NCERT Solutions for Class 9 Maths Chapter 10 Circles

Class 9 maths chapter 10 NCERT solutions - Exercise: 10.1

Fill in the blanks:

Q1 (i) The centre of a circle lies in _____________ of the circle. (exterior/ interior)

Answer:

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

Fill in the blanks:

Q1 (ii) A point, whose distance from the centre of a circle is greater than its radius lies in_____________ of the circle. (exterior/ interior)

Answer:

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

Fill in the blanks:

Q1 (iii) The longest chord of a circle is a ___________ of the circle.

Answer:

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

Fill in the blanks:

Q1 (iv) An arc is a ________________ when its ends are the ends of a diameter.

Answer:

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

Fill in the blanks:

Q1 (v) Segment of a circle is the region between an arc and ________________ of the circle.

Answer:

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

Fill in the blanks:

Q1 (vi) A circle divides the plane, on which it lies, in _______________ parts.

Answer:

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

Write True or False: Give reasons for your answers.

Q2 (i) Line segment joining the centre to any point on the circle is a radius of the circle.

Answer:

True. As line segment joining the centre to any point on the circle is a radius of the circle.

Write True or False: Give reasons for your answers.

Q2 (ii) A circle has only finite number of equal chords.

Answer:

False . As a circle has infinite number of equal chords.

Write True or False: Give reasons for your answers.

Q2 (iii) If a circle is divided into three equal arcs, each is a major arc.

Answer:

False. If a circle is divided into three equal arcs, each arc makes angle of 120 degrees whereas major arc makes angle greater than 180 degree at centre.

Write True or False: Give reasons for your answers.

Q2 (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

Answer:

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

Write True or False: Give reasons for your answers.

Q2 (v) Sector is the region between the chord and its corresponding arc.

Answer:

False. As the sector is the region between the radii and arc.

Write True or False: Give reasons for your answers.

Q2 (vi) A circle is a plane figure.

Answer:

True. A circle is a plane figure.

Circles class 9 NCERT solutions - Exercise: 10.2

Q1 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.

Answer:

Given: The two circles are congruent if they have the same radii.

To prove: The equal chords of congruent circles subtend equal angles at their centres i.e. $\angle$ BAC= $\angle$ QPR

Proof :

In $\triangle$ ABC and $\triangle$ PQR,

BC = QR (Given)

AB = PQ (Radii of congruent circle)

AC = PR (Radii of congruent circle)

Thus, $\triangle$ ABC $\cong$ $\triangle$ PQR (By SSS rule)

$\angle$ BAC= $\angle$ QPR (CPCT)

Q2 Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer:

Given : chords of congruent circles subtend equal angles at their centres,

To prove : BC = QR

Proof :

In $\triangle$ ABC and $\triangle$ PQR,

$\angle$ BAC= $\angle$ QPR (Given)

AB = PQ (Radii of congruent circle)

AC = PR (Radii of congruent circle)

Thus, $\triangle$ ABC $\cong$ $\triangle$ PQR (By SAS rule)

BC = QR (CPCT)

Class 9 circles ncert solutions - Exercise: 10.3

Q1 Draw different pairs of circles. How many points does each pair have in common? What ii the maximum number of common points?

Answer:

In (i) we do not have any common point.

In (ii) we have 1 common point.

In (iii) we have 1 common point.

In (iv) we have 2 common points.

The maximum number of common points is 2.

Q2 Suppose you are given a circle. Give a construction to find its centre.
Answer:

Given : Points P,Q,R lies on circle.

Construction :

1. Join PR and QR

2. Draw perpendicular bisector of PR and QR which intersects at point O.

3. Taking O as centre and OP as radius draw a circle.

4. The circle obtained is required.

Q3 If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Answer:

Given: Two circles intersect at two points.

To prove: their centres lie on the perpendicular bisector of the common chord.

Construction: Joinpoint P and Q to midpoint M of chord AB.

