In this chapter, we provide RD Sharma Solutions for Class 10 Circles Exercise 10.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 10 Circles Exercise 10.1 pdf, free RD Sharma Solutions for Class 10 Circles Exercise 10.1 book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 10 |

Chapter Name | Circles |

Exercise | 10.1 |

Category | RD Sharma Solutions |

**RD Sharma Class 10 Solutions Circles Exercise 10.1**

Question 1.

Fill in the blanks :

(i) The common point of a tangent and the circle is called ……….

(ii) A circle may have ………. parallel tangents.

(iii) A tangent to a circle intersects it in ……….. point(s).

(iv) A line intersecting a circle in two points is called a …………

(v) The angle between tangent at a point on a circle and the radius through the point is ………..

Solution:

(i) The common point of a tangent and the circle is called the point of contact.

(ii) A circle may have two parallel tangents.

(iii) A tangents to a circle intersects it in one point.

(iv) A line intersecting a circle in two points is called a secant.

(v) The angle between tangent at a point, on a circle and the radius through the point is 90°.

Question 2.

How many tangents can a circle have ?

Solution:

A circle can have infinitely many tangents.

Question 3.

O is the centre of a circle of radius 8 cm. The tangent at a point A on the circle cuts a line through O at B such that AB = 15 cm. Find OB.

Solution:

Radius OA = 8 cm, ST is the tangent to the circle at A and AB = 15 cm

OA ⊥ tangent TS

In right ∆OAB,

OB² = OA² + AB² (Pythagoras Theorem)

= (8)² + (15)² = 64 + 225 = 289 = (17)²

OB = 17 cm

Question 4.

If the tangent at a point P to a circle with centre O cuts a line through O at Q such that PQ = 24 cm and OQ = 25 cm. Find the radius of the circle.

Solution:

OP is the radius and TS is the tangent to the circle at P

OQ is a line

OP ⊥ tangent TS

In right ∆OPQ,

OQ² = OP² + PQ² (Pythagoras Theorem)

=> (25)² = OP² + (24)²

=> 625 = OP² + 576

=> OP² = 625 – 576 = 49

=> OP² = (7)²

OP = 7 cm

Hence radius of the circle is 7 cm

**All Chapter RD Sharma Solutions For Class10 Maths**

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**All Subject NCERT Solutions For Class 10**

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