly, we get,

∠ g = ∠ f ------------- equation 2

∠ e = ∠ d ------------- equation 3

∠ c = ∠ b ------------- equation 4

5) Adding all central angles, we get

∠ a + ∠ g + ∠ e + ∠ c + ∠ h + ∠ f + ∠ d + ∠ b = 360° ------------- equation 5

6) Rearranging the terms of equation 5, we get,

(∠ a + ∠ h) + (∠ g + ∠ f) + (∠ e + ∠ d) + (∠ c + ∠ b) = 360° ----- equation 6

7) Put ∠ h = ∠ a, ∠ g = ∠ f, ∠ d = ∠ e, ∠ c = ∠ b from 1, 2, 3, and 4, in 6,

(∠ a + ∠ a) + (∠ f + ∠ f) + (∠ e + ∠ e) + (∠ b + ∠ b) = 360°

2 ∠ a + 2 ∠ f + 2 ∠ e + 2 ∠ b = 360°

2 (∠ a + ∠ f + ∠ e + ∠ b) = 360°

(∠ a + ∠ f + ∠ e + ∠ b) = 180°

(∠ a + ∠ b) + (∠ e + ∠ f) = 180°

(∠ AOP + ∠ BOP) + (∠ COR + ∠ DOR) = 180°

(∠ AOB) + (∠ COD) = 180° ------------- equation 7

8) Simillarly, we can prove that,

(∠ AOD) + (∠ BOC) = 180° ------------- equation 8

9) From equations 7, and 8, we can say that,

**opposite sides of a quadrilateral subtend supplementary angles at the centre of the circle. ****Hence proved**.