Class 10 RD Sharma Solutions - Chapter 10 Circles - Exercise 10.1

In this section, we explore Chapter 10 of the Class 10 RD Sharma textbook, which focuses on Circles. Exercise 10.1 is designed to help students understand the fundamental properties and theorems related to circles, laying a strong foundation for more advanced geometric concepts.

Class 10 RD Sharma Solutions - Chapter 10 Circles - Exercise 10.1

This section provides detailed solutions for Exercise 10.1 from Chapter 10 of the Class 10 RD Sharma textbook. These solutions are intended to assist students in mastering the essential concepts of circles, ensuring a robust understanding of geometric principles.

Question 1. Fill in the blanks :

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

Solution:

Point of contact.

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

Solution:

 Two 

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

Solution:

One

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

Solution:

Secant

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

Solution:

Right angle (90°)

Question 2. How many tangents can a circle have?

Solution:

Tangent is a line that intersect a circle at only point. Since there are a infinite number of points on a circle, a circle can have infinite 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:

👁 Image

Radius OA = 8 cm

AB = 15 cm

OA ⊥ tangent AB

Therefore, In right ∆OAB, by applying Pythagoras Theorem:

      OB² = OA² + AB² 

=> OB² = (8)² + (15)² 

            = 64 + 225 = 289 = (17)²

=> OB = 17 cm

Thus, 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:

👁 Image

OP is the radius

OQ = 25 cm

PQ = 24 cm

OP ⊥ tangent PQ

therefore, In right ∆OPQ, by applying Pythagoras Theorem:

      OQ² = OP² + PQ² 

=> (25)² = OP² + (24)²

=> 625 = OP² + 576

=> OP² = 625 – 576 = 49

=> OP² = (7)²

     OP = 7 cm

Thus, radius of the circle is 7 cm

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• Class 10 RD Sharma Solutions - Chapter 10 Circles - Exercise 10.2
• Class 10 RD Sharma Solutions - Chapter 10 Circles - Exercise 10.3
• Class 10 RD Sharma Solutions - Chapter 10 Circles - Exercise 10.4