NCERT Solutions for Class 9 Maths Chapter 10: Circles - Exercise 10.6

EXERCISE 10.6

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

Ans:

(Image will be Uploaded Soon)

Given: Line of centres of two intersecting circles. 

To prove: Subtends equal angles at the two points of intersection. 

Let two circles having their centres as O and O’ intersect each other at point A and B respectively. Let us join O’.

(Image will be Uploaded Soon)

In $\Delta AOO'$ and $\Delta BOO'$,

OA = OB (Radius of circle 1)

O’A = O’B (Radius of circle 2)

OO’ = OO’ (Common)

 $\vartriangle {\text{AOO’}} \cong \Delta {\text{BOO’}}$ (SSS congruence rule)

$\angle OAO’ = \angle OBO’(By CPCT)$

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

2. Two chords AB and CD of lengths 5 cm 11cm 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 cm, find the radius of the circle.

Ans:

Given: Two chords of a circle, AB and CD, with lengths of 5 cm and 11 cm respectively, are parallel to one other and on opposing sides of the circle's centre. 

To Find: The radius of the circle if the distance between AB and CD is 6 cm.

(Image will be Uploaded Soon)

${\text{BM}} = \dfrac{{{\text{AB}}}}{2} = \dfrac{5}{2}\quad $ (Perpendicular from the centre bisects the chord) ${\text{ND}} = \dfrac{{{\text{CD}}}}{2} = \dfrac{{11}}{2}$

Let ON be ${\text{x}}$. Therefore, OM will be $6 - {\text{x}}$.

In $\vartriangle {\text{MOB}}$,

${\text{O}}{{\text{M}}^2} + {\text{M}}{{\text{B}}^2} = {\text{O}}{{\text{B}}^2}$

${(6 - x)^2} + {\left( {\dfrac{5}{2}} \right)^2} = {\text{O}}{{\text{B}}^2}$

$36 + {x^2} - 12x + \dfrac{{25}}{4} = {\text{O}}{{\text{B}}^2}\quad  \ldots $

In $\vartriangle {\text{NOD}}$,

${\text{O}}{{\text{N}}^2} + {\text{N}}{{\text{D}}^2} = {\text{O}}{{\text{D}}^2}$

${x^2} + {\left( {\dfrac{{11}}{2}} \right)^2} = {\text{O}}{{\text{D}}^2}$

${x^2} + \dfrac{{121}}{4} = O{D^2}\quad  \ldots $

We hate ${\text{OB}} = {\text{OD}}$ (Radii of the same circle)

Therefore, from Equations (1) and (2), $36 + {x^2} - 12x + \dfrac{{25}}{4} = {x^2} + \dfrac{{121}}{4}$

$12x = 36 + \dfrac{{25}}{4} - \dfrac{{121}}{4}$

$ = \dfrac{{144 + 25 - 121}}{4} = \dfrac{{48}}{4} = 12$

${\text{x}} = 1$

From Equation (2),

(1) $^2 + \dfrac{{121}}{4} = {\text{O}}{{\text{D}}^2}$

${\text{O}}{{\text{D}}^2} = \dfrac{{121}}{4} + 1 = \dfrac{{125}}{4}$

${\text{OD}} = \dfrac{5}{2}\sqrt 5 $

Therefore, the radius of the circle is $\dfrac{5}{2}\sqrt 5 \;{\text{cm  =  }}5.6\;{\text{cm}}$ (approx.)

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

Ans:

Given: Two parallel chords of a circle have lengths of 6 cm and 8 cm, respectively. 

To Find: What is the distance between the smaller and larger chords from the centre if the smaller chord is 4 cm away?

(Image will be Uploaded Soon)

Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.

