** Class 10 Maths NCERT Solutions of Chapter 10 Circle**

This blog brings the** Class 10 Maths NCERT Solutions of Chapter 10 Circle** to get ready for **board exams 2020-21**. **Solutions** to every one of the unsolved questions of **Class 10 NCERT Maths Chapter 10** are given here with suitable **answers** that offers you an idea to display how to solve **maths **questions in the most ideal manner in the **exam**.These **NCERT** **Solutions** of **Maths Class 10 Chapter 10 Circle** are created by an expert of **Maths** who has 20 years of experience in teaching **Maths** subject to 9 **,10**,11 and 12 Classes.

**Maths NCERT solutions for class 10 **are most valuable study inputs for achieving excellent marks in CBSE board exams, therefore all students are required to repeat NCERT solutions of maths cyclically . For achieving excellency in maths of class 10 60-70 % contribution goes to NCERT solutions and rest 30-40 % credit goes to study of last year’s maths unsolved or solved question papers. In studying maths as much as you practice you learn more and more,

**Introduction:** **The circle** is a collection of points that are at a constant distance from a fixed point known as the center of the **circle**. The terms related to the **circle** are the following.

**Secant:** The line which intersects the **circle** known as the secant of the **circle**.When such an intersecting line intersects the** circle** at one point or contacts the **circle** is known as the tangent of the **circle.** Tangent is called the special case of secant.

**NCERT Solutions Class 10 Science from chapter 1 to 16**

**Tangent:** Tangent to a **circle** is a line that contacts the **circle** at only one point,there is only one tangent at a point of the **circle.**

#### The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Let P is the contact point to the **circle** of the tangent RS and Q is a point on RS

Let if Q is inside the **circle** then RS would be secant of the **circle**

Therefore OQ must be longer than the radius of the **circle**

OQ> OP

It is true for every point which lies on RS except the point P

Since OP is the shortest distance from O to RS, therefore OP is perpendicular on RS

OP ⊥RS

#### Perpendicular drawn from the center to chord of a circle bisects the chords.

To prove this theorem drawing OP⊥ AB and joining both ends of the chord to the center

Considering both ∆AOP and ∆BOP

AO= OB (radii of the same circle)

OP = OP(common)

∠APO = ∠BPO (Each of 90º)

∆APO = ∆BPO( RHS rule)

AP = BP(By CPCT)

**Class 10 Maths NCERT Solutions of Chapter 10 Circle**

### Exercise 10.1

**Q1. How many tangents can a circle have?**

Ans.A circle has infinite points and through each point a tangents can be drawn so a circle can have infinite tangents.

**Q2.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…….**

Ans. (i) A tangent to a circle intersects it in one point.

(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 contact point.

**Q3.A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center 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**

Ans.

As we know OP⊥ PQ, because P is the contact point of the circle , PQ is the tangent on the circle and OP is the radius of circle.

∆OPQ is the right triangle

So, OQ² = OP² + PQ²

PQ ² = OQ² – OP²

PQ ² = 12² -5²

PQ² = 144 – 25 =119

**Class 10 Maths NCERT Solutions of Chapter 10 Circle**

**Exercise 10.2**

**In question number 1 to 3, choose the correct option and give justification.**

**Q1.From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is.**

**(A) 7 cm (B)12 cm**

**(C)15 cm (D)24.5**

Ans.

Let O is the centre of circle ,we are given that OQ = 25 cm, PQ = 24 cm

According to theorem OP ⊥ PQ, where OP is the radius and PQ is the tangent

Therefore ΔOPQ is the right triangle

Applying Pythagoras theorem

OQ² = PQ² + OP²

OP² = OQ² – PQ²

OP² = 25² – 24²

OP² = 625 – 576

OP² = 49

OP = 7

Therefore the length of the radius OP is 7 cm, so (A) is the right option.

**Q2.In the given fig. if TP and TQ are the two tangents to a circle with centre O so that ∠POQ=110º then ∠PTQ is equal to**

**(A) 60º (B)70º**

**(C)80 º (D)90º**

**Ans.**

As we are given TP and TQ are tangents of the circle ,therefore P and Q will be contact points of tangents.

O is the centre of circle , therefore OP and OQ will be the radii of the given circle

∵OP⊥ TP and OQ ⊥ TQ ( The radius is perpendicular to the tangent)

∴∠OQT = 90º and ∠ OPT = 90º

∠POQ + ∠PTQ + ∠OQT + ∠ OPT = 360º (Angle sum property of quadrilateral)

110º + ∠PTQ + 90º + 90º = 360º

∠PTQ = 360º – 290º

∠PTQ = 70º

Therefore right option is (B).

