Maths MCQ Class IX Ch-10 | Circle

      Mathematics MCQ | Class 09 | Chapter 10

CIRCLE

Multiple Choice Questions (MCQ)

• MCQ Based on the definition of circle.
• MCQ  Based on the properties of circle.
• MCQ Based on the cyclic quadrilateral.

Features

• In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations. 
• Solution Hints are also given to some difficult problems. 
• Each MCQ contains four options from which one option is correct. 

Action Plan

• First of all students should Learn and write all basic points and Formulas related to the Chapter 10 Circle.
• Start solving  the NCERT Problems with examples.
• Solve the important assignments on the Chapter 10 Class IX.
• Then start solving the following MCQ.

MCQ |Chapter 10 | Circle | Class IX

Question 1

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called
a) square
b) triangle
c) rectangle
d) circle

Answer d

Question 2

The centre of the circle lies in______ of the circle.
a) Interior
b) Exterior
c) Circumference
d) None of the above
Answer: a

Question 3

A circle divides the plane into __________ parts.
a) two
b) three
c) four
d) five
Answer b

Question 4

The longest chord of the circle is:
a) Radius
b) Arc
c) Diameter
d) Segment
Answer: c

Question 5

From the diagram given below, the shaded area is __________

a) major segment
b) minor segment
c) major sector
d) minor segment
Answer b

Question 6

How many circles can pass through three points which are not collinear?
a) One
b) Two
c) Three
d) Four
Answer a

Question 7

Equal _____ of the congruent circles subtend equal angles at the centres.

a) Segments
b) Radii
c) Arcs
d) Chords
Answer: d
Explanation: See the figure below:

Let ΔAOB and ΔCOD are two triangles inside the circle.

OA = OC and OB = OD (radii of the circle)

AB = CD (Given)

So, ΔAOB ≅ ΔCOD (SSS congruency)

∴  ∠AOB = ∠COD ……….. By CPCT

Hence, this prove the statement.

Question 8

The region between an arc and the two radii, joining the centre to the end points of the arc is called __________
a) arc
b) chord
c) sector
d) area of circle
Answer c

Question 9

When does both segments and both sectors become equal?
a) When two arcs are equal
b) When radius = 2 x chord
c) When two arcs are unequal
d) When chord = radius
Answer a

Question 10

If chords AB and CD of congruent circles subtend equal angles at their centres, then:
a) AB = CD
b) AB > CD
c) AB < AD
d) None of the above
Answer: a
Explanation:
In triangles AOB and COD,
∠AOB = ∠COD (given)
OA = OC and OB = OD (radii of the circle)
So, ΔAOB ≅ ΔCOD. (SAS congruency)
∴ AB = CD (By CPCT)

Question 11
From the diagram given below, if O is the centre of the circle and OL is perpendicular to chord AB, then which of the following is true?

a) AL > LB
b) AL < LB
c) AL ≠ LB
d) AL = LB
Answer d

Question 12

If there are two separate circles drawn apart from each other, then the maximum number of common points they have:
a) 0
b) 1
c) 2
d) 3
Answer: a

Question 13

What is the value of AC if radius of circle is 5cm and OB = 3cm?

a) 5cm

b) 8cm
c) 4cm
d) 2cm

Question 14

The angle subtended by the diameter of a semi-circle is:

a) 90^o
b) 45^o
c) 180^o
d) 60^o
Answer: c

Explanation: Angle in a semi circle is always right

Question 15

Find the value of AB if radius of circle is 10cm and PQ = 12cm and RS = 16cm?

a) 14cm
b) 16cm
c) 4cm
d) 2cm
Answer d

Question 16
If AB and CD are two chords of a circle intersecting at point E, as per the given figure. Then:

a) ∠BEQ > ∠CEQ
b) ∠BEQ = ∠CEQ
c) ∠BEQ < ∠CEQ
d) None of the above
Answer: b

Explanation:

OM = ON …………….. (Equal chords are always equidistant from the centre)

OE = OE  …………….. (Common)

∠OME = ∠ONE …… (Each = 90^o)

So, ΔOEM ≅ ΔOEN ……… (by RHS congruence rule)

Hence, ∠MEO = ∠NEO …….. (by CPCT)

∴ ∠BEQ = ∠CEQ

Question 17

Find the radius of circle if AB = 7cm, RS = 6cm and PQ = 8cm.

a) 5cm

b) 8cm
c) 4cm
d) 20cm
Answer a

Explanation:

Let r be the radius of circle.
Since perpendicular from centre to a chord bisects the chord,
AR = RS/2 = 3cm and BP = PQ/2 = 4cm
In ΔBOP, ∠OBP = 90°
By Pythagoras theorem, OP² = OB² + BP²
⇒ r² = x² + (4)²
⇒ r² = x² + 16  ——————- (i)
Similarly, In ΔAOR, ∠OAR = 90°
By Pythagoras theorem, OR² = OA² + AR²
⇒ r² = (7 – x)² + (3)²
⇒ r² = (7 – x)² + 9  ———————— (ii)
From equation i and ii, x² + 16 = (7 – x)² + 9
⇒ x² + 16 = 7² – 2 x 7 x x + x² + 9
⇒ 2 x 7 x x = 49 – 16 + 9
⇒ x = 3
Substituting value of x in equation i, r² = 3² + 16
⇒ r = 5cm.

Question 18
If a line intersects two concentric circles with centre O at A, B, C and D, then:
a) AB = CD
b) AB > CD
c) AB < CD
d) None of the above
Answer: a
Explanation: See the figure below:

From the above fig., OM ⊥ AD.

Therefore, AM = MD …….. (1)

Also, since OM ⊥ BC, OM bisects BC.

Therefore, BM = MC ……… (2)

From equation (1) and equation (2)

AM – BM = MD – MC

∴ AB = CD

Question 19
Two circles of radius 13cm and 17cm intersect at A and B. What is the distance between the centres of the circles if AB = 24cm?

a) 15cm
b) 25cm
c) 22cm
d) 14cm
Answer d

Explanation:
Since line joining the centres of the circles bisects the common chord, AC = BC = 12cm.
Also, In ΔAOC, ∠OCA = 90°
By Pythagoras theorem, OA² = OC² + AC²
⇒ 15² = OC² + 12²
⇒ OC² = 225 – 144
⇒ OC = 9cm
Similarly, In ΔAO’C, ∠O’CA = 90°
By Pythagoras theorem, O’A² = O’C² + AC²
⇒ 13² = O’C² + 12²
⇒ OC² = 169 – 144
⇒ OC = 5cm
Now, OO’ = OC + O’C
⇒ OO’ = 9 + 5 = 14cm.

Question 20

In the below figure, the value of ∠ADC is:

a) 60°
b) 30°
c) 45°
d) 55°
Answer: c

Explanation: ∠AOC = ∠AOB + ∠BOC

So, ∠AOC = 60° + 30°

∴ ∠AOC = 90°

An angle subtended by an arc at the centre of the circle is twice the angle subtended by that arc at any point on the rest part of the circle.

So, ∠ADC = 1/2∠AOC

= 1/2 × 90° = 45°

Question 21
In the given figure, find angle OPR.

a) 20°
b) 15°
c) 12°
d) 10°
Answer: d

Explanation: Major arc PR subtend Reflex ∠POR at the centre of the circle and ∠PQR in the remaining part of the circle

So, Reflex ∠POR = 2 × ∠PQR,

We know the values of angle PQR as 100°

So, Reflex ∠POR = 2 × 100° = 200°

∴ ∠POR = 360° – 200° = 160° 

Now, in ΔOPR,

OP and OR are the radii of the circle

So, OP = OR

Also, ∠OPR = ∠ORP = x

By angle sum property of triangle, we know:

∠POR + ∠OPR + ∠ORP = 180°

160° + x + x  = 180°

2x = 180° – 160°

2x = 20° ⇒ x = 10°

Question 22

In the given figure, ∠AOB = 90º and ∠ABC = 30º, then ∠CAO is equal to:

a) 30º
b) 45º
c) 60º
d) 90º
Answer: c
Explanation:

Given that ∠AOB = 90º and ∠ABC = 30º
OA = OB (Radii of the circle)
Let ∠OAB = ∠OBA = x
In the triangle OAB,
∠OAB + ∠OBA + ∠AOB = 180° (By using the angle sum property of triangle)
⇒ x + x + 90° = 180°
⇒ 2x = 180° – 90°
⇒ x = 90°/ 2 = 45°
Therefore, ∠OAB = 45° and ∠OBA = 45°
By using the theorem, “ the angles subtended by arcs at the centre of the circle double the angle subtended at the remaining part of the circle”, we can write
∠AOB = 2∠ACB
This can also be written as,
∠ACB = ½ ∠AOB = (½) × 90° = 45°
Now, apply the angle sum property of triangle on the triangle ABC,
∠ACB + ∠BAC + ∠CBA = 180°
45° + 30° + ∠BAC = 180°
75° + ∠BAC = 180° ⇒ ∠BAC = 180° - 75°
⇒ ∠BAC = 105°
∠CAO + ∠OAB = 105°
∠CAO + 45° = 105°
⇒ ∠CAO = 105° - 45° = 60°
Question 23

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to:


a) 30º
b) 40º
c) 50º
d) 80º
Answer: c

Explanation:

Given that ABCD is a cyclic quadrilateral and ∠ADC = 140º.