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

${\text{OM}} = 4\;{\text{cm}}$

${\text{MB}} = \dfrac{{{\text{AB}}}}{2} = \dfrac{6}{2} = 3\;{\text{cm}}$

In $\Delta {\text{OMB}}$,

${\text{O}}{{\text{M}}^2} + {\text{M}}{{\text{B}}^2} = {\text{O}}{{\text{B}}^2}$

${(4)^2} + {(3)^2} = {\text{O}}{{\text{B}}^2}$

$16 + 9 = {\text{O}}{{\text{B}}^2}$

${\text{OB}} = \sqrt {25} $

${\text{OB}} = 5\;{\text{cm}}$

In $\Delta {\text{OND}}$,

${\text{OD}} = {\text{OB}} = 5\;{\text{cm}}\quad $ (Radii of the same circle)

${\text{ND}} = \dfrac{{{\text{CD}}}}{2} = \dfrac{8}{2} = 4\;{\text{cm}}$

${\text{O}}{{\text{N}}^2} + {\text{N}}{{\text{D}}^2} = {\text{O}}{{\text{D}}^2}$

${\text{O}}{{\text{N}}^2} + {(4)^2} = {(5)^2}$

${\text{O}}{{\text{N}}^2} = 25 - 16 = 9$

${\text{ON}} = 3\;{\text{cm}}$

Therefore, the distance of the bigger chord from the centre is $3\;{\text{cm}}$.

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 $\angle ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Ans:

(Image will be Uploaded Soon)

Given: Allow the vertex of an angle ABC to be outside of a circle, and the sides of the angle to intersect the circle with equal chords AD and CE. 

To Prove: Half the difference between the angles spanned by the chords AC and DE in the centre.

In $\vartriangle {\text{AOD}}$ and $\vartriangle {\text{COE}}$

${\text{OA}} = {\text{OC}}$ (Radii of the same circle) ${\text{OD}} = {\text{OE}}$ (Radii of the same circle) ${\text{AD}} = {\text{CE}}$ (Given)

$\vartriangle {\text{AOD}} \cong \Delta {\text{COE}}$ (SSS congruence rule)

$\angle OAD = \angle OCE(ByCPCT)$

$\angle {\text{ODA}} = \angle {\text{OEC}}({\text{ByCPCT}})$

Also, $\angle OAD = \angle ODA($ As $OA = OD)$

From Equations (1), (2), and (3), we $\angle OAD = \angle OCE = \angle ODA = \angle OEC$

Let $\angle OAD = \angle OCE = \angle ODA = \angle OEC = x$

In $\vartriangle {\text{OAC}}$

${\text{OA}} = {\text{OC}}$

$\angle OCA = \angle OAC$ (Let a)

In $\Delta {\text{ODE}}$,

${\text{OD}} = {\text{OE}}$

$\angle {\text{OED}} = \angle {\text{ODE}}$ (Let ${\text{y}})$ is a cyclic $\angle {\text{CAD}} + \angle {\text{DEC}} = {180^\circ }$ (Opposite angles are supplementary) ${\text{x}} + {\text{a}} + {\text{x}} + {\text{y}} = {180^\circ }$

$2x + a + y = {180^\circ }$

$y = {180^\circ } - 2x - a$

However, $\angle {\text{DOE}} = {180^\circ } - 2{\text{y}}$

And, $\angle {\text{AOC}} = {180^\circ } - 2{\text{a}}$

$\angle {\text{DOE}} - \angle {\text{AOC}} = 2{\text{a}} - 2{\text{y}} = 2{\text{a}} - 2\left( {{{180}^\circ } - 2{\text{x}} - {\text{a}}} \right)$

$ = 4{\text{a}} + 4{\text{x}} - {360^\circ }\quad  \ldots (5)$

$\angle {\text{BAC}} + \angle {\text{CAD}} = {180^\circ }$ (Linear pair)

$\therefore \angle {\text{BAC}} = {180^\circ } - \angle {\text{CAD}} = {180^\circ } - ({\text{a}} + {\text{x}})$

Similarly, $\angle ACB = {180^\circ } - (a + x)$

In $\vartriangle {\text{ABC}}$,

$\angle {\text{ABC}} + \angle {\text{BAC}} + \angle {\text{ACB}} = {180^\circ }$ (Angle sum property of a triangle)

$\angle {\text{ABC}} = {180^\circ } - \angle {\text{BAC}} - \angle {\text{ACB}}$

$ = {180^\circ } - \left( {{{180}^\circ } - {\text{a}} - {\text{x}}} \right) - \left( {{{180}^\circ } - {\text{a}} - {\text{x}}} \right)$

$ = 2{\text{a}} + 2{\text{x}} - {180^\circ }$

$ = \dfrac{1}{2}\left[ {4a + 4x - {{360}^\circ }} \right]$

$\angle {\text{ABC}} = \dfrac{1}{2}[\angle {\text{DOE}} - \angle {\text{AOC}}]$ [Using Equation

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

Ans:

Given: A circle is given. 

