NCERT Solutions for Class 9 Maths Chapter 9 Circles

NCERT Solutions for Class 9 Maths Chapter 9, "Circles," contains six exercises that cover different topics related to circles. Here is a brief overview of the types of questions dealt with in each exercise:

• Exercise 9.1: This exercise consists of two questions that are based on the tangents and the properties of tangents of a circle. The questions include finding the length of the tangent, the angle between the tangent and the radius, and the distance of the point from the centre of the circle.

• Exercise 9.2: This exercise consists of three questions that are based on the secants of a circle. The questions include finding the length of the secant, the intersection point of two secants, and the angle between the secant and the tangent.

• Exercise 9.3: This exercise consists of six questions that are based on the chords of a circle. The questions include finding the length of the chord, the angle between the chords, and the perpendicular bisector of the chord.

Access NCERT Solutions for Class 9 Maths Chapter 9 – Circles

Exercise 9.1

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Ans:

As we know that a circle is a collection of points therefore, they are equidistant from a fixed point. Now, this fixed point will be the centre of the circle and the equal distance between these points will be the radius of the circle. Hence, the shape of a circle will depend on its radius. Therefore, when we superimpose two circles of equal radius, then both the circles will cover each other. Thus, these two circles will be congruent when they have equal radius. Now, let us assume that two congruent circles have a common centre: $O$ and $O^{\prime }$, $AB$ and $CD$ are the two chords of same length.

In $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }C{O}^{\prime }D$, we can observe that 

$AB=CD$ as they are chords of the same length. 

$OA=O^{\prime }C$ as they are radii of congruent circles, 

$$OB=O^{\prime }D$$ as they are radii of congruent circles.

Therefore, $\mathrm{\Delta }AOB\cong \mathrm{\Delta }C{O}^{\prime }D$ by the SSS congruence rule. This implies $\mathrm{\angle }AOB\cong \mathrm{\angle }C{O}^{\prime }D$ By CPCT. Hence, equal chords of congruent circles subtend equal angles at their centres.

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Ans:

Let us assume that there are two congruent circles with the same radii that have centres as $O$ and $O^{\prime }$.

In $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }C{O}^{\prime }D$, 

$\mathrm{\angle }AOB=\mathrm{\angle }C{O}^{\prime }D$ (Given) 

$OA=O^{\prime }C$ as they are radii of congruent circles 

$OB=O^{\prime }D$ as they are radii of congruent circles 

Therefore,

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }C{O}^{\prime }D$ by the SSS congruence rule. 

$\Rightarrow AB=CD$ (By CPCT) 

Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Exercise 9.2

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

Ans: Let us assume that the radius of the circle which is centred at $O$ and $O^{\prime }$ be $5cm$ and $3cm$.

Therefore,

$OA=OB$

$\Rightarrow 5cm$

Similarly,

$O^{\prime }A=O^{\prime }B$

$\Rightarrow 3cm$

Now, the line segment $O{O}^{\prime }$ will be the perpendicular bisector of the chord $AB$.True. 

We know that the points on the circle are always on equal distances from the centre of the circle and hence, this equal distance is defined as the radius of the circle. This is why a line segment joining the centre to any point on the circle is a radius of the 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 us assume that $PQ$ and $RS$ are the two chords of equal length of a circle and they are intersecting at a common point $T$.

So, let us draw two perpendicular bisectors $OV$ and $OU$ on these chords.

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

We have $OV=OU$ as they are equal chords of a circle and are equidistant from the centre.

Also, $\mathrm{\angle }OVT=\mathrm{\angle }OUT$.

Therefore,

$\mathrm{\Delta }OVT\cong \mathrm{\Delta }OUT$, by the RHS congruent rule.

$\Rightarrow VT=UT$ by CPCT.

Now, we have given that –

$PQ=RS$

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

$\Rightarrow PV=RU$.

Now, let us add both the conditions as –

$PV+VT=RU+UT$

$\Rightarrow PT=RT$.

On subtracting we get –

$PQ-PT=RS-RT$

This equation indicates that a corresponding segment of the chords are congruent to each other. Hence, proved.