We know that the sum of opposite angles of a cyclic quadrilateral is 180°.

Hence, ∠ADC + ∠ABC = 180° 

140° + ∠ABC = 180°

∠ABC = 180° – 140° = 40° 

Now  ∠ACB = 90° . . . . . . .  Angles in the semi circle is right

By using the angle sum property of triangle in the triangle, ABC, 

∠CAB + ∠ABC + ∠ACB = 180°

∠CAB + 40° + 90° = 180° 

∠CAB + 130°   = 180° 

∠CAB = 180° – 130°

∠CAB = 50° 

Therefore, ∠CAB or ∠BAC =50°.

Question 24

 In the given figure, if ∠OAB = 40º, then ∠ACB is equal to

a) 40º
b) 50º
c) 60º
d) 70º
Answer: b
Explanation:

Given that ∠OAB = 40º,

In  the triangle OAB, 

OA = OB (radii)

 ∠OAB = ∠OBA  = 40°  …… (∴ angles opposite to equal sides are equal)

Now, by using the angle sum property of triangle, we can write

∠AOB + ∠OBA + ∠BAO = 180° 

∠AOB + 40° + 40° = 180° 

∠AOB = 180 – 80° = 100° 

Since, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle

∠AOB = 2 ∠ACB 

∠ACB = ½ ∠AOB

Hence, ∠ACB = 100°/2 = 50°.

Question 25

In the given figure, if ∠ABC = 20º, then ∠AOC is equal to:


a) 10º
b) 20º
c) 40º
d) 60º
Answer: c
Explanation:
Given that, ∠ABC = 20º.
since the angle subtended by an arc at the centre of the circle is double the angle subtended at the remaining part.
∠AOC = 2∠ABC
∠AOC = 2 × 20° = 40°.

Question 26
In the given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to:

a) 2 cm
b) 3 cm
c) 4 cm
d) 5 cm
Answer: a

Explanation:

From the given diagram, we can observe that OC is perpendicular to chord AB. Therefore, OC bisects the chord AB

Hence. AC = CB

Also,  AC + CB = AB

AC + CB = 8

2AC = 8 (Since, AC = CB)

AC = 8/2 = 4 cm

As, the triangle OCA is a right-angled triangle, by using Pythagoras theorem, we can write 

AO² = AC² + OC²

5² = 4² + OC²

5² − 4² = OC²

OC² = 9

OC = 3 cm

As, OD is the radius of the circle, OA = OD = 5cm

CD = OD - OC

CD = 5 - 3 = 2 cm

Hence, the value of CD is equal to 2cm.

Question 27

Find the value of x if ABCD is a cyclic quadrilateral if ∠1 : ∠2 = 3 : 6.

a) 90°
b) 45°
c) 60°
d) 20°
Answer: d

Explanation: Since ABCD is a cyclic quadrilateral, Sum of either pair of opposite angles of cyclic quadrilateral is 180°
⇒ ∠1 + ∠2 = 180°
⇒ 3k + 6k = 180° ⇒ k = 20°
Now, x = ∠1 [Exterior angle formed when one side of cyclic quadrilateral is produced is equal to the interior opposite angle]
⇒ x = 3k ⇒ x = 60°.

Question 28

If PQRS is a cyclic quadrilateral and PQ is diameter, find the value of ∠PQS.

a) 45°
b) 110°
c) 20°
d) 80°
Answer: c
Explanation: Since PQRS is a cyclic quadrilateral, Sum of either pair of opposite angles of cyclic quadrilateral is 180°. ⇒ ∠QRS + ∠QPS = 180°

 ⇒ ∠QPS = 70°  ————–(i)
Also, ∠PSQ = 90°  [Angle in a semicircle] ———-(ii)
In ΔPSQ,  ∠PSQ + ∠SPQ + ∠SQP = 180°  [Angle sum property of triangle]

⇒ 90° + 70° + ∠PQS = 180°  [from equation i and ii]
⇒ ∠PQS = 20°.

Question 29

In the given figure, BC is the diameter of the circle and ∠BAO = 60º. Then ∠ADC is equal to

a) 30º
b) 45º
c) 60º
d) 120º
Answer: c
Explanation: 

Given that ∠BAO = 60° 

Since OA = OB, ∠OBA = 60°

Then ∠ADC = 60° ………. (Angles in the same segment of a circle are equal).