**Q3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80º then ∠POA is equal to**

**(A) 50º (B) 60º (C) 70º (D) 80º**

**Ans.**

∠PAO = ∠PBO = 90º ( The radius is perpendicular to the tangent)

∠PAO + ∠PBO + ∠AOB + ∠APB = 360º (Angle sum property of quadrilateral)

90º + 90º + 80º +∠AOB = 360º

∠AOB = 360º – 260º

∠AOB = 100º

Considering ΔPOA and ΔPOB

OP = OP (Common)

OA = OB (Radii of same circle)

∠PAO = ∠PBO = 90º ( The radius is perpendicular to the tangent)

ΔPOA ≅ ΔPOB (RHS rule )

∠POA = ∠POB (By CPCT)

∠POA +∠POB = 100º

2∠POA = 100º

∠POA = 50º

**Latest Laptops and Desktops on very small EMI:no extra cost**

**Q4. Prove that the tangents drawn at the ends of diameter of a circle are parallel to each other.**

Ans. Let PQ and RS are the tangents drawn on both the ends of the diameter MN

MO and NO are the radii of the circle

PM ⊥ MO and RN ⊥ NO ( Radius is perpendicular to the tangent)

∴∠PMO = 90º and ∠ RNO = 90º

∠PMO + ∠ RNO =90º + 90º = 180º

Since the angles, ∠PMO and ∠ RNO are co-interior angles, MN transversal

Here the sum of co-interior angles is 180º, therefore RS ∥ PQ

**Q5. Prove that perpendicular at the point of contact to the tangent to a circle passes through the centre.**

Ans.

Let PQ is the tangent to a circle with centre O where T is the contact point of the tangent

Let TR ⊥ PQ , doesn’t pass through the centre O

∠OTQ = 90º (The angle between tangent and radius)

∠RTQ = 90º(We have supposed)

∠OTQ = ∠RTQ

R and O coincide to each other, therefore TR must have to pass through the centre of the circle.

**Q6. The length of tangent AB from a point A at a distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.**

Ans.

Let B is the contact point and AB is the tangent

∠ABO = 90º (Radius OB⊥ AB)

∴ABO is right triangle

Therefore the radius of the circle will be = 3 cm

**Q7.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.**

Ans.

Let AB is the chord of the larger circle which touches smaller circle at C

OC ⊥ AB (The radius is perpendicular to the tangent)

ΔOCB will be a right triangle

BC² = OB² – OC²

BC² = 5² – 3²= 25 – 9 = 16

BC = 4

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

AB = 2 × 4 = 8

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

Click the link below and make your preparation easy

**CBSE Class lX , X Class, NCERT Solutions and Imporant questions of Science and Maths**

**Q8. A quadrilateral is drawn to circumscribe a circle (see fig.) . Prove that AB + CD = AD + BC**

Ans.

Let the circle touches the quadrilateral at P,Q,R and S points

Applying the theorem that tangent from an external point to the circle are equal

From the fig.we have

Hint: Keep the line segments of opposite sides at one side

AP = AS……….(i)

BP = BQ………..(ii)

DR = DQ………(iii)

CR = CS……….(iv)

Adding all of the equations

AP + BP + DR + CR = AS + BQ + DQ + CS

Arranging these line segments to get the sides

(AP + BP )+ (DR + CR) = (AS + CS) + (BQ + DQ )

AB + CD = AD + BC

**Q9. In fig. XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent with point of contact C intersecting XY at A and X’Y’ at B . Prove that ∠AOB = 90º.**

Ans. In the given fig. joining the points A to O, B to O and O to C

Let ∠PAC = 2x

∠OAC = x ( The line joining to centre of the circle bisects the angle between the tangents)

∠PAC + ∠QBC = 180 (Sum of co-interior angles)

∠QBC = 180 – 2x

∠OBC = 90 – x

Considering ΔAOB

∠AOB + ∠OBC + ∠OAC = 180 ( Angle sum property of triangle)

∠AOB + 90 – x + x = 180

∠AOB = 90

Hence ∠AOB =90º

**Q10. 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 point of contact at the centre.**

Ans.

