NCERT Solutions for Class 9 Maths Chapter 10 - Circles

NCERT Solutions for Class 9 Maths Chapter 10, "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 10.1: This exercise consists of only two questions that are aimed at testing the understanding of the definition and properties of a circle. The questions involve finding the centre and radius of a given circle.

• Exercise 10.2: 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 10.3: 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 10.4: 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.

• Exercise 10.5: This exercise consists of twelve questions that are based on the practical applications of circles. The questions include finding the radius of a wheel, the length of the belt, the distance between the two parallel tracks, and the number of revolutions made by the wheel.

• Exercise 10.6: This exercise consists of ten questions that are based on the problems that require the application of multiple concepts of circles. The questions include finding the area of the shaded region, the length of the tangent, and the equation of the circle.

Access NCERT solutions for Class 9 Maths Chapter 10 - Circles

Exercise 10.1

1. Fill in the blanks 

(i) The centre of a circle lies in __________ of the circle. (exterior/interior)

Ans:

The centre of a circle lies in the interior of the circle.

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/interior)

Ans:

A point, whose distance from the centre of a circle is greater than its radius lies in the exterior of the circle. 

(iii) The longest chord of a circle is a __________ of the circle.

Ans:

The longest chord of a circle is the diameter of the circle.

(iv) An arc is a __________ when its ends are the ends of a diameter.

Ans:

An arc is a semi-circle when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and __________ of the circle.

Ans:

Segment of a circle is the region between an arc and the chord of the circle.

(vi) A circle divides the plane, on which it lies, in __________ parts.

Ans:

A circle divides the plane, on which it lies, in three parts.

2. Write True or False: Give reasons for your answers. 

(i) Line segment joining the centre to any point on the circle is a radius of the circle.

Ans:

(Image will be uploaded soon)

3. If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

(Image will be uploaded soon)

Ans:

Let us assume that there are two circles centred at point $O$ and $O'$, intersecting each other at the point $A$ and $B$. Now, we will join the line segment $AB$. From the given figure we can observe that $AB$ is the chord of the circle centred at $O$. Therefore, when we draw a perpendicular bisector of $AB$, then it will pass through the centre of both the circles $O$ and $O'$. Hence, we can conclude that if two circles intersect at two points, then their centres lie on the perpendicular bisector of the common chord.

Exercise 10.4

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 $O'$ be $5cm$ and $3cm$.

(Image will be uploaded soon)

Therefore,

$OA=OB$

$\Rightarrow 5cm$

Similarly,

$O'A=O'B$

$\Rightarrow 3cm$

Now, the line segment $OO'$ 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.

(i) A circle has only a finite number of equal chords.

Ans:

False.

As we know that there are infinite points on a circle. Therefore, when we draw an infinite number of chords of given length on a circle, it has an infinite number of equal chords. This is why a circle is infinite instead of a finite number of equal chords.

(ii) If a circle is divided into three equal arcs, each is a major arc.

Ans:

False. 

Now, let us assume three arcs which have the same length as \[AB\], \[BC\], and \[CA\]. Now, we can observe here that for a minor arc \[BDC\], \[CAB\] is a major arc. Hence, \[AB\], \[BC\], and \[CA\] will be the minor arcs of the circle. This is why the statement is false.

(iii) A chord of a circle, which is twice as long as its radius, is the diameter of the circle.

Ans:

True. 

Let us consider a chord \[AB\] which is twice as long as its radius. Now, we can observe from here that this chord is to be passing through the centre of the circle. Hence, the chord will be the diameter of the circle. This is why the statement is true.

(iv) Sector is the region between the chord and its corresponding arc.

Ans:

False. 

We know that a sector is a region between the arc and the two radii which will be joining the centre to its end points. Hence, sector cannot be the region between chord and its corresponding. This is why this statement is false.

(v) A circle is a plane figure.

Ans:

True. 

We know that a circle is a two-dimensional figure, and hence it can also be referred to as a plane figure. Therefore, this statement holds true.

Exercise 10.2

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 $O'$, $AB$ and $CD$ are the two chords of same length.

(Image will be uploaded soon)

In $\Delta AOB$ and $\Delta CO'D$, we can observe that 

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

$OA=O'C$ as they are radii of congruent circles, 

\[OB=O'D\] as they are radii of congruent circles.

Therefore, $\Delta AOB\cong \Delta CO'D$ by the SSS congruence rule. This implies $\angle AOB\cong \angle CO'D$ 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 $O'$.

