Class 9 Maths Chapter 10 Circles

Exercise 10.1

Question 1.
Fill in the blanks.

1. The center of a circle lies in……………… of the circle. (exterior/interior)
2. A point, whose distance from the center of a circle is greater than its radius lies in…………… of the circle, (exterior/interior)
3. The longest chord of a circle is a…………………… of the circle.
4. An arc is a…………… when its ends are the ends of a diameter.
5. Segment of a circle is the region between an arc and………………….. of the circle.
6. A circle divides the plane, on which it lies, in…………………… parts.

Solution:

1. The center of a circle lies in interior of the circle.
2. A point, whose distance from the center of a circle is greater than its radius lies in exterior of the circle.
3. The longest chord of a circle is a diameter of the circle.
4. An arc is a semi-circle when its ends are the ends of a diameter.
5. Segment of a circle is the region between an arc and chord of the circle.
6. A circle divides the plane, on which it lies, in three parts.

Question 2.
Write True of False. Give reasons for your answers.

1. Line segment joining the center to any point on the circle is a radius of the circle.
2. A circle has only finite number of equal chords.
3. If a circle is divided into three equal arcs, each is a major arc.
4. A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
5. Sector is the region between the chord and its corresponding arc.
6. A circle is a plane figure.

Solution.

1. True. Because all points are equidistant from the center of the circle.
2. False. Because the circle has infinitely many equal chords.
3. False. Because all three arcs are equal, so there is no difference between the
   major and minor arcs.
4. True. By the definition of diameter that diameter is twice the radius.
5. False. Because the sector is the region between two radii and an arc.
6. True. Because the circle is a part of the plane figure.

 

Exercise 10.2

Question 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 centers.
Solution.

Let MN and PQ are two equal chords of two congruent circles with centers O and O’ respectively,
i.e. MN = PQ.
Δ MON and Δ PO’Q, we have
MO = PO’ [radii of congruent circles]
NO = QO’  [radii of congruent circles]
and MN   = PQ [given]
Δ MON    = Δ PO’Q [by SSS congruence rule]
So,  ∠MON    = ∠ PO’Q [by CPCT]
Hence, equal chords of congruent circles subtend equal angles at their centers.
Hence proved.

Question 2.
Prove that, if chords of congruent circles subtend equal angles at their centers, then the chords are equal.
Solution.
Given: MN and PQ are two chords of congruent circles such that angles subtended by these chords at the centers O and O’ of the circles are equal.

To prove :   MN   = PQ
Proof: In ∠MON apd ∠ PO’Q, we get
MO = PO’
NO = QO’
and ∠MON  = ∠PO’Q
∴ By SAS criteria, we get
Δ MON ≅ APO’Q
Hence Proved.

 

Exercise 10.3

Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution.
Different pairs of circles are given below :
(i) Two points are common

(ii) One point is common

(iii) No point is common

(iv) No point is common

(v) One point is common

From figures, it is obvious that these pairs may have 0 or 1 or 2 points in common. Hence, a pair of circles cannot intersect each other at more than two points.

Question 2.
Suppose, you are given a circle. Give a construction to find its center.
Solution.

Steps of Construction

• Take three points P, Q, and R on the circle.
• Join PQ and
• Draw MT and NS, which are respectively the perpendicular bisectors of PQ and RQ and intersecting each other at a point 0. Hence, 0 is the center of the circle.

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

Given: Two circles with centers O and O’ intersect at two points M and N, so that MN is the common chord of the two circles and 00′ is the line segment joining their centers. Let OO’ intersects MN at P.
To prove: OO’ is the perpendicular bisector of MN.

 

Exercise 10.4

Question 1.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.
Solution.
Let O and O’ be the centers of the circles of radii 5 cm and 3 cm, respectively. Let AB be their common chord

Given, OA = 5 cm, O’ A = 3 cm and 00′ = 4 cm
Then, AO’2 + OO’2 = 32 + 42 = 9 +16 = 25 = OA2
So, 00′ A is a right angled triangles and right angled at O’.
∴ Area of Δ OO’ A =  x O’A x OO’ =  x 3 x 4 = 6 cm2 …(i)
Also, we know that when two circles intersect at two points, then their centers lie on the perpendicular bisector of the common chord.
∴ Area of Δ OO’ A =  x OO’ x AM =  x 4 x AM=2AM ………(ii)
From eqs. (i) and (ii), we get
2AM = 6 ⇒ AM =3 cm
∴ AB = 2 x AM = 2 x 3 = 6 cm

Question 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.
Solution.