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 us assume that $PQ$ and $RS$ are the two chords of the same length of a circle which are intersecting at a common point $T$.

So, let us draw two perpendicular bisectors $OV$ and $OU$ on these chords.

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

We have $OV=OU$ as they are equal chords of a circle and are equidistant from the centre.

Also, $\mathrm{\angle }OVT=\mathrm{\angle }OUT$.

Therefore,

$\mathrm{\Delta }OVT\cong \mathrm{\Delta }OUT$, by the RHS congruence rule.

Therefore, we can conclude that $\mathrm{\angle }OVT=\mathrm{\angle }OUT$ by CPCT. Hence, if two equal chords of a circle intersect within the circle, then the line joining the point of intersection to the centre makes equal angles with the chords. Hence, proved.

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

Ans:

In the figure, let us draw a perpendicular $OM$ bisecting the chord $BC$ and $AD$.

We can observe from the figure that $BC<AD$.

Hence, we have –

$BM=MC$ and

$AM=MD$.

On subtracting both equations, we get –

$$AM-BM=MD-MC$$

$\Rightarrow AB=CD$.

Hence, proved.

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

Ans:

Let us assume that $OA$ and $OB$ are the two perpendiculars of $RS$ and $SM$ as shown in the figure below.

Hence, we have –

$AR=AS$

$\Rightarrow 3m$.

Also, $OR=OS=OM=5m$.

In $\mathrm{\Delta }OAR$,

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

$\Rightarrow O{A}^2=25-9$

$\Rightarrow OA=4m$.

As, from the figure we can observe that $ORSM$ is a kite. Now, we know that the diagonals of a kite are perpendicular.

Therefore,

$\mathrm{\angle }RCS=90{}^{\circ }$ and $RC=CM$.

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

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

$\Rightarrow RC=4.8$

Hence,

$RM=2RC$

$\Rightarrow RM=9.6m$.

Therefore, the distance between Reshma and Mandip will be $9.6m$.

6. A circular park of radius $20m$ 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 to each other. Find the length of the string of each phone.

Ans:

Let us draw a figure as –

From the figure, we can observe that $AS=SD=DA$.

Hence, $\mathrm{\Delta }ASD$ will be an equilateral triangle and $OA=20m$.

Now, we know that the medians of an equilateral triangle will pass through the centre. Also, the medians will intersect each other at the ratio $2:1$.

Therefore, the median $AB$ is –

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

$\Rightarrow \frac{20}{OB}=\frac{2}{1}$

$\Rightarrow OB=10m$

Hence, $AB=OA+OB$

$\Rightarrow AB=30m$.

In $\mathrm{\Delta }ABD$, we have –

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

$\Rightarrow A{D}^2=900+{\left(\frac{AD}{2}\right)}^2$

$\Rightarrow 3A{D}^2=3600$

$\Rightarrow AD=20\sqrt{3}$

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

Exercise 9.3

1. In the given figure, $A,B,$ and $C$ are three points on a circle with centre $O$ such that $\mathrm{\angle }BOC=30{}^{\circ }$ and $\mathrm{\angle }AOB=60{}^{\circ }$. If $D$ is a point on the circle other than the arc $ABC$, find $\mathrm{\angle }ADC$.

Ans:

From the figure, we can observe that –

$\mathrm{\angle }AOC=\mathrm{\angle }AOB+\mathrm{\angle }BOC$

$\Rightarrow \mathrm{\angle }AOC=60{}^{\circ }+30{}^{\circ }$

$\Rightarrow \mathrm{\angle }AOC=90{}^{\circ }$.

As, the angle subtended by the arc at the centre will be twice the angle on the remaining part. Therefore,

$\Rightarrow \mathrm{\angle }ADC=\frac{1}{2}(90{}^{\circ })$

$\Rightarrow \mathrm{\angle }ADC=45{}^{\circ }$.

2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Ans:

In $\mathrm{\Delta }OAB$,

We have –

$AB=OA=OB$ as radius.