(Image will be uploaded soon)

In $\Delta AOB$ and $\Delta CO'D$, 

$\angle AOB=\angle CO'D$ (Given) 

$OA=O'C$ as they are radii of congruent circles 

$OB=O'D$ as they are radii of congruent circles 

Therefore,

$\Delta AOB\cong \Delta CO'D$ 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 10.3

1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Ans:

Now, let us consider two circles as –

(Image will be uploaded soon)

As, we can observe that the two circles are not intersecting. Hence, they don’t have a point of intersection.

(Image will be uploaded soon)

From the above figure we can observe that the circles are intersecting each other at a common point $Y$.

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From the above figure we can observe that there is only a single point of intersection $X$.

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From the figure given above we can observe that the circles are intersecting each other at two points $G$ and $H$. Therefore, the circles have two points in common.

2. Suppose you are given a circle. Give a construction to find its centre.

Ans:

Below are the steps that will be followed to find the centre of the given circle.

Step 1: Take any two different chords $AB$ and $CD$ of the given circle and draw perpendicular bisectors of these chords. 

Step 2: Let the perpendicular bisectors intersect at point $O$. Hence, $O$ is the centre of the given circle.

Hence, $AC=CB$.

Now, we have $OO'=4cm$.

So, let us assume $OC$ be $x\ \text{cm}$. Hence, $O'C$ will be $4-x\ \text{cm}$.

Now, in $\Delta OAC$, we have by applying Pythagoras theorem –

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

$\Rightarrow 25=A{{C}^{2}}+{{x}^{2}}\ \ ......\text{(1)}$

Now, in $\Delta O'AC$,

$O'{{A}^{2}}=A{{C}^{2}}+O'{{C}^{2}}$

$\Rightarrow 9=A{{C}^{2}}+{{(4-x)}^{2}}\ \ ......\text{(2)}$

After equating both equations $(1)$ and $(2)$, we get –

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

$\Rightarrow 8x=32$

$\Rightarrow x=4$. 

Therefore, we have $OC=4cm$ and hence, the chord will pass through the centre $O'$ which will be the diameter of the smaller circle.

Thus,

$A{{C}^{2}}=25-16$

$\Rightarrow AC=3cm$.

Hence, the length of the chord $AB$ will be $2(AC)=2\times 3$

$\Rightarrow AB=6cm$.

3. 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 $T$.

(Image will be uploaded soon)

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

In $\Delta OVT$ and $\Delta OUT$,

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

Also, $\angle OVT=\angle OUT$.

Therefore,

$\Delta OVT\cong \Delta OUT$, 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.

4. 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 $T$.

(Image will be uploaded soon)

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

In $\Delta OVT$ and $\Delta OUT$,

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

Also, $\angle OVT=\angle OUT$.

Therefore,

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

Therefore, we can conclude that $\angle OVT=\angle OUT$ 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.

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

(Image will be uploaded soon)

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.

6. 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.

(Image will be uploaded soon)

Hence, we have –

$AR=AS$

$\Rightarrow 3m$.

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

In $\Delta OAR$,

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

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

$\Rightarrow OA=4m$.

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

Therefore,

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

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

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

$\Rightarrow RC=4.8$

Hence,

$RM=2RC$

$\Rightarrow RM=9.6m$.

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

7. 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 –

(Image will be uploaded soon)

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

Hence, $\Delta ASD$ 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 $\Delta ABD$, 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 10.5

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

(Image will be uploaded soon)

Ans:

From the figure, we can observe that –

$\angle AOC=\angle AOB+\angle BOC$

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

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

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

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

$\Rightarrow \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 $\Delta OAB$,

(Image will be uploaded soon)

We have –

$AB=OA=OB$ as radius.

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

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

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

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

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

In quadrilateral $ACBD$,

We have –

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

$\Rightarrow \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, $\angle PQR=100{}^\circ $, where $P,Q,$ and $R$ are points on a circle with centre $O$. Find $\angle OPR$.

(Image will be uploaded soon)

Ans:

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

(Image will be uploaded soon)

$PQRS$ is a cyclic quadrilateral.

Hence, we have –

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

$\Rightarrow \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,

$\angle PQR=2\angle PSR$

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

In $\Delta POR$,

We can observe that –

$OP=PR$.