Let MN and AB are two chords of a circle with center 0. AB and MN intersect at P and MN = AB
To prove : MP = PB and PN = AP
Construction: Draw OD  ⊥ L MN and OC ⊥ AB. Join OP.

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

Let RQ and MN be two equal chords of a circle with center O.
MN and RQ intersect at P and MN = RQ.
To prove : ∠OPM = ∠OPQ

Question 4.
If a line intersects two concentric circles (circles with the same center) with center O at A, B, C and D, then prove that AB = CD (See figure).

Solution.

Let a line l intersects two concentric circles with center O at A, B, C and D.
To prove  : AB   = CD
Construction: Draw OP perpendicular from O on line l.
Proof: We know that the perpendicular drawn from the center of a circle to a chord bisects the chord. Here, BC is the chord of the smaller circle and OP ⊥ BC.
∴  BP  = PC
and AD is a chord of the larger circle and OP ⊥ AD.
∴   AP    = PD              …(ii)
On subtracting eq. (i) from eq. (ii), we get
AP -BP = PD -PC
⇒ AB = CD
Hence proved.

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

Let 0 be the center of the circle and Reshma,
Salma and Mandip are represented by the points R, S, and M, respectively.
Draw OP ⊥ RM and ON ⊥ RS.

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

Let Ankur, Syed and David be sitting on the points P, Q, and R, respectively on the boundary of circular park.
Clearly, PQ = QR = PR, as they sitting at an equal distance.
Thus, ΔPQR is an equilateral triangle.
Let  PQ = QR = PR = x m
Now, draw altitudes PC, QD and RN from vertices to .the sides of a triangle and these altitudes intersect at the center of circle M.
[∵  altitudes of equilateral triangle passes through the circumcenter of the equilateral triangle.]
As Δ PQR is an equilateral triangle, therefore these altitudes bisect their sides.

 

Exercise 10.5

Question 1.
In the figure, A, B, and C are three points on a circle with center O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution.

Here, ∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°
Since, arc ABC makes an angle of 90° at the center  the circle.
∴ ∠ADC =   ∠AOC
[since, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle]
=  x 90° = 45°

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.
Solution.

Let PQ be chord. Join OP and OQ.
It is given that PQ = OP = OQ (∵  Chord = radius
Δ OPQ is equilateral.
⇒ ∠POQ = 60°
Since arc PBQ makes reflex ∠POQ = 360 °-60 ° = 300 ° at the center of the circle and ∠PBQ at a point in the minor arc of the circle.
∠PBC =  (reflex ∠POQ)

Question 3.
In the figure, ∠PQR = 100°, where P, Q, and R are points on a circle with center O. Find ∠OPR.
Solution.

Question 4.
In figure, ∠ABC = 69° and ∠ACB = 31°. Find ∠BDC.
Solution.

Question 5.
In the figure, A, B, C, and D are four points on a circle. AC and BD intersect at a point E such that
∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Solution.


Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find q ∠ECD.

Solution.

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

Given: Diagonals NP and QM of a cyclic
quadrilateral NQPM are diameters of the circle passing through the vertices M, P, Q, and N.
To prove : Quadrilateral NQPM is a rectangle.
Here, ON = OP = OQ = OM [radii of same circle]
Then,  ON = OP =  NP
and  OM = OQ =  MQ
∴  NP = MQ
Hence, the diagonals of the quadrilateral NQPM are equal and bisect each other. So, quadrilateral NQPM is a rectangle.
Hence proved

Question 8.
If the non-parallel sides of a trapezium are equal, then prove that it is cyclic.
Solution.
Given: Non-parallel sides PS and QR of a trapezium PQRS are equal. To prove: PQRS is a cyclic trapezium.
Construction: Draw SM ⊥  PQ and RN ⊥  PQ.