GIVEN. Two tangents AB and BC inclined with angle ∠ABC and their contact points on the circle are A and C, AC subtends ∠AOC to the centre.

TO PROVE. ∠AOC + ∠ABC = 180º

CONSTRUCTION. Join A to C

PROOF. ∠OAB = 90º and ∠OCB = 90º ( The radius is perpendicular to tangent)

In quadrilateral OABC

∠OAB + ∠OCB + ∠AOC + ∠ABC = 360º ( Angle sum property of quadrilateral)

90 + 90 + ∠AOC + ∠ABC = 360º

∠AOC + ∠ABC = 360º – 180 = 180

Therefore the angle between two tangents is supplementary to the angle subtended by the line segment joining the point of contact at the centre.

**Q11. Prove that the parallelogram circumscribing a circle is a rhombus.**

Answer.

Given → Parallelogram ABCD circumscribing a circle with center o which is touching it at P, Q, R and S points.

To Prove → The given parallelogram ABCD is a rhombus

Proof → AB = DC……….(i)(Opposite side of the parallelogram)

AD = BC……….(ii)(Opposite side of the parallelogram)

AP = AS ………(iii) [Tangents drown on a circle from a same external point are equal]

BP = BQ ………(iv)[Tangents drown on a circle from a same external point are equal]

DR = DS ………(v)[Tangents drown on a circle from a same external point are equal]

CR = CQ ……….(vi)[Tangents drown on a circle from a same external point are equal]

**HINT: Write the tangents corresponding to opposite sides same side as written above.**

Adding (iii) to (vi) all equation we get

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

Arranging these line segments in order to get the side of paralellogram

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

AB + DC = AD + BC

From equation number (i) and (ii)

DC + DC = AD + AD

2DC = 2AD

DC = AD

It is clear that AB = BC = DC= AD

All sides of the given parallelogram are equal therefore it is a rhombus.

**Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segment 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.). Find the sides AB and AC.**

Ans.

Let AF = AE = x (Tangents drawn from an external point to the circle)

Drawing OF ⊥ AC, OE ⊥ AB and OD ⊥ BC

Area of ΔABC = Sum of area of all Δ’s inside it

arΔABC = 4x +24 +32 = 4x + 56…………(i)

AC = (6 + x) cm, BC = 8+6= 14 cm, AB = (8+x) cm

From (i)

Squiring both sides

48x(x +14) = (4x + 56)²

48 x² + 672 x = 16 x² + 3136 + 448 x

48 x² – 16x² + 672x – 448 x – 3136 = 0

32 x² + 224 x – 3136 = 0

x² + 7 x – 98= 0

x² + 14 x – 7x -98 = 0

x (x + 14) – 7( x + 14) = 0

(x + 14)( x – 7) = 0

x = 7, x = -14,neglecting x = -14,since length of a line segment can’t be negative

Hence AC = AF+ CF = x + 7 = 6 + 7 =13 cm and AB = AE+ BE = x + 8 = 7 + 8 = 15 cm

Study the following post and clear your A to Z concept of electric current and its heating effect

**Electric Current and its Heating Effect**

**Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplimentry angles at the centre of the circle.**

Ans.

GIVEN. ABCD is a quadrilateral

TO PROVE. ∠AOB + ∠DOC = 180º

∠AOD + ∠BOC = 180º

PROOF. Considering the Δ’s AOC and DOC

The line joining to the centre from the point tangents drawn bisect the angle between the tangents.

Therefore we have

Adding both the equation

∠A + ∠B + ∠C + ∠D = 360º ( Angle sum property of triangle)

Similarly

∠AOD + ∠BOC = 180º

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

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

**NCERT Solutions of class 9 science **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

**NCERT Solutions of class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

**50 important science questions for cbse 10 class**

**Chemical properties of acid and base**

**Chemistry viva voce questions & answers**

**Chemistry practical based questions**

**Solutions of previous years science & maths question papers**

**Solutions of class 10 Science question paper 2019 CBSE**

**Solutions of class 10 CBSE maths question paper 2020 **

**CBSE Class 10 science question paper 2020 SET -3 solutions**

**Download e-book of Science NCERT 634 questions with solutions for Polytechnic and NDA entrance exams**

platforms says

Hі! Someone in my Facebook group ѕhared this

site with us so I came to check it oᥙt.

I’m definitely loѵing thе information. I’m book-markіng and will be tweeting tһis to

my followers! Superb blog and great ѕtyle and design.