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

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

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

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

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

4. In fig. 10.38, $\angle ABC=69{}^\circ $, $\angle ACB=31{}^\circ $, find $\angle BDC$?

(Image will be uploaded soon)

Ans:

From the given figure, we have –

$\angle BAC=\angle BDC$.

In $\Delta ABC,$

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

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

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

Therefore, we have $\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 $\angle BEC=130{}^\circ $ and $\angle ECD=20{}^\circ $. Find $\angle BAC$.

(Image will be uploaded soon)

Ans:

From the given figure, we have –

In $\Delta CDE,$

$\angle CDE+\angle DCE=\angle CEB$

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

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

But we know that $\angle CDE=\angle BAC$

Therefore,

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

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

Ans:

The figure will be as –

(Image will be uploaded soon)

From figure, we can observe that –

$\angle CBD=\angle CAD$

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

$\Rightarrow \angle BAD=\angle BAC+\angle CAD$

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

Therefore, we have –

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

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

Now, in $\Delta ABC,$ we have –

$AB=BC$

$\Rightarrow \angle BCA=\angle CAB$

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

Also, we have –

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

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

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

\[\Rightarrow \angle ACD=50{}^\circ \]

\[\Rightarrow \angle ECD=50{}^\circ \].

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 $O$.

(Image will be uploaded soon)

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

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

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

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

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

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

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

$\Rightarrow \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.

(Image will be uploaded soon)

From the figure, we can observe that $AM\bot CD$ and $BN\bot CD$.

Therefore, in $\Delta AMD$ and $\Delta BNC$,

$AD=BC$.

$\Rightarrow \angle AMD=\angle BNC$

$AM=BN$

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

Therefore, $\angle ADC=\angle BCD$.

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

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

$\angle BAD+\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 $\angle ACP=\angle QCD$.

(Image will be uploaded soon)

Ans:

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

Therefore,

$\angle PBA=\angle ACP$,

Also, $\angle DBQ=\angle QCD$.

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

Therefore,

$\angle PBA=\angle DBQ$

Hence, we have $\angle ACP=\angle QCD$.

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 $\Delta ABC$ in the figure given below –

(Image will be uploaded soon)

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 –

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

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

$\Rightarrow \angle BDC=\angle ADB+\angle ADC$

$\Rightarrow \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 $\angle CAD=\angle CBD$.

Ans:

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

(Image will be uploaded soon)

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

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

In $\Delta ADC,$

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

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

After adding both the conditions, we get –

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

\[\Rightarrow \left( \angle BCA+\angle ACD \right)+\left( \angle CAB+\angle DAC \right)=180{}^\circ \]

\[\Rightarrow \angle BCD+\angle DAB=180{}^\circ \].

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

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

Hence,

$\angle CAD=\angle CBD$.

12. Prove that a cyclic parallelogram is a rectangle.

Ans:

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

(Image will be uploaded soon)

We have –

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

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

Therefore,

$\angle A=\angle C$ and

$\angle B=\angle D$.

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

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

Similarly,

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

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

Exercise 10.6

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

Ans:

Let us assume that two circles which have centre as $O$ and $O'$ are intersecting each other at points $A$ and $B$ as shown in the figure –

(Image will be uploaded soon)

Now, let us join both the centres.

Hence, in $\Delta AOO'$ and $\Delta BOO'$ we have –

$OA=OB$

$\Rightarrow O'A=O'B$

$\Rightarrow \Delta AOO'\cong \Delta BOO'$

Therefore,

$\angle OAO'=\angle OBO'$.

Hence, proved that 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 $5cm$, $11cm$ 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 $6cm$, find the radius of the circle.

Ans:

Let us draw two perpendicular bisectors as $OM\bot AB$ and $ON\bot CD$. So, let us join both the points $OB$ and $OD$ as shown in the figure below –

(Image will be uploaded soon)

Therefore,

$BM=\frac{AB}{2}$

$\Rightarrow BM=\frac{5}{2}$

$ND=\frac{CD}{2}$

$\Rightarrow ND=\frac{11}{2}$.

Now, let us take $ON$ as $x$. Therefore, $OM$ will become $6-x$.

In $\Delta MOB,$ we have –

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

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

$\Rightarrow 36+{{x}^{2}}-12x+\frac{25}{4}=O{{B}^{2}}$.

Similarly, in $\Delta NOD,$ we have –

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

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

$\Rightarrow {{x}^{2}}+\frac{121}{4}=O{{D}^{2}}$

From the figure we have $OD=OB$.