Proof: AB is a chord of circle C(Q,r) and QM is the bisector of chord AB.

$\therefore PM\perp AB$

Hence, $\angle PMA =90 ^\circ$

Similarly, AB is a chord of circle(Q,r' ) and QM is the bisector of chord AB.

$\therefore QM\perp AB$

Hence, $\angle QMA =90 ^\circ$

Now, $\angle QMA +\angle PMA=90 ^\circ+90 ^\circ= 180 ^\circ$

$\angle$ PMA and $\angle$ QMA are forming linear pairs so PMQ is a straight line.

Hence, P and Q lie on the perpendicular bisector of common chord AB.

NCERT Solutions for Class 9 Maths Chapter 10 Circles - Exercise: 10.4

Q1 Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points and the distance between their centres is $\small 4\hspace{1mm}cm$ . Find the length of the common chord.
Answer:

Given: Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points and the distance between their centres is $\small 4\hspace{1mm}cm$ .

To find the length of the common chord.

Construction: Join OP and draw $OM\perp AB\, \, and \, \, \, ON\perp CD.$

Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.

$\therefore PM\perp AB$

$\angle PMA=90 ^\circ$

Let, PM = x , so QM=4-x

In $\triangle$ APM, using Pythagoras theorem

$AM^2=AP^2-PM^2$ ...........................1

Also,

In $\triangle$ AQM, using Pythagoras theorem

$AM^2=AQ^2-MQ^2$ ...........................2

From 1 and 2, we get

$AP^2-PM^2=AQ^2-MQ^2$

$\Rightarrow 3^2-x^2=5^2-(4-x)^2$

$\Rightarrow 9-x^2=25-16-x^2+8x$

$\Rightarrow 9=9+8x$

$\Rightarrow 8x=0$

$\Rightarrow x=0$

Put,x=0 in equation 1

$AM^2=3^2-0^2=9$

$\Rightarrow AM=3$

$\Rightarrow AB=2AM=6$

Q2 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.
Answer:

Given: two equal chords of a circle intersect within the circle

To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP=DP.

Construction : Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$

Proof :

In $\triangle$ OMP and $\triangle$ ONP,

AP = AP (Common)

OM = ON (Equal chords of a circle are equidistant from the centre)

$\angle$ OMP = $\angle$ ONP (Both are right angled)

Thus, $\triangle$ OMP $\cong$ $\triangle$ ONP (By SAS rule)

PM = PN..........................1 (CPCT)

AB = CD ............................2(Given )

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AM = CN$ ......................3

Adding 1 and 3, we have

AM + PM = CN + PN

$\Rightarrow AP = CP$

Subtract 4 from 2, we get

AB-AP = CD - CP

$\Rightarrow PB = PD$

Q3 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.
Answer:

Given: two equal chords of a circle intersect within the circle.

To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. $\angle$ OPM= $\angle$ OPN

Proof :

Construction: Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$

In $\triangle$ OMP and $\triangle$ ONP,

AP = AP (Common)

OM = ON (Equal chords of a circle are equidistant from the centre)

$\angle$ OMP = $\angle$ ONP (Both are right-angled)

Thus, $\triangle$ OMP $\cong$ $\triangle$ ONP (By RHS rule)

$\angle$ OPM= $\angle$ OPN (CPCT)

Q4 If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that $\small AB=CD$ (see Fig. $\small 10.25$ ).

Answer:

Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.

To prove : AB = CD

Construction: Draw $OM\perp AD$

Proof :

BC is a chord of the inner circle and $OM\perp BC$

So, BM = CM .................1

(Perpendicular OM bisect BC)

Similarly,

AD is a chord of the outer circle and $OM\perp AD$

So, AM = DM .................2

(Perpendicular OM bisect AD )

Subtracting 1 from 2, we get

AM-BM = DM - CM

$\Rightarrow AB = CD$

Q5 Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius $\small 5m$ 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 $\small 6m$ each, what is the distance between Reshma and Mandip?

