NCERT Solutions for Class 9 Maths Exercise 10.6 Chapter 10 Circles

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.

NCERT Solutions for Class 9 Maths Chapter 10 – Circles Exercise 10.6 is provided here, which can be downloaded in PDF format just by clicking on the link given below. Our subject experts have designed them considering the NCERT syllabus and guidelines (2023-24). These are helpful for students who want to score good marks in Maths.

Students can use NCERT Solutions for Class 9 Maths subject as reference material to do their homework and prepare for the examination. It will help them to understand the concepts in a deeper way.

NCERT Solutions for Class 9 Maths Chapter 10 – Circles Exercise 10.6

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Access Other Exercise Solutions of Class 9 Maths Chapter 10- Circles

Exercise 10.1 Solutions 2 Question ( 2 Short)

Exercise 10.2 Solutions 2 Question ( 2 long)

Exercise 10.3 Solutions 3 Question ( 3 long)

Exercise 10.4 Solutions 6 Question ( 6 long)

Exercise 10.5 Solutions 12 Questions (12 long)

Access Answers to NCERT Class 9 Maths Chapter 10 – Circles Exercise 10.6

1.  Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution:

Consider the following diagram

In ΔPOO’ and ΔQOO’

OP = OQ          (Radius of circle 1)

O’P = O’Q        (Radius of circle 2)

OO’ = OO’        (Common arm)

So, by SSS congruency, ΔPOO’ ≅ ΔQOO’

Thus, ∠OPO’ = ∠OQO’ (proved).

2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 , find the radius of the circle.

Solution:

Here, OM ⊥ AB and ON ⊥ CD are drawn, and OB and OD are joined.

We know that AB bisects BM as the perpendicular from the centre bisects the chord.

Since AB = 5 so,

BM = AB/2 = 5/2

Similarly, ND = CD/2 = 11/2

Now, let ON be x.

So, OM = 6−x.

Consider ΔMOB,

OB² = OM²+MB²

Or,

Consider ΔNOD,

OD² = ON² + ND²

Or

We know, OB = OD (radii)

From equation 1 and equation 2, we get

Now, from equation (2), we have,

OD²= 1² +(121/4)

Or OD = (5/2)×√5 cm

3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance 4 cm from the centre, what is the distance of the other chord from the centre?

Solution:

Consider the following diagram

Here AB and CD are 2 parallel chords. Now, join OB and OD.

Distance of smaller chord AB from the centre of the circle = 4 cm

So, OM = 4 cm

MB = AB/2 = 3 cm

Consider ΔOMB

OB² = OM²+MB²

Or, OB = 5cm

Now, consider ΔOND,

OB = OD = 5 (since they are the radii)

ND = CD/2 = 4 cm

Now, OD²= ON²+ND²

Or, ON = 3 cm.

4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Solution:

Consider the diagram

Here AD = CE

We know any exterior angle of a triangle is equal to the sum of interior opposite angles.

So,

∠DAE = ∠ABC+∠AEC (in ΔBAE) ——————-(i)

DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.

So,

∠DAE = (½)∠DOE ——————-(ii)

Similarly, ∠AEC = (½)∠AOC  ——————-(iii)

Now, from equations (i), (ii), and (iii), we get,

(½)∠DOE = ∠ABC+(½)∠AOC

Or, ∠ABC = (½)[∠DOE-∠AOC]  (hence proved).

5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Solution:

To prove: A circle drawn with Q as the centre will pass through A, B and O (i.e. QA = QB = QO)

Since all sides of a rhombus are equal,

AB = DC

Now, multiply (½) on both sides

(½)AB = (½)DC

So, AQ = DP

BQ = DP

Since Q is the midpoint of AB,

AQ= BQ

Similarly,

RA = SB

Again, as PQ is drawn parallel to AD,

RA = QO

Now, as AQ = BQ and RA = QO we get,

QA = QB = QO (hence proved).

6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE, = AD.

Solution:

Here, ABCE is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of the opposite angles is 180°.

So, ∠AEC+∠CBA = 180°

As ∠AEC and ∠AED are a linear pair,

∠AEC+∠AED = 180°

Or, ∠AED = ∠CBA … (1)

We know in a parallelogram, opposite angles are equal.

So, ∠ADE = ∠CBA … (2)

Now, from equations (1) and (2), we get,

∠AED = ∠ADE

Now, AD and AE are angles opposite to equal sides of a triangle,

∴ AD = AE (proved).

7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.

Solution:

Here chords AB and CD intersect each other at O.

Consider ΔAOB and ΔCOD,

∠AOB = ∠COD (They are vertically opposite angles)

OB = OD (Given in the question)

OA = OC (Given in the question)

So, by SAS congruency, ΔAOB ≅ ΔCOD

Also, AB = CD (By CPCT)

Similarly, ΔAOD ≅ ΔCOB

Or, AD = CB (By CPCT)

In quadrilateral ACBD, opposite sides are equal.

So, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal.

So, ∠A = ∠C

Also, as ABCD is a cyclic quadrilateral,

∠A+∠C = 180°

⇒∠A+∠A = 180°

Or, ∠A = 90°

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

∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.

8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°–(½)A, 90°–(½)B and 90°–(½)C.

Solution:

Consider the following diagram

Here, ABC is inscribed in a circle with centre O and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F, respectively.

Now, join DE, EF and FD

As angles in the same segment are equal, so,

∠EDA = ∠FCA ————-(i)

∠FDA = ∠EBA ————-(i)

By adding equations (i) and (ii), we get,

∠FDA+∠EDA = ∠FCA+∠EBA

Or, ∠FDE = ∠FCA+∠EBA = (½)∠C+(½)∠B

We know, ∠A +∠B+∠C = 180°

So, ∠FDE = (½)[∠C+∠B] = (½)[180°-∠A]

∠FDE = [90-(∠A/2)]

In a similar way,

∠FED = [90° -(∠B/2)] °

And,

∠EFD = [90° -(∠C/2)] °

9. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Solution:

The diagram will be

Here, ∠APB = ∠AQB (as AB is the common chord in both the congruent circles.)

Now, consider ΔBPQ,

∠APB = ∠AQB

So, the angles opposite to equal sides of a triangle.

∴ BQ = BP

10. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Solution:

Consider this diagram

Here, join BE and CE.

Now, since AE is the bisector of ∠BAC,

∠BAE = ∠CAE

Also,

∴arc BE = arc EC

This implies that chord BE = chord EC

Now, consider triangles ΔBDE and ΔCDE,

DE = DE     (It is the common side)

BD = CD     (It is given in the question)

BE = CE      (Already proved)

So, by SSS congruency, ΔBDE ≅ ΔCDE.

Thus, ∴∠BDE = ∠CDE

We know, ∠BDE = ∠CDE = 180°

Or, ∠BDE = ∠CDE = 90°

∴ DE ⊥ BC (hence proved).

Exercise 10.6 has ten questions in total. It is mentioned as an optional exercise in the book, which consists of problems based on all the topics covered in the chapter. Though it is an optional exercise, students should solve them as well to have good practice. Also, solve problems for each exercise of chapter 10 of Maths Class 9 here with detailed answers.

We provide here some advanced learning materials such as notes, tips and tricks to prepare for exams. For all the classes from 6 to 12, PDFs of chapter-wise and exercise-wise NCERT solutions are provided here. Students can download them and learn offline as well.