Therefore, $36+{{x}^{2}}-12x+\frac{25}{4}={{x}^{2}}+\frac{121}{4}$

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

$\Rightarrow x=1$.

Hence,

$1+\frac{121}{4}=O{{D}^{2}}$

$OD=\frac{5}{2}\sqrt{5}$

Therefore, the radius will be $\frac{5}{2}\sqrt{5}cm$.

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

Ans:

Let us assume that $AB$ and $CD$ are the two parallel chords in a circle which are centred at $O$. So, we will join $OB$ and $OD$. Hence, the distance of smaller chord $AB$ from the centre of the circle will be of length $4cm$ as shown in the figure below –

(Image will be uploaded soon)

Therefore, we have –

$MB=\frac{AB}{2}$

$\Rightarrow MB=3cm$.

Now, in $\Delta OMB,$

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

$\Rightarrow {{4}^{2}}+{{3}^{2}}=O{{B}^{2}}$

$\Rightarrow OB=5cm$.

Hence, in $\Delta OND,$ we have –

$OD=OB=5cm$.

$\Rightarrow ND=\frac{CD}{2}$

$\Rightarrow ND=4cm$.

Therefore,

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

$\Rightarrow O{{N}^{2}}={{5}^{2}}-{{4}^{2}}$

$\Rightarrow ON=3cm$.

Therefore, we have the distance of the bigger chord from the centre of the circle as $3cm$.

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:

From the figure we have –

(Image will be uploaded soon)

In $\Delta AOD$ and $\Delta COE$,

$OA=OC$

$\Rightarrow OD=OE$

$\Rightarrow AD=CE$.

$\Rightarrow \Delta AOD\cong \Delta COE$ by the SSS congruence rule.

Hence,

$\angle OAD=\angle OCE$

Similarly, $\angle ODA=\angle OEC$.

But also,

$\angle OAD=\angle ODA$

Therefore, we have –

$\angle OAD=\angle OCE=\angle ODA=\angle OEC$.

Now, let the above equality be equal to an unknown variable $x$.

So, in $\Delta ODE,$ we have –

$OD=OE$

$\Rightarrow \angle OED=\angle ODE$

This implies $ADEC$ is a cyclic quadrilateral.

$\Rightarrow \angle CAD+\angle DEC=180{}^\circ $

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

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

Thus, $\angle DOE=180{}^\circ -2y$.

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

This implies –

$\angle DOE-\angle AOC=2a-2y$

$\Rightarrow \angle DOE-\angle AOC=2a-2(180{}^\circ -2x-a)$

$\Rightarrow \angle DOE-\angle AOC=4a+4x-360{}^\circ $

Now, $\angle BAC+\angle CAD=180{}^\circ $

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

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

Therefore, in $\Delta ABC,$ we have –

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

$\Rightarrow \angle ABC=180{}^\circ -(180{}^\circ -a-x)-(180{}^\circ -a-x)$

$\Rightarrow \angle ABC=2a+2x-180{}^\circ $

$\Rightarrow \angle ABC=\frac{1}{2}(4a+4x-360{}^\circ )$

$\Rightarrow \angle ABC=\frac{1}{2}[\angle DOE-\angle AOC]$.

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.

Ans:

Let us assume that $ABCD$ is a rhombus where all its diagonals will be intersecting at a common point $O$ and a circle is being drawn with assuming its diagonal as the diameter of the circle as shown in the figure –

(Image will be uploaded soon)

Now, we know that a diameter will subtend a $90{}^\circ $ on the arc.

Therefore, we have $\angle COD=90{}^\circ $.

Now, we know that in a rhombus, its each diagonal will intersect with a making angle of $90{}^\circ $.

Hence,

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

From the above point mentioned, we have made it clear that the point $O$ lies 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:

From the figure given below, we can observe that $ABCD$ is a cyclic quadrilateral. Hence, the sum of all the opposite angles will be $180{}^\circ $.

(Image will be uploaded soon)

Therefore, we have –

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

$\Rightarrow \angle AEC+\angle AED=180{}^\circ $

$\Rightarrow \angle AED=\angle CBA$

As, opposite sides of a parallelogram are equal and so are their angles, therefore,

$\angle ADE=\angle CBA$

$\Rightarrow \angle AED=\angle ADE$

$\Rightarrow AD=AE$.

Hence, proved.

7. $AC$ and $BD$ are chords of a circle which bisect each other. Prove that 

(i) $AC$ and $BD$ are diameters.

