## RD Sharma Class 10 Solutions Chapter 10 Circles Ex 10.1

### 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