NCERT Solutions for Class 9 Maths Chapter 9 - Circles Exercise 9.2

Exercise 9.2

1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. 

Ans:

Consider the radius of the circle with centre as $\text{O}$ and ${{\text{O}}^{\prime }}$ be $5~\text{cm}$ and $3~\text{cm}$ respectively.

$\text{OA}=\text{OB}=5~\text{cm}$   (Radius of same circle)

${{\text{O}}^{\prime }}\text{A}={{\text{O}}^{\prime }}\text{B}=3~\text{cm}$   (Radius of same circle)

OO' will be the perpendicular bisector of chord $\text{AB}$.

$\therefore \text{AC}=\text{CB}$

It is given that,

$\text{O}{{\text{O}}^{\prime }}=4~\text{cm}$

Let the length of OC be $\text{x}$ cm. Therefore, O'C will be ($4-\text{x}$) cm.

In $\Delta \text{OAC}$,

$\angle ACO$ is a right angle. Therefore,

Using Pythagoras theorem,

$\text{O}{{\text{A}}^{2}}=\text{A}{{\text{C}}^{2}}+\text{O}{{\text{C}}^{2}}$

$\Rightarrow {{5}^{2}}=\text{A}{{\text{C}}^{2}}+{{\text{x}}^{2}}$

$\Rightarrow 25-{{\text{x}}^{2}}=\text{A}{{\text{C}}^{2}}\quad \ldots \left( 1 \right)$

In $\Delta {{\text{O}}^{\prime }}\text{AC}$,

$\angle ACO'$ is a right angle. Therefore,

Using Pythagoras theorem,

${{\text{O}}^{\prime }}{{\text{A}}^{2}}=\text{A}{{\text{C}}^{2}}+{{\text{O}}^{\prime }}{{\text{C}}^{2}}$

$\Rightarrow {{3}^{2}}=A{{C}^{2}}+{{(4-x)}^{2}}$

$\Rightarrow 9=\text{A}{{\text{C}}^{2}}+16+{{\text{x}}^{2}}-8\text{x}$

$\Rightarrow \text{A}{{\text{C}}^{2}}=-{{\text{x}}^{2}}-7+8\text{x}\,\,\,\,\,...\left( 2 \right)$

From equations (1) and (2), we get

$25-{{x}^{2}}=-{{x}^{2}}-7+8x$

$8x=32$

$\text{x}=4$

Putting the value of $x$ in equation (1), we get

$\begin{align} & \text{A}{{\text{C}}^{2}}=25-{{4}^{2}} \\ & \text{A}{{\text{C}}^{2}}=25-16 \\ & \text{A}{{\text{C}}^{2}}=9 \\ & \text{A}{{\text{C}}^{2}}=\sqrt{9}=3\,\text{cm} \\ \end{align}$

Since,

$\begin{align} & AB=2\times AC \\ & \,\,\,\,\,\,\,\,=2\times 3=6\text{ cm} \\ \end{align}$

Therefore, the common chord of both the circle will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.

2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. 

Ans: Let RS and PQ be two chords of equal lengths, of a given circle and they are intersect each other at point T.

Construct two perpendicular lines $\text{OV}$ and $\text{OU}$ on these chords.

In $\Delta \text{OVT}$ and $\Delta \text{OUT}$,

Since equal chords of a circle are equidistant from the centre. Therefore,

$\text{OV}=\text{OU}$

$\angle OVT=\angle OUT={{90}^{{}^\circ }}$

The side OT is present in both triangles. Therefore,

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

$\therefore \Delta \text{OVT}\cong \Delta $ OUT (By RHS axiom of congruency)

Hence $\text{VT}=\text{UT}$     …(1) 

As they are corresponding parts of the corresponding triangles.

It is also given that $\text{PQ}=\text{RS}$     …(2)

$\Rightarrow \frac{1}{2}\text{PQ}=\frac{1}{2}\text{RS}$

$\Rightarrow \text{PV}=\text{RU}\quad \ldots \left( 3 \right)$

On adding Equations (1) and (3), we obtain

$\text{PV}+\text{VT}=\text{RU}+\text{UT}$

$\Rightarrow \text{PT}=\text{RT}$         …(4)

On subtracting Equation (4) from Equation (2), we obtain

$\text{PQ}-\text{PT}=\text{RS}-\text{RT}$

$\Rightarrow \text{QT}=\text{ST}$        …(5)

From equation (4) and (5) we can conclude that the corresponding segments of chords PQ and RS are congruent to each other.

3.If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. 

Ans: Let RS and PQ be two chords of equal lengths, of a given circle and they are intersect each other at point T.

Construct two perpendicular lines $\text{OV}$ and $\text{OU}$ on these chords.

In $\Delta \text{OVT}$ and $\Delta \text{OUT}$,

Since equal chords of a circle are equidistant from the centre. Therefore,

$\text{OV}=\text{OU}$

$\angle OVT=\angle OUT={{90}^{{}^\circ }}$

The side OT is present in both triangles. Therefore,

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

$\therefore \Delta \text{OVT}\cong \Delta $ OUT (By RHS axiom of congruency)

Hence $\angle \text{OTV}=\angle \text{OTU}$  

As they are corresponding parts of the corresponding triangles.