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, P and Q respectively (see figure). Prove that ∠ACP = ∠QCD.

Solution.
Given: Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn which intersect the circles at A, D, P and Q, respectively

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

Given: Two circles are drawn with sides AC and
AB of Δ ABC as diameters. Both circles intersect each other at D.
To prove : D lies on BC.
Construction: Join AD.
Proof : Since, AC and AB are the diameters of the two circles.
∴  ∠ADB = 90°     …(i) [angle in semi-circle]
and  ∠ADC = 90°  …(ii) [angle in semi-circle]
On adding Eqs. (i) and (ii), we get
∠ADB + ∠ADC = 90°+ 90° = 180°
∠BDC = 180°
Hence, BDC is a straight line.
So, point of intersection D lies on the third side.
Hence proved.

Question 11.
ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution.
Since Δ ADC and Δ ABC are right angled triangles with common hypotenuse.

Draw a circle with AC as diameter passing through B and D. Join BD.
∵ Angles in the same segment are equal.
∴ ∠CBD = ∠CA

Ex 10.5 Class 9 Maths Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution.
Given : PQRS is a parallelogram inscribed in a circle.
To prove: PQRS is a rectangle.
Proof : Since, PQRS is a cyclic quadrilateral.
∠P + ∠R = 180°   …(i)
[since, sum of pair of opposite angles in a cyclic quadrilateral is 180°]
∠P=∠R  …(ii) [since, in a parallelogram, opposite angles are equal]

Exercise 10.6

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

Given Two circles with centers O and O’ which intersect each other at C and D.
To prove: ∠OCO’ = ∠ODO’
Construction: Join OC, OD, O’C and O’ D.
Proof: In ∠OCO’ and ∠ODO’, we have
OC = OD    [radii of the same circle]
O’C = O’D  [radii of the same circle]
OO’ = OO’  [common sides]
∴ Δ OCO’ = Δ ODO’  [by SSS congruence rule]
Then, ∠OCO’ = ∠ODO’  [by CPCT]

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, then find the radius of the circle.
Solution.

Let 0 be the center of the given circle and its radius be b cm.
AB = 5 cm and CD = 11 cm are two chords on opposite sides of its center.
Draw ON ⊥ AB and OM ⊥ CD.
Since, ON ⊥  AB, OM ⊥  CD, and AB||CD, therefore the points N, O and M are collinear.
Given, distance between AB and CD = MN = 6 cm
Let ON=a cm
OM=(6-a) cm

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

Let PQ and RS be two parallel chords of a circle with center O such that PQ = 6 cm and RS = 8 cm.
Let a be the radius of the circle. Draw ON ⊥ RS and OM ⊥ PQ.
Since, PQ || RS, and ON ⊥ RS and OM ⊥ PQ.
Therefore, points O, N, and M are collinear. v OM = 4 cm and M and N are the mid-points of PQ and RS, respectively.

Question 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 ∠ABC is equal to half of the difference of the angles subtended by the chords AC and DE at the center.
Solution.

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Solution.

Given: Let PQRS be a rhombus and PR and SQ are its two diagonals which bisect each other at right angles at 0.
To prove: A circle drawn on PQ as diameter will pass through O.
Construction: Through O, draw MN || PS and EF || PQ.

Ex 10.6 Class 9 Maths Question 6.
ABCD is a parallelogram. The circle through A, B, and C intersect CD (produced, if necessary) at E. Prove that AE = AD.
Solution.

Question 7.
AC and BD are chords of a circle which bisects each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.
Solution.
(i) Let BD and AC be two chords of a circle bisect at P.
In Δ ASP and  Δ  CPD, we have PA = PC

Question 8.
Bisectors of angles A, B and C of the triangle ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the Δ DEF are 90° –  A, 90° – B and 90° – C.
Solution.

Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn, so that P and Q lie on the two circles. Prove that BP = BQ.
Solution.

Question 10.
In any Δ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the Δ ABC.
Solution.
(i) Let bisector of ∠A meet the circumcircle of Δ ABC at M.
Join BM and CM.