Answer:

Given: From the figure, R, S, M are the position of Reshma, Salma, Mandip respectively.

So, RS = SM = 6 cm

Construction : Join OR,OS,RS,RM and OM.Draw $OL\perp RS$ .

Proof:

In $\triangle$ ORS,

OS = OR and $OL\perp RS$ (by construction )

So, RL = LS = 3cm (RS = 6 cm )

In $\triangle$ OLS, by pytagoras theorem,

$OL^2=OS^2-SL^2$

$\Rightarrow OL^2=5^2-3^2=25-9=16$

$\Rightarrow OL=4$

In $\triangle$ ORK and $\triangle$ OMK,

OR = OM (Radii)

$\angle$ ROK = $\angle$ MOK (Equal chords subtend equal angle at centre)

OK = OK (Common)

$\triangle$ ORK $\cong$ $\triangle$ OMK (By SAS)

RK = MK (CPCT)

Thus, $OK\perp RM$

area of $\triangle$ ORS = $\frac{1}{2}\times RS\times OL$ ...............................1

area of $\triangle$ ORS = $\frac{1}{2}\times OS\times KR$ .............................2

From 1 and 2, we get

$\frac{1}{2}\times RS\times OL$ $=\frac{1}{2}\times OS\times KR$

$\Rightarrow RS\times OL=OS\times KR$

$\Rightarrow 6\times 4=5\times KR$

$\Rightarrow KR=4.8 cm$

Thus, $RM =2 KR=2\times 4.8 cm=9.6 cm$

Q6 A circular park of radius $\small 20m$ 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 each other. Find the length of the string of each phone.

Answer:

Given: In the figure, A, S, D are positioned Ankur, Syed and David respectively.

So, AS = SD = AD

Radius of circular park = 20 m

so, AO=SO=DO=20 m

Construction: AP $\perp$ SD

Proof :

Let AS = SD = AD = 2x cm

In $\triangle$ ASD,

AS = AD and AP $\perp$ SD

So, SP = PD = x cm

In $\triangle$ OPD, by Pythagoras,

$OP^2=OD^2-PD^2$

$\Rightarrow OP^2=20^2-x^2=400-x^2$

$\Rightarrow OP=\sqrt{400-x^2}$

In $\triangle$ APD, by Pythagoras,

$AP^2=AD^2-PD^2$

$\Rightarrow (AO+OP)^2+x^2=(2x)^2$

$\Rightarrow (20+\sqrt{400-x^2})^2+x^2=4x^2$

$\Rightarrow 400+400-x^2+40\sqrt{400-x^2}+x^2=4x^2$

$\Rightarrow 800+40\sqrt{400-x^2}=4x^2$

$\Rightarrow 200+10\sqrt{400-x^2}=x^2$

$\Rightarrow 10\sqrt{400-x^2}=x^2-200$

Squaring both sides,

$\Rightarrow 100(400-x^2)=(x^2-200)^2$

$\Rightarrow 40000-100x^2=x^4-40000-400x^2$

$\Rightarrow x^4-300x^2=0$

$\Rightarrow x^2(x^2-300)=0$

$\Rightarrow x^2=300$

$\Rightarrow x=10\sqrt{3}$

Hence, length of string of each phone $= 2x=20\sqrt{3}$ m

NCERT Solutions for Class 9 Maths Chapter 10 Circles - Exercise: 10.5

Q1 In Fig. $\small 10.36$ , A,B and C are three points on a circle with centre O such that $\small \angle BOC=30^{\circ}$ and $\small \angle AOB=60^{\circ}$ . If D is a point on the circle other than the arc ABC, find $\small \angle ADC$ .

Answer:

$\angle$ AOC = $\angle$ AOB + $\angle$ BOC= $60 ^\circ+30 ^\circ=90 ^\circ$

$\angle$ AOC = 2 $\angle$ ADC (angle subtended by an arc at the centre is double the angle subtended by it at any)

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

$\Rightarrow \angle ADC=\frac{1}{2}90 ^\circ= 45 ^\circ$

Q2 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.