Ans:

Let there be two chords $AB$ and $CD$ such that they are intersecting at a common point $O$ as shown in the figure below –

(Image will be uploaded soon)

Therefore,

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

We have $OA=OC$

$\Rightarrow OB=OD$

\[\Rightarrow \angle AOB=\angle COD\]

\[\Rightarrow \Delta AOB\cong \Delta COD\]

\[\Rightarrow AB=CD\]

Similarly, we can prove that \[\Delta AOD\cong \Delta COB\]

\[\Rightarrow AD=CB\].

Since it is a quadrilateral, opposite sides will be equal and \[ACBD\] will be a parallelogram which will have opposite angles equal.

Hence,

\[\angle A=\angle C\].

And as we proved earlier, here \[\angle A=90{}^\circ \]

Therefore, \[BD\] and \[AC\] should be the diameter of the circle.

(ii) $ABCD$ is a rectangle.

Ans:

As, \[\angle A=90{}^\circ \]

\[\Rightarrow \angle C=90{}^\circ \]

Hence, it will form a rectangle having each angle of \[90{}^\circ \]. Therefore, $ABCD$ will be a rectangle.

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 $90{}^\circ -\frac{1}{2}A$, $90{}^\circ -\frac{1}{2}B$, and $90{}^\circ -\frac{1}{2}C$.

Ans:

From the figure given below, we can observe that $BE$ is the bisector of $\angle B$.

(Image will be uploaded soon)

Therefore, we have –

$\angle ABE=\frac{\angle B}{2}$.

But $\angle ADE=\angle ABE$,

$\Rightarrow \angle ADE=\frac{\angle B}{2}$.

Also, $\angle ACF=\angle ADF=\frac{\angle C}{2}$

$\Rightarrow \angle D=\angle ADE+\angle ADF$

$\Rightarrow \angle D=\frac{\angle B}{2}+\frac{\angle C}{2}$

$\Rightarrow \angle D=\frac{1}{2}(180{}^\circ -\angle A)$

$\Rightarrow \angle D=90{}^\circ -\frac{\angle A}{2}$

In the similar manner, we have $\angle E=90{}^\circ -\frac{\angle B}{2}$ and

$\angle F=90{}^\circ -\frac{\angle C}{2}$.

Hence, proved.

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:

From the figure given below, we can observe that $AB$ is the common chord for both the congruent circles.

(Image will be uploaded soon)

Hence, $\angle APB=\angle AQB$

In $\Delta BPQ$,

$\angle APB=\angle AQB$

$\Rightarrow BQ=BP$ as the sides of the opposite equal angles are equal.

Hence, proved.

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

Ans:

Let us consider a perpendicular bisector of side $BC$ and angle bisector of $\angle A$ both intersect at a common point $D$. And let us consider the perpendicular bisector of side $BC$ intersect with it at point $E$ as given in the figure below –

(Image will be uploaded soon)

Now, we know that $\angle BAC$ are the angles subtended by arc $BC$.

Therefore, $\angle BOC=2\angle BAC$

$\angle BOC=2\angle A$.

In \[\Delta BOE\] and \[\Delta COE\],

\[OB=OC\],

\[\Rightarrow \angle OEB=\angle OEC\]

\[\Rightarrow \Delta BOE\cong \Delta COE\]

\[\Rightarrow \angle BOE=\angle COE\]

Also, \[\angle BOE+\angle COE=\angle BOC\]

\[\Rightarrow \angle BOE+\angle BOE=2\angle A\]

\[\Rightarrow \angle BOE=\angle A\]

\[\Rightarrow \angle BOE=\angle COE=\angle A\]

Therefore, the perpendicular bisector of side $BC$ and $\angle A$ intersect at $D$.

\[\Rightarrow \angle BOD=\angle BOE=\angle A\]

As, from the figure we have $AD$ as the bisector of $\angle A$.

Hence, $\angle BAD=\frac{\angle A}{2}$

\[\Rightarrow 2\angle BAD=\angle A\].

From both the conditions, we have –

\[\angle BOD=2\angle BAD\].

Now, we know that this will be possible only when the point $BD$ will also be a chord of the circle. Therefore, the perpendicular bisector of the side $BC$ and $\angle A$ intersect on the circumcircle of triangle $ABC$. Hence, proved.

