NCERT Solutions for Class 9 Maths Chapter 10 - Circles
Score marks in short answer questions easily by practising with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 10 Circles. Write and practise with TopperLearning’s expert solutions for long answer questions based on chords. By revising a variety of textbook questions, you’ll learn how to prove chord properties in a step-wise format.
With CBSE Class 9 Maths chapter solutions, learn the application of important concepts such as RHS congruence rule, angle sum property of a triangle etc. This knowledge will help you think logically and arrive at correct answers in your Maths exam. Other than expert solutions, you may want to use sample papers, online mock tests, concept videos etc. to revise circles for exam preparation.
Chapter 10 - Circles Exercise Ex. 10.1
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.
Chapter 10 - Circles Exercise Ex. 10.2
So, if we try to superimpose two circles of equal radius, one each other both circles will cover each other.
So, two circles are congruent if they have equal radius.


AB = CD (chords of same length)
OA = O'C (radii of congruent circles)
OB = O'D (radii of congruent circles)






Hence equal chords of congruent circles subtend equal angles at their centres.




OA = O'C (radii of congruent circles)
OB = O'D (radii of congruent circles)



Hence, if chords of congruent circles subtend equal angles at their centres then chords are equal.
Chapter 10 - Circles Exercise Ex. 10.3
(i) circles don't intersect each other at any point, so circles are not having any point in common.
(iv) These circles intersect each other at two points P and Q. So the circles have two points in common. We may observe that there can be maximum 2 points in common.
We can have a situation in which two congruent circles are superimposed on each other, this situation can be referred as if we are drawing circle two times.
Step1. Take the given circle centered at point O.
Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these
Step3. Let these perpendicular bisectors meet at point O. Now, O is the centre of given circle.
Join AB. AB is the chord for circle centered at O, so perpendicular bisector of AB will pass through O.
Again AB is also chord of circle centered at O', so, perpendicular bisector of AB will also pass through O'.
Clearly centres of these circles lie on the perpendicular bisector of common chord.
Chapter 10 - Circles Exercise Ex. 10.4
OA = OB = 5 cm
O'A = O'B = 3 cm
OO' will be the perpendicular bisector of chord AB.
Given that OO' = 4 cm
Let OC be x. so, O'C will be 4 - x
In

OA2 = AC2 + OC2


In

O'A2 = AC2 + O'C2



From equations (1) and (2), we have
25 - x2 = - x2 - 7 + 8x
8x = 32
x = 4
So, the common chord will pass through the centre of smaller circle i.e. O'. and hence it will be diameter of smaller circle.
The length of the common chord AB = 2 AC = (2

In


OV = OU (Equal chords of a circle are equidistant from the centre)


OT = OT (common)



It is given that
PQ = RS ... ... ... ... (2)



On adding equations (1) and (3), we have
PV + VT = RU + UT

On subtracting equation (4) from equation (2), we have
PQ - PT = RS - RT

Equations (4) and (5) shows that the corresponding segments of
chords PQ and RS are congruent to each other.
Draw perpendiculars OV and OU on these chords.
In


OV = OU (Equal chords of a circle are equidistant from the centre)


OT = OT (common)




Here, BC is chord of smaller circle and AD is chord of bigger circle.
We know that the perpendicular drawn from centre of circle bisects the chord.
And AM = MD ... (2)
Subtracting equations (2) from (1), we have
AM - BM = MD - MC

Let R, S and M be the position of Reshma, Salma and Mandip respectively.

In OAR
OA2 + AR2 = OR2
OA2 + (3 m)2 = (5 m)2
OA2 = (25 - 9) m2 = 16 m2
OA = 4 m
We know that in an isosceles triangle altitude divides the base, so in


Area of





RC = 4.8
So, ASD is a equilateral triangle
OA (radius) = 20 m.
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.
We also know that median intersect each other at the 2: 1. As AB is the median of equilateral triangle ABC, we can write

In

AD2 = AB2 + BD2
AD2 = (30)2 +


So, length of string of each phone will be

Chapter 10 - Circles Exercise Ex. 10.5



= 60o + 30o
= 90o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.


AB = OA = OB = radius

So, each interior angle of this triangle will be of 60o





So, angle subtended by this chord at a point on major arc and minor arc are 30o and 150o respectively.
Consider PR as a chord of circle.
Take any point S on major arc of circle.
Now PQRS is a cyclic quadrilateral.



We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.



OP = OR (radii of same circle)






2

















But













In




We have



30o +










90o +


Here, each interior angle of cyclic quadrilateral is of 90o. Hence it is a rectangle.


In


AD = BC (Given)


AM = BM (Perpendicular distance between two parallel lines is same)










This equation shows that the opposite angles are supplementary.
So, ABCD is a cyclic quadrilateral.

For chord AP


For chord DQ


ABD and PBQ are line segments intersecting at B.


From equations (1), (2) and (3), we have



Two circles are drawn while taking AB and AC as diameter.
Let they intersect each other at D and let D does not lie on BC.
Join AD





Hence BDC is straight line and our assumption was wrong.
Thus, Point D lies on third side BC of











In










Adding equations (1) and (2), we have











But it is given that


From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180o.
So, it is a cyclic quadrilateral.
Consider chord CD.
Now,




We know that opposite angles of a parallelogram are equal




From equation (1)







Parallelogram ABCD is having its one of interior angles as 90o, so, it is a rectangle.
Chapter 10 - Circles Exercise Ex. 10.6
Construction: Let us join OO',


OA = OB (radius of circle 1)
O'A = O'B (radius of circle 2)
OO' = OO' (common)




So, line of centres of two intersecting circles subtends equal angles at the two points of intersection.




In




So, from equation (1) and (2)



OM = 4 cm







OA = OC (radii of same circle)
OD = OE (radii of same circle)
AD = CE (given)






We also have


From equations (1), (2) and (3), we have




Let




In

OA = OC



OD = OE


ADEC is a cyclic quadrilateral


2x + a + y = 180o
y = 180 - 2x - a ... (4)
But

And

Now,


= 4a + 4x - 360o ... (5)
Now,




Similarly,

Now, in







= 180 - (180 - a - x) - (180 - a -x)
= 2a + 2x - 180
=





We know that angle in a semicircle is of 90o.

Also in rhombus the diagonals intersect each other at 90o




So, point O has to lie on the circle.






For a parallelogram opposite angles are equal.


From (1) and (2)


AD = AE (angles opposite to equal sides of a triangle)
In


OA = OC (given)
OB = OD (given)




AB = CD (by CPCT)
Similarly, we can prove


Hence, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal








As ACBD is a parallelogram and one of its interior angles is 90o, so it is a rectangle.


It is given that BE is the bisector of



But




Similarly,



Now,







Now in




Let perpendicular bisector of side BC intersects it at E.
Perpendicular bisector of side BC will pass through circum centre O of circle. Now,


We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.



In


OB = OC (radii of same circle)







But























From equations (3) and (4), we have


It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.
Therefore, the perpendicular bisector of side BC and angle bisector of

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