Answer:

Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.

To find: ADB and $\angle$ ACB.

Solution :

In $\triangle$ OAB,

OA = AB (Given )

OA = OB (Radii of circle)

So, OA=OB=AB

$\Rightarrow$ ABC is a equilateral triangle.

So, $\angle$ AOB = $60 ^\circ$

$\angle$ AOB = 2 $\angle$ ADB

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

$\Rightarrow \angle ADB=\frac{1}{2}60 ^\circ=30$

ACBD is a cyclic quadrilateral .

So, $\angle$ ACB+ $\angle$ ADB = $180 ^\circ$

$\Rightarrow \angle ACB+30 ^\circ= 180 ^\circ$

$\Rightarrow \angle ACB= 180 ^\circ-30 ^\circ=150 ^\circ$

Q3 In Fig. $\small 10.37$ , $\small \angle PQR=100^{\circ}$ , where P, Q and R are points on a circle with centre O. Find $\small \angle OPR$ .

Answer:

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, $\angle$ PSR + $\angle$ PQR = $180 ^\circ$

$\Rightarrow \angle PSR+100 ^\circ=180 ^\circ$

$\Rightarrow \angle PSR=180 ^\circ-100 ^\circ=80 ^\circ$

Here, $\angle$ POR = 2 $\angle$ PSR

$\Rightarrow \angle POR=2\times 80 ^\circ=160 ^\circ$

In $\triangle$ OPR ,

OP=OR (Radii )

$\angle$ ORP = $\angle$ OPR (the angles opposite to equal sides)

In $\triangle$ OPR ,

$\angle$ OPR+ $\angle$ ORP+ $\angle$ POR= $180 ^\circ$

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

$\Rightarrow 2\angle OPR=180 ^\circ- 160 ^\circ$

$\Rightarrow 2\angle OPR=20 ^\circ$

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

Q4 In Fig. $\small 10.38$ , $\small \angle ABC=69^{\circ}, \angle ACB=31^{\circ},$ find $\small \angle BDC$

Answer:

In $\triangle$ ABC,

$\angle$ A+ $\angle$ ABC+ $\angle$ ACB= $180^\circ$

$\Rightarrow \angle A+69 ^\circ+31 ^\circ=180^\circ$

$\Rightarrow \angle A+100 ^\circ=180^\circ$

$\Rightarrow \angle A=180 ^\circ-100^\circ$

$\Rightarrow \angle A=80 ^\circ$

$\angle$ A = $\angle$ BDC = $80 ^\circ$ (Angles in same segment)

Q5 In Fig. $\small 10.39$ , A, B, C and D are four points on a circle. AC and BD intersect at a point E such that $\small \angle BEC=130^{\circ}$ and $\small \angle ECD=20^{\circ}$ . Find $\small \angle BAC$

.

Answer:

$\angle$ DEC+ $\angle$ BEC = $180 ^\circ$ (linear pairs)

$\Rightarrow$ $\angle$ DEC+ $130 ^\circ$ = $180 ^\circ$ ( $\angle$ BEC = $130 ^\circ$ )

$\Rightarrow$ $\angle$ DEC = $180 ^\circ$ - $130 ^\circ$

$\Rightarrow$ $\angle$ DEC = $50 ^\circ$

In $\triangle$ DEC,

$\angle$ D+ $\angle$ DEC+ $\angle$ DCE = $180 ^\circ$

$\Rightarrow \angle D+50 ^\circ+20 ^\circ= 180 ^\circ$

$\Rightarrow \angle D+70 ^\circ= 180 ^\circ$

$\Rightarrow \angle D= 180 ^\circ-70 ^\circ=110 ^\circ$

$\angle$ D = $\angle$ BAC (angles in same segment are equal )

$\angle$ BAC = $110 ^\circ$

Q6 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $\small \angle DBC=70^{\circ}$ , $\small \angle BAC$ is $\small 30^{\circ}$ , find $\small \angle BCD$ . Further, if $\small AB=BC$ , find $\small \angle ECD$ .