NCERT Solutions Class 9 Maths Chapter 10 Circles Free PDF

You can opt for Chapter 10 - Circles NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid’s Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Herons formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

The students have another opportunity to get the entire NCERT Solutions Class 9 Maths Chapter 10 Circles in the PDF format and can download them from the official website of Vedantu for free. This helps the students to get rid of bothering about internet connections. This can be stored as a soft copy as well as a hard copy for Future. Well-experienced mathematics teachers prepared these materials as per the latest curriculum.

We Cover all the Exercises in the Chapter Given Below:

Exercise 10.1 - 2 Questions with Solutions in PDF.

Exercise 10.2 - 2 Questions with Solutions in PDF.

Exercise 10.3 - 3 Questions with Solutions in PDF.

Exercise 10.4 - 6 Questions with Solutions in PDF.

Exercise 10.5 - 12 Questions with Solutions in PDF.

Exercise 10.6 - 10 Questions with Solutions in PDF.

Maths NCERT Solutions Class 9 Chapter 10

10.1 Introduction

The class 9 maths chapter 10  explains to all the students that the circle is a round shape that we can find many things in our daily life. To understand the figure of a circle, we can see many things like a wall clock, wheels of the vehicle, buttons of your shirt, fruits like oranges, coins, CDs, etc. circles solutions class 9 explains more about circles, it's terminology,y, and their properties, etc.

10.2 Circles and It's Related Terms - Review

In this section, NCERT Solutions of Chapter 10 Maths Class 9 defines circle as, The collection of every point in a plane, which is at a fixed distance from a fixed point in the plane, is known as a circle.

The entire circle is divided into two - inside of the circle is the interior region and outside of the circle is the exterior region. If you're lying down inside the circle by touching two points of its surface, it is called a chord. If the chord cuts the second into two halves exactly, then it is known as diameter. The longest chord in the circle is equal to the diameter. Another related term to the circle is the sector.

Exercise 10.1 has six fill in the blanks and six true or false.

10.3 Angle Subtended by a Chord at a Point

Students can refer to class 9 chapter 10 maths to get a better understanding of this section. Students are asked to draw a card in a circle. Then extend that line to another point which joins two line segments. It forms a triangle inside the circle. This is known as an angle subtended by a chord at a point. Based on these two theorems were explained in the PDF. If the lengths of chords are the same, then their angles are the same and vice versa.

Exercise 10.2 has two theorems to prove.

10.4 Perpendicular From the Centre of a Chord

In this section, students can learn about the perpendicular drawn from the centre of the chord by making an activity. Here also two theorems were given.

• The perpendicular drawn from the centre can bisect the chord.

• The line is drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

10.5 Circles Through Three Points

Circles Class 9 NCERT Solutions

Chapter 10 Class 9 specifies the students to learn about the number of circles that can be drawn from a single point. They have already learned the basics of this concept in the 6th class. Here a slight difference to this concept is, if the circles can be drawn from multiple points, these points are known as collinear points centre.

• If we have more than two collinear points, then there is no chance to get more than a single circle.

Exercise 10.3 with three short answer type questions and one among them is a theorem.

10.6 Equal Chords and Their Distance From the Centre

In this section, the NCERT Solutions for Class 9 Maths Chapter 10 defines. The length of the perpendicular is between the point to a line and the distance of the line. And if the point lies on the line, the distance of the line from the point is zero.

• If the lengths of chords are the same, then their distance is also the same from the centre.

• Chords equidistant from the centre of a circle are equal in length.

These are the solved theorems for students to understand.

Exercise 10.4 with six long answer type questions.

10.7 Angle Subtended by the Area of Circle

Chapter 10 Maths Class 9 NCERT Solutions completely deals with circles. If we draw two congruent arcs from the centre, they must be equal.

• The  angle subtended by it at any point on the remaining part of the circle is twice of the angle subtended.

• In a circle, it is proved that the angles lie on a segment may not differ.

10.8 Cyclic Quadrilaterals

The NCERT Solutions for Class 9 Maths Chapter 10 let the students know that, if all the vertices of a quadrilateral lie in Circle, then it is called the cyclic quadrilateral.

Exercise 10.5 with 12 Long answer questions.

Exercise 10.6 with 10 questions.

Key Takeaways of NCERT Solutions Class 9 Maths Chapter 10 Free PDF

The circles class 9 questions with solutions allow the students to take several benefits. They are:

• Students can score better marks and gain good knowledge.

• These PDFs are available for free.

• Students can take either a soft copy or hard copy.