 Question 30
If RS is equal to the radius of circle, find the value of ∠PTQ.

a) 90°
b) 60°
c) 45°
d) 30°
Answer: b
Explanation: Join OS, OR and RQ
From figure, ∠PRQ = 90° as angle in a semicircle is right angle.
Now, ∠QRT = 180° – ∠PRQ  [Linear Pair]
⇒ ∠QRT = 180° – 90° = 90° ————–(i)
In ΔROS, OS = OR = RS ⇒ ΔROS is an equilateral triangle.
⇒ ∠ROS = 60°
Now, ∠SQR = ½ ∠ROS  [angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference]
⇒ ∠SQR = 30° —————-(ii)
In ΔTQR, ∠TRQ + ∠QTR + ∠TQR = 180°  [Angle sum property of triangle]
⇒ 90° + ∠PQR + 30° = 180°  [from equation i and ii]
⇒ ∠PQR = 60°.
Question 31

In the given figure, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to:

a) 50º
b) 60º
c) 70º
d) 80º
Answer: c

Explanation:

∠DAB = 60º, ∠ABD = 50º

By using the angle sum property in the triangle ABD

∠DAB + ∠ABD + ∠ADB = 180º

60º + 50º + ∠ADB = 180º

∠ADB = 180º - 110º

∠ADB = 70º

∠ADB = ∠ACB ……… (Angles in the same segment of a circle are equal).

∠ACB =70º

Question 32

In the given figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to:

a) 30º
b) 45º
c) 60º
d) 90º
Answer: b
Explanation:

We know that the angle in a semi circle is = 90°

Hence, ∠ACB = 90° 

Also, given that AC = BC 

∠CAB = ∠CBA = x …..  (let) (As, the angles opposite to equal sides are also equal) 

Now, by using the angle sum property of triangle in ∆ACB, we can write

 ∠CAB + ∠ABC + ∠BCA = 180°

x + x + 90° = 180° 

2x = 180° – 90° ⇒ 2x = 90° ⇒ x = 45°

∠CAB = x = 45°

Question 33

Find the value of ∠ABC if O is the centre of circle.

a) 60°
b) 120°
c) 240°
d) 180°
Answer: a
Explanation: Reflex ∠AOC = 240°

⇒ ∠AOC = 360° – Reflex ∠AOC = 360° – 240° = 120°
Since angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference, ∠AOC = 2∠ABC
⇒ ∠ABC = ½ ∠AOC = 60°.

Question 34 
If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:

a) 6 cm
b) 8 cm
c) 10 cm
d) 12 cm
Answer: c

Explanation:

Given that AB = 12 cm, BC = 16 cm and AB is perpendicular to BC.

Hence, AC is the diameter of the circle passing through points A, B and C.

Hence, ABC is a right-angled triangle. 

Thus by using the Pythagoras theorem: 

AC² = (CB)² + (AB)² 

⇒ AC² = (16)² + (12)² 

⇒ AC² = 256 + 144 

⇒ AC² = 400 

Hence, the diameter of the circle, AC = 20 cm.

Thus, the radius of the circle is 10 cm.

Question 35

Find the value of ∠QOR if O is the centre of circle.

a) 50°
b) 65°
c) 130°
d) 30°
Answer c
Explanation: Join OP

In ΔPOQ, OP = OQ  [Radii of same circle]
⇒ ∠OPQ = ∠OQP = 20°  [Angles opposite to equal sides are equal]
In ΔPOR, OP = OR  [Radii of same circle]
⇒ ∠OPR = ∠ORP = 45°  [Angles opposite to equal sides are equal]
Now, ∠QPR = ∠OPR + ∠OPQ = 45° + 20° = 65°
Angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference ⇒ ∠ROQ = 2∠QPR = 2 x 65° = 130°.

Question 36

AD is the diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is


a) 4 cm
b) 8 cm
c) 15 cm
d) 17 cm
Answer: b

Explanation:

Given that, Diameter, AD = 34 cm.

Chord, AB = 30 cm.

Hence, the radius of the circle, OA = 17 cm

Now, consider the figure.

From the centre “O”. OM is perpendicular to the chord AB.

(i.e) OM ⊥ AM 

AM =  ½  AB 

AM =  ½ (30) = 15 cm

 Now by using the Pythagoras theorem in the right triangle AOM,

AO² = OM² + AM² 

OM² = AO²– AM² 

OM²= 17² – 15² 

OM² = 64 

OM = √64 

OM = 8 cm

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