Answer:

$\angle BDC=\angle BAC$ (angles in the same segment are equal )

$\angle BDC= 30 ^\circ$

In $\triangle BDC,$

$\angle BCD+\angle BDC+\angle DBC= 180 ^\circ$

$\Rightarrow \angle BCD+30 ^\circ+70 ^\circ= 180 ^\circ$

$\Rightarrow \angle BCD+100 ^\circ= 180 ^\circ$

$\Rightarrow \angle BCD=180 ^\circ- 100 ^\circ=80 ^\circ$

If AB = BC ,then

$\angle BCA=\angle BAC$

$\Rightarrow \angle BCA=30 ^\circ$

Here, $\angle ECD+\angle BCE=\angle BCD$

$\Rightarrow \angle ECD+30 ^\circ=80 ^\circ$

$\Rightarrow \angle ECD=80 ^\circ-30 ^\circ=50 ^\circ$

Q7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer:

AC is the diameter of the circle.

Thus, $\angle ADC=90 ^\circ$ and $\angle ABC=90 ^\circ$ ............................1(Angle in a semi-circle is a right angle)

Similarly, BD is the diameter of the circle.

Thus, $\angle BAD=90 ^\circ$ and $\angle BCD=90 ^\circ$ ............................2(Angle in a semi-circle is a right angle)

From 1 and 2, we get

$\angle BCD=\angle ADC=\angle ABC=\angle BAD =90 ^\circ$

Hence, ABCD is a rectangle.

Q8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer:

Given: ABCD is a trapezium.

Construction: Draw AD || BE.

Proof: In quadrilateral ABED,

AB || DE (Given )

AD || BE ( By construction )

Thus, ABED is a parallelogram.

AD = BE (Opposite sides of parallelogram )

AD = BC (Given )

so, BE = BC

In $\triangle$ EBC,

BE = BC (Proved above )

Thus, $\angle C = \angle 2$ ...........1(angles opposite to equal sides )

$\angle A= \angle 1$ ...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

$\angle 1+\angle 2=180 ^\circ$ (linear pair)

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

Thus, ABED is a cyclic quadrilateral.

Q9 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 (see Fig. $\small 10.40$ ). Prove that $\small \angle ACP=\angle QCD$ .

Answer:

$\angle ABP=\angle QBD$ ................1(vertically opposite angles)

$\angle ACP=\angle ABP$ ..................2(Angles in the same segment are equal)

$\angle QBD=\angle QCD$ .................3(angles in the same segment are equal)

From 1,2,3 ,we get

$\angle ACP=\angle QCD$

Q10 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.

Answer:

Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

Proof: AB is the diameter of the circle and $\angle$ ADB is formed in a semi-circle.

$\angle$ ADB = $90 ^\circ$ ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and $\angle$ ADC is formed in a semi-circle.

$\angle$ ADC = $90 ^\circ$ ........................2(angle in a semi-circle)

From 1 and 2, we have

$\angle$ ADB+ $\angle$ ADC= $90 ^\circ$ + $90 ^\circ$ = $180 ^\circ$

$\angle$ ADB and $\angle$ ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

Q11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that
$\small \angle CAD =\angle CBD$ .

Answer:

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove : $\small \angle CAD =\angle CBD$

Proof :

Triangle ABC and ADC are on common base BC and $\angle$ BAC = $\angle$ BDC.

Thus, point A,B,C,D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)

$\angle$ CAD = $\angle$ CBD (Angles in same segment are equal)

Q12 Prove that a cyclic parallelogram is a rectangle.
Answer:

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

In cyclic quadrilateral ABCD.

