**NCERT Class 10 Mathematics Chapter 10 Circles** **Solutions,** **NCERT Solutions For Class 10 Maths, CBSE Solutions For Class 10 Mathematics **to each chapter is provided in the list so that you can easily browse throughout different chapter **NCERT Class 10 Mathematics Chapter 10 Circles****Notes** and select needs one.

**NCERT Class 10 Mathematics Chapter 10 Circles**

**NCERT Class 10 Mathematics Chapter 10 Circles**Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. **NCERT****Class 10 Mathematics Textual Question Answer. **These solutions are part of NCERT All Subject Solutions. Here we have given

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**Solutions****NCERT Class 10 Mathematics Chapter 10 Circles****Circles**

**Circles****Chapter – 10**

Exercise 10.1 |

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

Ans: A circle can have infinitely many tangents.

**2. Fill in the blanks:**

**(i) A tangent to a circle intersects it in _______________ point (s).**

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

**(ii) A line intersecting a circle in two points is called ________________.**

Ans: A line intersecting a circle in two points is called a secant.

**(iii) A circle can have _______________ parallel tangents at the most.**

Ans: 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 ______________.**

Ans: The common point of a tangent to a circle and the circle is called point of contact.

**3. Tangent PQ at a point P of a circle of radius 5cm meets a line through the centre O at a point Q so that OQ = 12cm Length PQ is:**

(a) 12cm

(b) 13cm

(c) 8.5 cm

(d) sqrt(119)

Ans: (d) sqrt(119)

**4. Draw a circle and and two lines parallel to a given line such that one is a tangent and the other a secant to the circle.**

Ans:

Exercise 10.2 |

**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

Ans:

Let O be the centre of the circle.

Given that,

OQ = 25cm 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} + 242 = 252

OP^{2} = 625 − 576

OP^{2} = 49

OP = 7

Therefore, the radius of the circle is 7 cm.

Hence, alternative (A) is correct

**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

Ans: (A) **Reason: **angle OPT = 90 deg angle TQO = 90 deg In Quad OPTQ Tangent is perpendicular to the tangent through the point of contact. angle OPT + angle PTQ+ angle TQO + angle POQ = 360 deg [Sum of all angles o quad. is 360°]

90^ + angle PTQ + 90 deg + 110 deg = 360 deg

angle PTQ = 360 deg – 110 deg – 90 deg – 90 deg

.. angle PTQ = 70

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

(a) 50

(b) 60

(c) 70

(d) 80

Ans: 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 = 90º and ∠OAP = 90º

In AOBP,

Sum of all interior angles = 360

∠OAP + ∠APB + ∠PBO + ∠BOA = 360

90º + 80º +90º + ∠BOA = 360

∠BOA = 100

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

Hence, alternative (A) is correct.

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

Ans: 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 ⊥ RS and OB ⊥ PQ

くOAR = 90° くOAS

= 90°

くOBP = 90°

く OBQ = 90°

It can be observed that

くOAR =くOBQ ( Alternate interior angles)

くOAS =く(Alternate interior angles)

(Since alternate interior angles)

Since alternatives interior angles are equal, lines PQ and RS Will parallel.

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

Ans: 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 pass through another point O’. Join OP and O’P.

As perpendicular to AB at passes through o, therefore

くOPB = 90 …………………………..(1)

O Is the centre of the circle and p is the points of contrat. We know the line joining the centre and the points of contract to the tangent of the circle perpendicular to each other

∴くOPB = 90 ………………………(2)

Comparing equations (1) and (2) we obtain

くO’PB = くOPB ………………………(3)

From the figure, it can be observed that

くOBP =くOPB ……………………….(4)

Therefore, くOPB = くOPB is not possible. It is only possible, when the line O’P Coincides with op

Therefore, the perpendicular to ab through centre O

**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.**

Ans:

∠OPA = 90° [Tangent is to the radius through the pt. of Contact]

In rt ∆OPA, OP + PA2 = OA2 [Pythagoras theorem]

OP² + 42=52

OP² = 25-16 = 9

Radius OP = √9 = 3

**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.**

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

Ans: 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 obtain

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

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

CD + AB = 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.**

**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.**

Ans:

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

Ans:

**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.**

Ans: 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 ABC,

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

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x s = 14 + x

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

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