NCERT Solutions for Class 9 Maths Chapter 9 Circles

Question 1: 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 ON, draw $OM\perp AB\

Proof: AB is a chord of the circle, and OM is the bisector of chord AB.
$\therefore OM\perp AB$
$\angle OMA = 90 ^\circ$
Let, OM = x , so O'M = 4 - x
In $\triangle$ AOM, using Pythagoras' theorem
$AM^2=AO^2 - OM^2$ ...........................(1)
Also,
In $\triangle$ AO'M, using Pythagoras' theorem
$AM^2=AO'^2 - MO'^2$ ...........................(2)
From (1) and (2), we get
$AO^2 - OM^2 = AO'^2 - MO'^2$
$\Rightarrow 5^2 - x^2 = 3^2 - (4 - x)^2$
$\Rightarrow 25 - x^2 = 9 - 16 - x^2 + 8x$
$\Rightarrow 32 = 8x$
$\Rightarrow x = 4$
Put x = 4 in equation (1)
$AM^2 = 5^2 - 4^2 = 9$
$\Rightarrow AM = 3$
$\Rightarrow AB = 2AM = 6$

Question 2: If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the 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,
OP = OP (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 equations (1) and (3), we have
AM + PM = CN + PN
$\Rightarrow AP = CP$ .................(4)
Subtract equation (4) from (2), we get
$ AB - AP = CD - CP$
$\Rightarrow PB = PD$

Question 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.

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,
OP = OP (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)

Question 4: 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 $OM\perp AD$

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 equation (1) from (2), we get
$ AM-BM = DM - CM$
$\Rightarrow AB = CD$

Question 5: 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, and M are the positions of Reshma, Salma, and Mandip, respectively.
So, RS = SM = 6 cm
Construction: Join ON, OS, OR 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 Pythagoras' theorem,
$OL^2 = OS^2 - SL^2$
$\Rightarrow OL^2 = 5^2 - 3^2 = 25 - 9 = 16$
$\Rightarrow OL = 4$
In $\triangle$ ORN and $\triangle$ OMN,
OR = OM (Radii)
$\angle$ RON = $\angle$ MON (Equal chords subtend equal angles at the centre)
ON = ON (Common)
$\triangle$ ORN $\cong$ $\triangle$ OMN (By SAS)
RN = MN (CPCT)
Thus, $ON\perp RM$
Area of $\triangle$ ORS = $\frac{1}{2}\times RS\times OL$ ..................(1)
Area of $\triangle$ ORS = $\frac{1}{2}\times OS\times NR$ ..................(2)
From 1 and 2, we get
$\frac{1}{2}\times RS\times OL$ $ = \frac{1}{2}\times OS\times NR$
$\Rightarrow RS\times OL= OS\times KR$
$\Rightarrow 6\times 4 = 5\times KR$
$\Rightarrow NR = 4.8 cm$
Thus, $RM =2 NR = 2\times 4.8 cm = 9.6 cm$

Question 6: 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, B, and C are positioned as Ankur, Syed and David, respectively.
So, AB = BC = CD
Radius of circular park = 20m
So, AO = OB = OC = 20m
Construction: AF $\perp$ BC
Proof:

Question 1: 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)

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

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

Question 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.

Answer:

Given: A chord of a circle is equal to the radius of the circle, i.e. OA = AB.
To find: $\angle$ADB and $\angle$ ACB.

In $\triangle$ OAB,OA = AB (Given)

OA = OB (Radii of circle)

So, OA = OB = AB

$\Rightarrow$ AOB is an 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 = $180 ^\circ$

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

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

Question 3: 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 PO and OR.
PQSR 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 = $180 ^\circ$

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

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

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

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

Question 4: 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)

Question 5: 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 the same segment are equal )

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

Question 6: 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, $\angle ECD+\angle BCE=\angle BCD$

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

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

Question 7: If the 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.

Question 8: 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 a 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.

Question 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 (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

$\angle ACP=\angle QCD$

Question 10: If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies 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.

Question 11: 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 a common base BC and $\angle$ BAC = $\angle$ BDC.

Thus, points A, B, C, and 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 the line containing the line segment, four points lie on the circle.)

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

Question 12: 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 a right angle is a rectangle.

Hence, ABCD is a rectangle.