$\angle A + \angle C = 180 ^\circ$ .......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

$\angle A = \angle C$ ........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

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

$\Rightarrow 2\angle A = 180 ^\circ$

$\Rightarrow \angle A = 90 ^\circ$

We know that a parallelogram with one angle right angle is a rectangle.

Hence, ABCD is a rectangle.

NCERT maths chapter 10 class 9 - Exercise: 10.6

Q1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Answer:

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : $\angle$ PAQ = $\angle$ PBQ

Proof : In $\triangle$ APQ and $\triangle$ BPQ,

PA = PB (radii of same circle)

PQ = PQ (Common)

QA = QB (radii of same circle)

So, $\triangle$ APQ $\cong$ $\triangle$ BPQ (By SSS)

$\angle$ PAQ = $\angle$ PBQ (CPCT)

Q2 Two chords AB and CD of lengths $\small 5\hspace {1mm}cm$ and $\small 11\hspace {1mm}cm$ 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 $\small 6\hspace {1mm}cm$ , find the radius of the circle.

Answer:

Given : AB = 5 cm, CD = 11 cm and AB || CD.

To find Radius (OA).

Construction: Draw $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

Proof: CD is a chord of circle and $OM \perp CD$

Thus, CM = MD = 5.5 cm (perpendicular from centre bisects chord)

and AN = NB = 2.5 cm

Let OM be x.

So, ON = 6 - x (MN = 6 cm )

In $\triangle$ OCM , using Pythagoras,

$OC ^2=CM^2+OM^2$ .............................1

and

In $\triangle$ OAN , using Pythagoras,

$OA ^2=AN^2+ON^2$ .............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$ (OC=OA =radii)

$5.5 ^2+x^2=2.5^2+(6-x)^2$

$\Rightarrow 30.25+x^2=6.25+36+x^2-12x$

$\Rightarrow 30.25-42.25=-12x$

$\Rightarrow -12=-12x$

$\Rightarrow x=1$

From 2, we get

$OC^2=5.5^2+1^2=30.25+1=31.25$

$\Rightarrow OC=\frac{5}{2}\sqrt{5} cm$

OA = OC

Thus, the radius of the circle is $\frac{5}{2}\sqrt{5} cm$

Q3 The lengths of two parallel chords of a circle are $\small 6\hspace {1mm}cm$ and $\small 8\hspace {1mm}cm$ . If the smaller chord is at distance $\small 4\hspace {1mm}cm$ from the centre, what is the distance of the other chord from the centre?

Answer:

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

Proof: CD is a chord of circle and $OM \perp CD$

Thus, CM = MD = 3 cm (perpendicular from centre bisects chord)

and AN = NB = 4 cm

Let MN be x.

So, ON = 4 - x (MN = 4 cm )

In $\triangle$ OCM , using Pythagoras,

$OC ^2=CM^2+OM^2$ .............................1

and

In $\triangle$ OAN , using Pythagoras,

$OA ^2=AN^2+ON^2$ .............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$ (OC=OA =radii)

$\Rightarrow 3 ^2+4^2=4^2+(4-x)^2$

$\Rightarrow 9+16=16+16+x^2-8x$

$\Rightarrow 9=16+x^2-8x$

$\Rightarrow x^2-8x+7=0$

$\Rightarrow x^2-7x-x+7=0$

$\Rightarrow x(x-7)-1(x-7)=0$

$\Rightarrow (x-1)(x-7)=0$

$\Rightarrow x=1,7$

So, x=1 (since $x\neq 7> OM$ )

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

Q4 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 $\small \angle ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer:

Given : AD = CE

To prove : $\angle ABC = \frac{1}{2}(\angle AOC-\angle DOE)$

Construction: Join AC and DE.