Hence, $\mathrm{\Delta }OAB$ will be an equilateral triangle.

This implies that each interior angle of the equilateral triangle will be $60{}^{\circ }$.

$\Rightarrow \mathrm{\angle }AOB=60{}^{\circ }$

$\Rightarrow \mathrm{\angle }ACB=\frac{1}{2}\mathrm{\angle }AOB$

$\Rightarrow \frac{1}{2}(60{}^{\circ })=30{}^{\circ }$.

In quadrilateral $ACBD$,

We have –

$\mathrm{\angle }ACB+\mathrm{\angle }ADB=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }ADB=150{}^{\circ }$.

Therefore, the angle subtended by the chord on the major and minor arc will be $30{}^{\circ }$ and $150{}^{\circ }$.

3. In the given figure, $\mathrm{\angle }PQR=100{}^{\circ }$, where $P,Q,$ and $R$ are points on a circle with centre $O$. Find $\mathrm{\angle }OPR$.

Ans:

Let us assume that $PR$ is a chord of the circle and $S$ is any point on the major arc.

$PQRS$ is a cyclic quadrilateral.

Hence, we have –

$\mathrm{\angle }PQR+\mathrm{\angle }PSR=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }PSR=80{}^{\circ }$

Now, we know that the angle subtended by the arc at centre will be double the angle subtended by it.

Therefore,

$\mathrm{\angle }PQR=2\mathrm{\angle }PSR$

$\Rightarrow \mathrm{\angle }POR=160{}^{\circ }$

In $\mathrm{\Delta }POR$,

We can observe that –

$OP=PR$.

$\Rightarrow \mathrm{\angle }OPR=\mathrm{\angle }ORP$ as they are opposite angles of equal sides of a triangle.

$\Rightarrow \mathrm{\angle }OPR+\mathrm{\angle }ORP+\mathrm{\angle }POR=180{}^{\circ }$ which is the angle sum property of a triangle.

$\Rightarrow 2\mathrm{\angle }OPR+160{}^{\circ }=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }OPR=10{}^{\circ }$

Therefore, $\mathrm{\angle }OPR=10{}^{\circ }$.

4. In figure, $\mathrm{\angle }ABC=69{}^{\circ }$, $\mathrm{\angle }ACB=31{}^{\circ }$, find $\mathrm{\angle }BDC$?

Ans:

From the given figure, we have –

$\mathrm{\angle }BAC=\mathrm{\angle }BDC$.

In $\mathrm{\Delta }ABC,$

$\mathrm{\angle }BAC+\mathrm{\angle }ABC+\mathrm{\angle }ACB=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }BAC=180{}^{\circ }-69{}^{\circ }-31{}^{\circ }$

$\Rightarrow \mathrm{\angle }BAC=80{}^{\circ }$.

Therefore, we have $\mathrm{\angle }BDC=80{}^{\circ }$.

5. In the given figure, $A,B,C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\mathrm{\angle }BEC=130{}^{\circ }$ and $\mathrm{\angle }ECD=20{}^{\circ }$. Find $\mathrm{\angle }BAC$.

Ans:

From the given figure, we have –

In $\mathrm{\Delta }CDE,$

$\mathrm{\angle }CDE+\mathrm{\angle }DCE=\mathrm{\angle }CEB$

$\Rightarrow \mathrm{\angle }CDE=130{}^{\circ }-20{}^{\circ }$

$\Rightarrow \mathrm{\angle }CDE=110{}^{\circ }$.

But we know that $\mathrm{\angle }CDE=\mathrm{\angle }BAC$

Therefore,

$\mathrm{\angle }BAC=110{}^{\circ }$.

6. $ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\mathrm{\angle }DBC=70{}^{\circ }$, $\mathrm{\angle }BAC=30{}^{\circ }$, find $\mathrm{\angle }BCD$. Further, if $AB=BC$, find $\mathrm{\angle }ECD$.

Ans:

The figure will be as –

From figure, we can observe that –

$\mathrm{\angle }CBD=\mathrm{\angle }CAD$

$\Rightarrow \mathrm{\angle }CAD=70{}^{\circ }$.