To Prove: The circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

 (Image will be Uploaded Soon)

Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends $90^\circ $ on the arc.

$\therefore \angle COD{\text{ }} = {\text{ }}90^\circ $

Also, in rhombus, the diagonals intersect each other at $90^\circ $.

$\angle AOB{\text{ }} = \angle BOC{\text{ }} = \angle COD{\text{ }} = \angle DOA{\text{ }} = {\text{ }}90^\circ $

Clearly, point O has to lie on the circle.

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

Ans:

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To Prove:  AE = AD.

(Image will be Uploaded Soon)

It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is $180^\circ $.

$\angle AEC{\text{ }} + \angle CBA{\text{ }} = {\text{ }}180^\circ $

$\angle AEC{\text{ }} + \angle AED{\text{ }} = {\text{ }}180^\circ $ (Linear pair)

$\angle AED{\text{ }} = \angle CBA{\text{ }}...{\text{ }}\left( 1 \right)$

For a parallelogram, opposite angles are equal.

$\angle ADE{\text{ }} = \angle CBA{\text{ }}...{\text{ }}\left( 2 \right)$

From (1) and (2),

$\angle AED{\text{ }} = \angle ADE$

AD = AE (sides opposite to equal Angles of a triangle are equal).

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.

Ans:

Given: AC and BD are chords of a circle which bisect each other. 

To Prove: (i) AC and BD are diameters; (ii) ABCD is a rectangle.

(Image will be Uploaded Soon)

Let two chords AB and CD are intersecting each other at point O.

In $\Delta AOB$ and $\Delta COD$,

OA = OC (Given)

OB = OD (Given)

$\angle AOB{\text{ }} = \angle COD$ (Vertically opposite angles)

$\therefore \Delta AOB \cong \Delta COD$ (SAS congruence rule)

AB = CD (By CPCT)

Similarly, it can be proved that $\Delta AOD \cong \Delta COB$

$\therefore AD{\text{ }} = {\text{ }}CB$ (By CPCT)

Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal.

$\therefore \angle A{\text{ }} = \angle C$

This shows that, $\angle A{\text{ }} + \angle C{\text{ }} = {\text{ }}180^\circ $ (ABCD is a cyclic quadrilateral)

$ \angle{A} + \angle{A} = 180^{\circ}$

$2 \angle{A} = 180^{\circ}$

$\therefore \angle{A} = 90^{\circ}$

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

$\angle A$ is the angle subtended by chord BD.

And as $\angle A{\text{ }} = {\text{ }}90^\circ $, 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 ${90^\circ } - \dfrac{1}{2}\;{\text{A}},{90^\circ } - \dfrac{1}{2}\;{\text{B}}$, and ${90^\circ } - \dfrac{1}{2}{\text{C}}$

Ans:

Given: Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. 

To Prove: The angles of the triangle DEF are ${90^\circ } - \dfrac{1}{2}\;{\text{A}},{90^\circ } - \dfrac{1}{2}\;{\text{B}}$, and ${90^\circ } - \dfrac{1}{2}{\text{C}}$

(Image will be Uploaded Soon)

It is given that ${\text{BE}}$ is the bisector of $\angle {\text{B}}$. $\angle {\text{ABE}} = \dfrac{{\angle {\text{B}}}}{2}$

However, $\angle ADE = \angle ABE$ (Angles in the same segment for chord AE) $\therefore \angle {\text{ADE}} = \dfrac{{\angle {\text{B}}}}{2}$