Proof :

Let $\angle$ ADC = x , $\angle$ DOE = y and $\angle$ AOD = z

So, $\angle$ EOC = z (each chord subtends equal angle at centre)

$\angle$ AOC + $\angle$ DOE + $\angle$ AOD + $\angle$ EOC = $360 ^\circ$

$\Rightarrow x+y+z+z=360 ^\circ$

$\Rightarrow x+y+2z=360 ^\circ$ .........................................1

In $\triangle$ OAD ,

OA = OD (Radii of the circle)

$\angle$ OAD = $\angle$ ODA (angles opposite to equal sides )

$\angle$ OAD + $\angle$ ODA + $\angle$ AOD = $180 ^\circ$

$\Rightarrow 2\angle OAD+z=180 ^\circ$

$\Rightarrow 2\angle OAD=180 ^\circ-z$

$\Rightarrow \angle OAD=\frac{180 ^\circ-z}{2}$

$\Rightarrow \angle OAD=90 ^\circ-\frac{z}{2}$ .............................................................2

Similarly,

$\Rightarrow \angle OCE=90 ^\circ-\frac{x}{2}$ .............................................................3

$\Rightarrow \angle OED=90 ^\circ-\frac{y}{2}$ ..............................................................4

$\angle$ ODB is exterior of triangle OAD . So,

$\angle$ ODB = $\angle$ OAD + $\angle$ ODA

$\Rightarrow \angle ODB=90 ^\circ-\frac{z}{2}+z$ (from 2)

$\Rightarrow \angle ODB=90 ^\circ+\frac{z}{2}$ .................................................................5

similarly,

$\angle$ OBE is exterior of triangle OCE . So,

$\angle$ OBE = $\angle$ OCE + $\angle$ OEC

$\Rightarrow \angle OEB=90 ^\circ-\frac{z}{2}+z$ (from 3)

$\Rightarrow \angle OEB=90 ^\circ+\frac{z}{2}$ .................................................................6

From 4,5,6 ;we get

$\angle$ BDE = $\angle$ BED = $\angle$ OEB - $\angle$ OED

$\Rightarrow \angle BDE=\angle BED=90 ^\circ+\frac{z}{2}-(90-\frac{y}{2})=\frac{y+z}{2}$

$\Rightarrow \angle BDE+\angle BED=y+z$ ..................................................7

In $\triangle$ BDE ,

$\angle$ DBE + $\angle$ BDE + $\angle$ BED = $180 ^\circ$

$\Rightarrow \angle DBE +y+z=180 ^\circ$

$\Rightarrow \angle DBE =180 ^\circ-(y+z)$

$\Rightarrow \angle ABC =180 ^\circ-(y+z)$ ...................................................8

Here, from equation 1,

$\frac{x-y}{2}=\frac{360 ^\circ-y-2x-y}{2}$

$\Rightarrow \frac{x-y}{2}=\frac{360 ^\circ-2y-2x}{2}$

$\Rightarrow \frac{x-y}{2}=180 ^\circ-y-x$ ...................................9

From 8 and 9,we have

$\angle ABC=\frac{x-y}{2}=\frac{1}{2}(\angle AOC-\angle DOE)$

Q5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Answer:

Given : ABCD is a rhombus.

To prove: the circle drawn with AB as diameter passes through the point O.

Proof :

ABCD is rhombus.

Thus, $\angle AOC = 90 ^\circ$ (diagonals of a rhombus bisect each other at $90 ^\circ$ )

So, a circle drawn AB as diameter will pass through point O.

Thus, the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Q6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that $\small AE=AD$ .

Answer:

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

To prove : AE = AD

Proof :

$\angle$ ADC = $\angle$ 3 , $\angle$ ABC = $\angle$ 4, $\angle$ ADE = $\angle$ 1 and $\angle$ AED = $\angle$ 2

$\angle 3+\angle 1=180 ^\circ$ .................1(linear pair)

$\angle 2+\angle 4=180 ^\circ$ ....................2(sum of opposite angles of cyclic quadrilateral)

$\angle$ 3 = $\angle$ 4 (oppsoite angles of parallelogram )

From 1 and 2,

$\angle$ 3+ $\angle$ 1 = $\angle$ 2 + $\angle$ 4

From 3, $\angle$ 1 = $\angle$ 2

From 4, $\triangle$ AQB, $\angle$ 1 = $\angle$ 2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

Q7 (i) AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters

Answer:

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

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA.

