**Download NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2 – Circles. This Exercise contains 13 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.**

### NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2 – Circles

**NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2 – Circles**

**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) 5 cm**

Let **P** be the point of contact and **O** be the centre of the circle with radius **r **∴ PQ=24 cm

OQ=25 cm

∵ Tangent and radius are perpendicular to each other at point of contact

∴ In △OPQ, ∠P=90°

OP^{2} + PQ^{2} = OQ^{2} (By Pythagorean Theorem)

r^{2} + 24^{2} = 25^{2 }r^{2} = 25^{2} – 24^{2 }r^{2} = (25 – 24)×(25 + 24)

r^{2 }= 49

r = 7 cm

Therefore, the correct option is (A)

**2. In the given figure, 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°**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠TPO = ∠TQO = 90°

∵ Sum of angles of a quadrilateral = 360°

∴ In quadrilateral TPOQ

∠TPO + ∠POQ + ∠OQT + ∠QTP = 360°

90° + 110° + 90° + ∠QTP = 360°

290° + ∠QTP = 360°

∠QTP = 360° – 290°

∠QTP = 70°

Therefore, correct option is (B)

**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°**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠PAO = ∠OBP = 90°

∵ Sum of angles of a quadrilateral = 360°

∴ In quadrilateral PAOB

∠PAO + ∠AOB + ∠OBP + ∠BPA = 360°

90° + ∠AOB + 90° + 80° = 360°

260° + ∠AOB = 360°

∠AOB = 360° – 260°

∠AOB = 100°

In △OPA and △OPB

∠PAO = ∠OBP = 90°

PO = PO [Common]

AO = OB [Radius]

∴ △OPA ≅ △OPB (RHS Congruency Rule)

⇒ ∠AOP = ∠BOP = ∠AOB = x 100° = 50°

Therefore, correct option is (A)

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

**Given:** **PQ** and **RS** be the two tangents drawn at the end points of the diameter **AB** of circle with centre **O**

**To Prove: PQ **and **RS **are parallel to each other

**Proof:**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠PBO = ∠QBO = 90°

∠RAO = ∠SAO = 90°

⇒ ∠RAO = ∠QBO

∠PBO = ∠SAO

∵ Alternate angles are equal

∴ PQ and RS are parallel to each other

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

**Given:** A circle with **O** as the centre of the circle and **AB** is a tangent to the circle with point of contact at **P**

**To Prove: **Perpendicular to the tangent **AB **passes through the centre **O**

**Proof (by contradiction):**

Let the perpendicular to the tangent AB at P does not pass through O but passes through a point O’, not lying on the line joining O and P

∴ ∠O’PB = 90° **…(i)**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠OPB = 90° **…(ii)**

From equation (i) and (ii)

∠O’PB = 90° = ∠OPB

This is possible only if O’, O and P lie on the same line

It is contradiction to the earlier assumption

Therefore, O’ lies on the line joining O and P

Hence, the perpendicular to the tangent AB at P passes through O

Therefore, the perpendicular at the point of contact to the tangent to a circle passes through the centre

**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.**

Let **B** be the point of contact and **O** be the centre of the circle with radius **r **∴ BA = 4 cm

OA = 5 cm

∵ Tangent and radius are perpendicular to each other at point of contact

∴ In △OAB, ∠B=90°

OB^{2} + BA^{2} = OA^{2} (By Pythagorean Theorem)

r^{2} + 4^{2} = 5^{2 }r^{2} = 5^{2} – 4^{2 }r^{2} = 25 – 16

r^{2 }= 9

r = 3 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.**

Let **AC **be the chord of the bigger circle and is a tangent to the smaller circle with point of contact **B**, both circles have common centre **O **∴ OA = 5 cm

OB = 3 cm

∵ Tangent and radius are perpendicular to each other at point of contact

∴ In △OBA, ∠B=90°

OB^{2} + BA^{2} = OA^{2} (By Pythagorean Theorem)

3^{2} + BA^{2} = 5^{2 }BA^{2} = 5^{2} – 3^{2 }BA^{2} = 25 – 9

BA^{2 }= 16

BA = 4 cm

AC = 2×AB (Perpendicular to the chord from the centre bisects it)

AC = 2×4 = 8 cm

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

**8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC**

The points **P**, **Q**, **R** and **S** are the point of where the circle is touched by the sides **AB**, **BC**, **CD** and **DA**, respectively.