Therefore, it can be concluded from above that the line joining the point of intersection to the centre makes equal angles with the chords.

4. If a line intersects two concentric circles (circles with the same centre) with centre $\text{O}$ at $\text{A},\text{B},\text{C}$ and $D$, prove that $AB=CD$ (see figure).

Ans: Construct a perpendicular line OM on line AD. 

It can be observed that the chord of the smaller circle is BC and the chord of the bigger circle is AD.

Perpendicular drawn from the centre of the circle bisects the chord. 

\[\begin{align} & \therefore \,\,\,\,\,\,\,\,\,BM=MC\,\,\,\,\,\,\,\,\text{ }.\,..\text{ }\left( 1 \right) \\ & \text{and, }AM=MD\,\,\,\,\,\,\,\,\text{ }...\text{ }\left( 2 \right) \\ \end{align}\]

On subtracting Equation (2) from (1), we obtain

AM − BM = MD − MC 

∴ AB = CD

5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip? 

Ans: Construct two perpendicular line OA and OB on line RS and SM respectively.

$\text{AR}=\text{AS}=\frac{6}{2}=3\,\text{m}$

$\text{OR}=\text{OS}=\text{OM}=5~\text{m}$. (Radii of the same circle)

In $\Delta \text{OAR}$,

Using Pythagoras theorem,

$\text{O}{{\text{A}}^{2}}+\text{A}{{\text{R}}^{2}}=\text{O}{{\text{R}}^{2}}$

$\Rightarrow$  $\text{O}{{\text{A}}^{2}}+{{(3~\text{m})}^{2}}={{(5~\text{m})}^{2}}$

$\Rightarrow$  $\text{O}{{\text{A}}^{2}}=(25-9){{\text{m}}^{2}}=16~{{\text{m}}^{2}}$

$\Rightarrow$  $\text{OA}=4~\text{m}$

ORSM will be a kite as pair of adjacent sides are equal $(\text{OR}=\text{OM}$ and $\text{RS}=\text{SM}$ ). Since the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.

$\angle $ RCS will be of ${{90}^{{}^\circ }}$ and $\text{RC}=\text{CM}$

Area of $\Delta \text{ORS}=\frac{1}{2}\times \text{OA}\times \text{RS}$

$\Rightarrow$  $\frac{1}{2}\times \text{RC}\times \text{OS}=\frac{1}{2}\times 4\times 6$

$\Rightarrow$  $\text{RC}\times 5=24$

$\Rightarrow$  $\text{RC}=4.8$

$\text{RM}=2\text{RC}=2(4.8)=9.6$ m

Therefore, Reshma and Mandip are $9.6~\text{m}$ apart.

6. A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. 

Ans:

As all the sides of the triangle are equal. Therefore,

$\Delta ASD$ is an equilateral triangle.

OA (radius) $=20~\text{m}$

Since circumcentre(O) is the point of intersection of all the medians of equilateral triangle ASD. We also know that medians intersect each other in the ratio \[2:1\]. Since $\text{AB}$ is the median of equilateral triangle ASD, we can write it as,

$\Rightarrow \frac{\text{OA}}{\text{OB}}=\frac{2}{1}$

$\Rightarrow \frac{20~\text{m}}{\text{OB}}=\frac{2}{1}$

$\Rightarrow \text{OB}=\left( \frac{20}{2} \right)=10~\text{m}$

$\text{AB}=\text{OA}+\text{OB}=(20+10)\text{m}=30~\text{m}$

In $\Delta \text{ABD}$,

Using Pythagoras theorem,

$\text{A}{{\text{D}}^{2}}=\text{A}{{\text{B}}^{2}}+\text{B}{{\text{D}}^{2}}$

$\Rightarrow$ $\text{A}{{\text{D}}^{2}}={{(30)}^{2}}+{{\left( \frac{\text{AD}}{2} \right)}^{2}}$

$\Rightarrow$ $\text{A}{{\text{D}}^{2}}=900+\frac{1}{4}\text{A}{{\text{D}}^{2}}$

$\Rightarrow$  $\frac{3}{4}\text{A}{{\text{D}}^{2}}=900$

$\Rightarrow$  $\text{A}{{\text{D}}^{2}}=1200$

$\Rightarrow$  $\text{AD}=20\sqrt{3}\,\,\text{m}$

Therefore, the length of the string of each phone will be $20\sqrt{3}~\text{m}$.

Conclusion

NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 - Circles provides detailed explanations and solutions to all exercises in circle geometry. This exercise has a total of 6 questions with fully solved solutions. It covers important topics like tangents, angles, and circle properties in an easy-to-understand way. Students can learn how to calculate the circumference, area, and other geometric dimensions of circles. Students can use Class 9 Maths Chapter 9 Exercise 9.2 resources to understand circle properties and their real-world applications while improving their geometry knowledge.

Class 9 Maths Chapter 9: Exercises Breakdown

CBSE Class 9 Maths Chapter 9 Other Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.