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# Class 9 RD SHARMA Solutions Maths Chapter 15 - Circles

With TopperLearning’s chapter-wise, R.D Sharma solutions, you have the best-ever resources to learn from. While Class 9 students usually regard Maths as challenging, things become easy with the right examples at your convenience. These class 9 solutions for chapter 15 - Circles have step-by-step answers for all the exercises from the main book rightfully explained.

While devising the solutions about circles, experienced subject matter experts did follow the latest CBSE class 9 Maths guidelines. Every answer is set chronologically according to the exercise question numbers. Moreover, its detailed elaborations help you comprehensively understand various theorems and problems related to circles. Overall, aiming to make learning easy and fun for students.

Also, at TopperLearning, students can score the highest grades with interactive video lessons about circles and amplify their preparation schedule with rightfully conducted mock tests. For classes 1 to 12, you have all the necessary resources rightfully available. Besides offering exclusive R.D Sharma solutions for class 9, the website also lets you avail of R.D Sharma Solutions for class 10R.D Sharma solutions for class 11, and R.D Sharma solutions for class 12.

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## Circles Exercise Ex. 15.1

### Solution 1

(i) interior/exterior

(ii) concentric

(iii) the exterior

(iv) arc

(v) diameter

(vi) semi-circle

(vii) centre

(viii) three

(i) T

(ii) T

(iii) T

(iv) F

(v) T

(vi) T

(vii) F

(viii) T

## Circles Exercise Ex. 15.2

### Solution 4

Steps of construction:

(1) Take three point A, B and C on the given circle.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of chord AB and BC which interesect each other at O.

(4) Point O will be the required circle because we know that the perpendicular bisector of a chord always passes through the centre.

### Solution 11

Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm

In OMB
In OND

So, distance of bigger chord from centre is 3 cm.

### Solution 14

Suppose two different circles can intersect each other at three points then they will pass through the three common points but we know that there is one and only one circle with passes through three non-collinear points, which contradicts our supposition.

Hence, two different circles cannot intersect each other at more than two points.

### Solution 15

Draw OM  AB and ON  CD. Join OB and OD

(Perpendicular from centre bisects the chord)
Let ON be x, so OM will be 6 - x
In MOB
In NOD

We have OB = OD             (radii of same circle)
So, from equation (1) and (2)

From equation (2)

So, radius of circle is found to be  cm.

## Circles Exercise Ex. 15.5

### Solution 25

Let O be the centre of the circle circumscribing the cyclic rectangle ABCD. Since ABC = 90o and AC is a chord of the circle, so, AC is a diameter of the circle. Similarly, BD is a diameter.

Hence, point of intersection of AC and BD is the centre of the circle.

## Circles Exercise 15.111

### Solution 18

The greatest chord of the circle is diameter of the circle.

Hence, correct option is (c).

### Solution 19

Angle formed in a minor segment is always a obtuse angle.

Hence, correct option is (b).

### Solution 20

Three non-collinear points make a triangle and there is only one circle that can pass through all three points,

i.e. circumcircle of that triangle.

Hence, correct option is (a).