∵ Tangents from an external point to a circle are of equal length

∴ AP = AS …(i)

BP = BQ …(ii)

DR = DS …(iii)

CR = CQ …(iv)

Adding equations (i), (ii), (iii) and (iv)

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

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

⇒ AB + CD = AD + BC

**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°.**

Join the centre of the circle with the point of contact of the tangent **AB**

In △OPA and △OCA

∠OPA = ∠OCA = 90° (Tangent and radius are perpendicular to each other at point of contact)

AO = AO [Common]

PO = CO = Radius

∴ △OPA ≅ △OCA (RHS Congruency Rule)

⇒ ∠POA = ∠COA **…(i)**(By CPCT)

Similarly, △OQB ≅ △OCB

⇒ ∠QOB = ∠COB **…(ii)**(By CPCT)

∵ PQ is a straight line

∴ ∠QOB = 180°

∠POA + ∠COA + ∠QOB + ∠COB = 180°

Substituting values from equation (i) and (ii)

2∠COA + 2∠COB = 180°

⇒ ∠AOB = 90°

**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.**

**Given: ****A** and **B** are the points of contact of the tangents from the external point **P** to the circle with centre **O**

**To Prove: **∠APB + ∠AOB = 180°

**Proof:**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠PAO = ∠PBO = 90°

∵ Sum of angles of a quadrilateral = 360°

∴ In quadrilateral PAOB

∠PAO + ∠AOB + ∠OBP + ∠APB = 360°

90° + ∠AOB + 90° + ∠APB = 360°

180° + ∠AOB + ∠APB = 360°

∠AOB + ∠APB = 360° – 180°

∠APB + ∠AOB = 180°

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.

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

**ABCD** is a parallelogram and the points **P**, **Q**, **R** and **S** are the point of where the circle is touched by the sides **AB**, **BC**, **CD** and **DA**, respectively

∴ AB = CD …(i)

BC = AD …(ii)

∵ Tangents from an external point to a circle are of equal length

∴ AP = AS …(iii)

BP = BQ …(iv)

DR = DS …(v)

CR = CQ …(vi)

Adding equations (iii), (iv), (v) and (vi)

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

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

⇒ AB + CD = AD + BC

Substituting values from equation (i) and (ii)

AB + AB = BC + BC

2AB = 2BC

AB = BC …(vii)

From equation (i), (ii) and (vii)

AB = BC = CD = DA

⇒ All sides of the parallelogram circumscribing a circle are equal

Since, a parallelogram with sides of equal length is a rhombus

Therefore, a parallelogram circumscribing a circle is a rhombus.

**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. Find the sides AB and AC.**

Let the sides AB and AC touch the circle at E and F respectively

Let AF = x

∵ Tangents from an external point to a circle are of equal length

∴ AE = AF = x

BD = BE = 8 cm

CF = CD = 6 cm

AB = AE + EB = x + 8

BC = BD + CD = 8 + 6 = 14 cm

CA = CF + FA = 6 + x

Semi-perimeter of △ABC (s) =(AB + BC + CA) = (x + 8 + 14 + 6 + x) = x + 14

Area of △ABC = (Heron’s Formula)

Area of △ABC =

Area of △ABC =

Area of △ABC = …(i)

Area of △ABC = ar(△AOB) + ar(△BOC) + ar(△COA)

= (AB×OE) + (BC×OD) + CA×OF)

= (x + 8)×4 + (14×4) + (x + 6)×4 (OD = OE = OF = 4 cm)

= 2×(x + 8) + 2×14 + 2×(x + 6)

= 2×(x + 8 + 14 + x + 6)

= 2×(2x + 28)

= 4(x + 14) …(ii)

From equation (i) and (ii)

48x(x + 14) = 16(x + 14)^{2} **(Both sides squared)**

Dividing by common factors

3x = (x + 14)

3x – x = 14

2x = 14

x = 7 cm

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

BC = 14 cm

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

Therefore, AB = 15 cm and AC = 13 cm.

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

Consider the quadrilateral ABCD circumscribing a circle with centre O. AB, BC, CD and DA touch the circle at points P, Q, R and S respectively

Join the centre of the circle to the points of contact and the vertices of the quadrilateral

In △POA and △SOA

∠OPA = ∠OSA = 90° (Tangent and radius are perpendicular to each other at point of contact)

AO = AO [Common]

OP = OS [Radius]

∴ △POA ≅ △SOA (RHS Congruency Rule)

⇒ ∠POA = ∠SOA (By CPCT) **…(i)**

Similarly, △SOD ≅ △ROD

⇒ ∠SOD = ∠ROD **…(ii)**

△ROC ≅ △QOC

⇒ ∠ROC = ∠QOC **…(iii)**

△QOB ≅ △POB

⇒ ∠QOB = ∠POB **…(iv)**

∠POA + ∠SOA + ∠SOD + ∠ROD + ∠ROC + ∠QOC + ∠QOB + ∠POB = 360°

Substituting values from equation (i), (ii), (iii) and (iv)

2∠POA + 2∠POB + 2∠ROD + 2∠ROC = 360°

2[(∠POA + ∠POB) + 2(∠ROD + ∠ROC)] = 360°

2(∠AOB + ∠COD) = 360°

⇒ ∠AOB + ∠COD = 180°

Similarly, ∠AOD + ∠BOC = 180°

Hence, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 – Circles****, has been designed by the NCERT to test the knowledge of the student on the topic – Number of Tangents from a Point on a Circle**

** Maths – NCERT Solutions Class 10**

**Download NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2 – Circles**

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