$\Rightarrow \mathrm{\angle }BAD=\mathrm{\angle }BAC+\mathrm{\angle }CAD$

$\Rightarrow \mathrm{\angle }BAD=100{}^{\circ }$

Therefore, we have –

$\mathrm{\angle }BCD+\mathrm{\angle }BAD=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }BCD=80{}^{\circ }$.

Now, in $\mathrm{\Delta }ABC,$ we have –

$AB=BC$

$\Rightarrow \mathrm{\angle }BCA=\mathrm{\angle }CAB$

$\Rightarrow \mathrm{\angle }BCA=30{}^{\circ }$.

Also, we have –

$\mathrm{\angle }BCD=80{}^{\circ }$

$\Rightarrow \mathrm{\angle }BCA+\mathrm{\angle }ACD=80{}^{\circ }$

$\Rightarrow \mathrm{\angle }ACD=80{}^{\circ }-30{}^{\circ }$

$$\Rightarrow \mathrm{\angle }ACD=50{}^{\circ }$$

$$\Rightarrow \mathrm{\angle }ECD=50{}^{\circ }$$.

7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans:

Let us assume a cyclic quadrilateral $ABCD$ having diagonals $BD$ and $AC$, intersecting at a common point $O$.

$\mathrm{\angle }BAD=\frac{1}{2}\mathrm{\angle }BOD$

$\Rightarrow \mathrm{\angle }BAD=90{}^{\circ }$

Now, $\mathrm{\angle }BCD+\mathrm{\angle }BAD=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }BCD=90{}^{\circ }$.

$\mathrm{\angle }ADC=\frac{1}{2}\mathrm{\angle }AOC$

$\Rightarrow \mathrm{\angle }ADC=90{}^{\circ }$

$\Rightarrow \mathrm{\angle }ADC+\mathrm{\angle }ABC=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }ABC=90{}^{\circ }$.

Therefore, each interior angle of the quadrilateral is $90{}^{\circ }$ which implies that $ABCD$ is a rectangle.

8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Ans:

Let us assume a trapezium $ABCD$ with $AB\parallel CD$ and $BC=AD$ as shown in the figure below.

From the figure, we can observe that $AM\mathrm{\perp }CD$ and $BN\mathrm{\perp }CD$.

Therefore, in $\mathrm{\Delta }AMD$ and $\mathrm{\Delta }BNC$,

$AD=BC$.

$\Rightarrow \mathrm{\angle }AMD=\mathrm{\angle }BNC$

$AM=BN$

$\Rightarrow \mathrm{\Delta }AMD\cong \mathrm{\Delta }BNC$ by the RHS congruence rule.

Therefore, $\mathrm{\angle }ADC=\mathrm{\angle }BCD$.

$\mathrm{\angle }BAD$ and $\mathrm{\angle }ADC$ are on the same side.

Therefore, $\mathrm{\angle }BAD+\mathrm{\angle }ADC=180{}^{\circ }$

$\mathrm{\angle }BAD+\mathrm{\angle }BCD=180{}^{\circ }$

Hence, the angles are supplementary. Therefore, $ABCD$ is a cyclic quadrilateral.

9. Two circles intersect at two points $B$ and $C$. Through $B$, two-line segments $ABD$ and $PBQ$ are drawn to intersect the circles at $A$, $D$ and $P$, $Q$ respectively. Prove that $\mathrm{\angle }ACP=\mathrm{\angle }QCD$.

Ans:

Let us join the chords $AP$ and $DQ$.

Therefore,

$\mathrm{\angle }PBA=\mathrm{\angle }ACP$,

Also, $\mathrm{\angle }DBQ=\mathrm{\angle }QCD$.

Now, we know that $ABD$ and $PBQ$ are the line segments intersecting at common point $B$.

Therefore,

$\mathrm{\angle }PBA=\mathrm{\angle }DBQ$

Hence, we have $\mathrm{\angle }ACP=\mathrm{\angle }QCD$.