Proof :

In $\triangle$ ABD and $\triangle$ CDO,

AO = OC (Given )

$\angle$ AOB = $\angle$ COD (Vertically opposite angles )

BO = DO (Given )

So, $\triangle$ ABD $\cong$ $\triangle$ CDO (By SAS)

$\angle$ BAO = $\angle$ DCO (CPCT)

$\angle$ BAO and $\angle$ DCO are alternate angle and are equal .

So, AB || DC ..............1

Also AD || BC ...............2

From 1 and 2,

$\angle A+\angle C=180 ^\circ$ ......................3(sum of opposite angles)

$\angle$ A = $\angle$ C ................................4(Opposite angles of the parallelogram )

From 3 and 4,

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

$\Rightarrow 2\angle A=180 ^\circ$

$\Rightarrow \angle A=90 ^\circ$

BD is a diameter of the circle.

Similarly, AC is a diameter.

Q7 (ii) AC and BD are chords of a circle which bisect each other. Prove that ABCD is a rectangle.

Answer:

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

To prove: ABCD is a rectangle.

Construction : Join AB,BC,CD,DA.

Proof :

ABCD is a parallelogram. (proved in (i))

$\angle A=90 ^\circ$ (proved in (i))

A parallelogram with one angle $90 ^\circ$ , is a rectangle )

Thus, ABCD is rectangle.

Q8 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 $\small 90^{\circ}-\frac{1}{2}C$ , $\small 90^{\circ}-\frac{1}{2}B$ and $\small 90^{\circ}-\frac{1}{2}A$

Answer:

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

Proof :

$\angle$ 1 and $\angle$ 3 are angles in same segment.therefore,

$\angle$ 1 = $\angle$ 3 ................1(angles in same segment are equal )

and $\angle$ 2 = $\angle$ 4 ..................2

Adding 1 and 2,we have

$\angle$ 1+ $\angle$ 2= $\angle$ 3+ $\angle$ 4

$\Rightarrow \angle D=\frac{1}{2}\angle B+\frac{1}{2}\angle C$ ,

$\Rightarrow \angle D=\frac{1}{2}(\angle B+\angle C)$

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

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

$\Rightarrow \angle D=90 ^\circ-\frac{1}{2}\angle A$

Similarly, $\Rightarrow \angle E=90 ^\circ-\frac{1}{2}\angle B$ and $\angle F=90 ^\circ-\frac{1}{2}\angle C$

Q9 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 $\small BP=BQ$ .

Answer:

Given: 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.

To prove : BP = BQ

Proof :

AB is a common chord in both congruent circles.

$\therefore \angle APB = \angle AQB$

In $\triangle BPQ,$

$\angle APB = \angle AQB$

$\therefore BQ = BP$ (Sides opposite to equal of the triangle are equal )

Q10 In any triangle ABC, if the angle bisector of $\small \angle A$ and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answer:

Given :In any triangle ABC, if the angle bisector of $\small \angle A$ and perpendicular bisector of BC intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

Let $\angle$ ABD = $\angle$ 1 , $\angle$ ADC = $\angle$ 2 , $\angle$ DCB = $\angle$ 3 , $\angle$ CBD = $\angle$ 4

$\angle$ 1 and $\angle$ 3 lies in same segment.So,

$\angle$ 1 = $\angle$ 3 ..........................1(angles in same segment)

similarly, $\angle$ 2 = $\angle$ 4 ......................2

also, $\angle$ 1= $\angle$ 2 ..............3(given)

From 1,2,3 , we get

$\angle$ 3 = $\angle$ 4

Hence, BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.