Similarly, $\angle {\text{ACF}} = \angle {\text{ADF}} = \dfrac{{\angle {\text{C}}}}{2}$ (Angle in the same segment for chord ${\text{AF}}$ ) $\angle {\text{D}} = \angle {\text{ADE}} + \angle {\text{ADF}}$

$ = \dfrac{{\angle {\text{B}}}}{2} + \dfrac{{\angle {\text{C}}}}{2}$

$ = \dfrac{1}{2}(\angle {\text{B}} + \angle {\text{C}})$

$ = \dfrac{1}{2}\left( {{{180}^\circ } - \angle {\text{A}}} \right)$

$ = {90^\circ } - \dfrac{1}{2}\angle {\text{A}}$

Similarly, it can be proved that $\angle {\text{E}} = {90^\circ } - \dfrac{1}{2}\angle {\text{B}}$

$\angle {\text{F}} = {90^\circ } - \dfrac{1}{2}\angle {\text{C}}$

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.

Ans:

Given: At the locations A and B, two congruent circles cross. Any line segment PAQ is drawn through A, so that P and Q are on the two circles.

To Prove: BP = BQ.

(Image will be Uploaded Soon)

AB is the common chord in both the congruent circles.

$\therefore \angle APB{\text{ }} = \angle AQB$

In $\Delta BPQ$,

$\angle APB{\text{ }} = \angle AQB$

$\therefore BQ{\text{ }} = {\text{ }}BP$ (Sides opposite to equal angles of a triangle are equal)

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

Ans:

Given: If the angle bisector of A and the perpendicular bisector of BC cross in any triangle ABC, 

To Prove: They intersect on the circum-circle of the triangle ABC.

(Image will be Uploaded Soon)

Let perpendicular bisector of side BC and angle bisector of $\angle A$ meet at point D. Let the perpendicular bisector of side BC intersect it at E.

Perpendicular bisector of side BC will pass through circumcenter O of the circle. $\angle BOC$ and $\angle BAC$ are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively. 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.

$\angle {\text{BOC}} = 2\angle {\text{BAC}} = 2\angle {\text{A}}$

In $\Delta {\text{BOE}}$ and $\Delta {\text{COE}}$,

${\text{OE}} = {\text{OE}}$ (Common)

${\text{OB}} = {\text{OC}}$ (Radii)

$\angle {\text{OEB}} = \angle {\text{OEC}}$ (Each ${90^\circ }$)

$\angle {\text{BOE}} = \angle {\text{COE}}({\text{ByCPCT}})$

However, $\angle {\text{BOE}} + \angle {\text{COE}} = \angle {\text{BOC}}$

$\therefore \angle {\text{BOE}} + \angle {\text{BOE}} = 2\angle {\text{A}}[{\text{Using}}$ Equations (1) and (2)]

$2\angle {\text{BOE}} = 2\angle {\text{A}}$

$\angle {\text{BOE}} = \angle {\text{A}}$

$\angle {\text{BOE}} = \angle {\text{COE}} = \angle {\text{A}}$

The perpendicular bisector of side ${\text{BC}}$ and angle bisector of $\angle {\text{A}}$ meet at point ${\text{D}}$. $\therefore \angle {\text{BOD}} = \angle {\text{BOE}} = \angle {\text{A}}$

Since ${\text{AD}}$ is the bisector of angle $\angle {\text{A}}$, $\angle {\text{BAD}} = \dfrac{{\angle {\text{A}}}}{2}$

$\therefore 2\angle {\text{BAD}} = \angle {\text{A}}$

From Equations (3) and (4), we obtain $\angle {\text{BOD}} = 2\angle {\text{BAD}}$ 

Only when point BD is a chord of the circle is this feasible. The point D is on the circumcircle for this. As a result, on the circumcircle of triangle ABC, the perpendicular bisector of side BC and the angle bisector of A meet.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.6

Opting for the NCERT solutions for Ex 10.6 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.6 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Subject Circles textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 10 Exercise 10.6 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 10 Exercise 10.6, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 10 Exercise 10.6 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.