10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans:

Let us consider a triangle $\mathrm{\Delta }ABC$ in the figure given below –

We can observe that two circles are drawn by taking the diameters $AB$ and $AC$. We will let the points $B$ and $C$ intersect each other at a common point $D$ which does not lie on the line segment $BC$.

Therefore, after joining $AD$ we have –

$\mathrm{\angle }ADB=90{}^{\circ }$

$\Rightarrow \mathrm{\angle }ADC=90{}^{\circ }$

$\Rightarrow \mathrm{\angle }BDC=\mathrm{\angle }ADB+\mathrm{\angle }ADC$

$\Rightarrow \mathrm{\angle }BDC=180{}^{\circ }$.

Hence, we have a straight line as $BDC$. This implies that the assumption that we considered was wrong.

Therefore, the point of intersection $D$ will lie on the line segment $BC$.

11. $ABC$ and $ADC$ are two right triangles with common hypotenuse $AC$. Prove that $\mathrm{\angle }CAD=\mathrm{\angle }CBD$.

Ans:

From the figure we know that in $\mathrm{\Delta }ABC$,

$\mathrm{\angle }ABC+\mathrm{\angle }BCA+\mathrm{\angle }CAB=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }BCA+\mathrm{\angle }CAB=90{}^{\circ }$.

In $\mathrm{\Delta }ADC,$

$\mathrm{\angle }CDA+\mathrm{\angle }ACD+\mathrm{\angle }DAC=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }ACD+\mathrm{\angle }DAC=90{}^{\circ }$

After adding both the conditions, we get –

$\mathrm{\angle }BCA+\mathrm{\angle }CAB+\mathrm{\angle }ACD+\mathrm{\angle }DAC=180{}^{\circ }$

$$\Rightarrow (\mathrm{\angle }BCA+\mathrm{\angle }ACD)+(\mathrm{\angle }CAB+\mathrm{\angle }DAC)=180{}^{\circ }$$

$$\Rightarrow \mathrm{\angle }BCD+\mathrm{\angle }DAB=180{}^{\circ }$$.

Now, we know that $\mathrm{\angle }B+\mathrm{\angle }D=180{}^{\circ }$.

Therefore, we can observe from the sum of each interior angle that it is a cyclic quadrilateral.

Hence,

$\mathrm{\angle }CAD=\mathrm{\angle }CBD$.

12. Prove that a cyclic parallelogram is a rectangle.

Ans:

Let us assume a cyclic parallelogram $ABCD$ as shown in the figure below –

We have –

$\mathrm{\angle }A+\mathrm{\angle }C=180{}^{\circ }$.

Now, we know that in a parallelogram opposite angles are always equal.

Therefore,

$\mathrm{\angle }A=\mathrm{\angle }C$ and

$\mathrm{\angle }B=\mathrm{\angle }D$.

$\Rightarrow \mathrm{\angle }A+\mathrm{\angle }C=180{}^{\circ }$

$\Rightarrow \mathrm{\angle }A=90{}^{\circ }$.

Similarly,

$\Rightarrow \mathrm{\angle }B=90{}^{\circ }$.

Therefore, all the interior angles of the parallelogram are $90{}^{\circ }$ which implies it is a rectangle. Hence, proved.

Overview of Deleted Syllabus for CBSE Class 9 Maths Circles

Class 9 Maths Chapter 9: Exercises Breakdown

Conclusion 

NCERT Chapter 9 of Class 9 Maths, "Circles," covers essential concepts and theorems related to circles. Key topics include properties of chords, angles subtended by chords, and cyclic quadrilaterals. Important theorems to focus on are those involving the perpendicular from the centre of a chord, equal chords and their distances from the centre, and the unique circle passing through three non-collinear points. Understanding these concepts and practising the related exercises can enhance logical thinking and problem-solving skills. NCERT solutions offer detailed explanations and a variety of solved and unsolved problems, which are crucial for thorough preparation.

Other Study Material for CBSE Class 9 Maths Chapter 9

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.

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