# RD Sharma Class 10 Ex 10.1 Solutions Chapter 10 Circles

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

# Chapter 10: Circles Exercise – 10.1

### Question: 1

Fill in the blanks:

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

(ii) A circle may have _____ parallel tangent.

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

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

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

### Solution:

(i) point of contact.

(ii) two

(iii) one

(iv) secant.

(v) 90°.

### Question: 2

How many tangents can a circle have?

### Solution:

Tangent: Aline intersecting circle in one point is called a tangent As there are infinite number of points on the circle, a circle has many (infinite) tangents.

### Question: 3

O is the centre the circle shown below with a radius of 8 cm. The circle cuts the tangent AB through O at B such that AB = 15 cm. Find OB.

### Solution:

Given data: AB = 15 cm OA = 8 cm (radius of the circle) we know that: the tangent cuts the circle at 90 degrees. Therefore, OA is the hypotenuse of the triangle OAB. Hence, the longest side can be found by using Pythagoras Theorem. We have, OB = 17 cm Therefore, OB = 17 cm

### Question: 4

If the tangent at point P to the 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:

Given data: PQ = 24 cm OQ = 25 cm OP = radius = ?

P is a point of contact, at point of contact, tangent and radius are perpendicular to each other. Therefore triangle is right angled triangle angle OPQ = 90° BY Pythagoras theorem, OP = 7 cm Therefore, OP = radius = 7 cm

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