RD Sharma Solutions Class 10 Circles Exercise 10.1

RD Sharma Class 10 Solutions Chapter 10 Ex 10.1 PDF Free Download

Exercise 10.1


1. Fill in the blanks:


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


Soln: point of contact.


(ii) A circle may have _____ parallel tangent.


Soln : two


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


Soln: one


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


Soln: secant.


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


Soln: 90°.


Q 2. How many tangents can a circle have?



Tangent: a line 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.


Q 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.




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


Q 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.


Ans :

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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