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CBSE Class 9  >  Class 9 Test  >  Mathematics (Maths)   >  Test: Circles- 1 - Class 9 MCQ

Circles- 1 - Free MCQ Practice Test with solutions, Class 9 Maths


MCQ Practice Test & Solutions: Test: Circles- 1 (25 Questions)

You can prepare effectively for Class 9 Mathematics (Maths) Class 9 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Circles- 1". These 25 questions have been designed by the experts with the latest curriculum of Class 9 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 25 minutes
  • - Number of Questions: 25

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Test: Circles- 1 - Question 1
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In the given figure, ∠BPC = 19o, arc AB = arc BC = arc CD. Then, the measure of ∠APD is

Detailed Solution: Question 1

If arc AB = arc BC = arc CD
then , angle suspended on circle is also equal
So, Sum of all angle is equal to angle APD
Angle APB+ BPC + CPD = angle APD
19 + 19 +19 = 57o

Test: Circles- 1 - Question 2
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ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. If ∠BEF = 80o, then ∠ABC is equal to

Detailed Solution: Question 2

Given, ABCD is a parallelogram and AEFD is a cyclic quadrilateral.
∠BEF=80∘
Now, ∠ADC=∠BEF=80∘ (Angle of a cyclic quadrilateral is equal to the opposite exterior angle )
Also, now in parallelogram ABCD,
∠ABC=∠ADC=80∘ (Opposite angles of a parallelogram are equal)

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Test: Circles- 1 - Question 3
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In the given figure, ABCD is a quadrilateral inscribed in circle with centre O. CD is produced to E. If ∠ADE = 95o and ∠OBA = 30o, then ∠OAC is equal to

Detailed Solution: Question 3

Given:

  • ABCD is a quadrilateral inscribed in a circle with centre O.
  • CD is produced to E.
  • ∠ADE = 95°
  • ∠OBA = 30°
  • We need to find ∠OAC.

Solution:

1. Since CD is produced to E, angles ∠ADC and ∠ADE are a linear pair:

∠ADC + ∠ADE = 180°

⇒ ∠ADC + 95° = 180°

⇒ ∠ADC = 180° - 95° = 85°

2. Quadrilateral ABCD is cyclic, so the arc ABC subtends angles ∠AOC at the centre and ∠ADC at the circumference.

According to the circle theorem:

∠AOC = 2 × ∠ADC = 2 × 85° = 170°

3. In triangle OAC, points O, A, and C form a triangle with O as the centre of the circle.

Since OA and OC are radii of the circle, OA = OC, so triangle OAC is isosceles.

Sum of angles in triangle OAC:

∠OAC + ∠OCA + ∠AOC = 180°

Let ∠OAC = ∠OCA = x (since OA = OC)

⇒ x + x + 170° = 180°

⇒ 2x = 180° - 170° = 10°

⇒ x = 10° / 2 = 5°

Therefore, ∠OAC = 5°.

Test: Circles- 1 - Question 4
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The given figures show two congruent circles with centre O and O’. Arc AXB subtends an angle of 75o at the centre and arc A’YB’ subtends an angle of 25o at the centre O’. Then, the ratio of arcs AXB to A ‘YB’ is

Detailed Solution: Question 4

Because the circles are congruent, therefore their radius are same

Length of arc AXB = [θ/(360°)] × 2πr = [(75°)/(360°)] × 2πr

Length of arc A’YB’ = [θ/(360°)] × 2πr = [(25°)/(360°)] × 2πr

Required Ratio = ([(75°)/(360°)] × 2πr) : ([(25°)/(360°)] × 2πr) = 3 : 1

Test: Circles- 1 - Question 5
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If a chord of a circle is equal to its radius, then the angle subtended by this chord in major segment is

Detailed Solution: Question 5

Let AB be the chord of the circle with center O

.

Given that AB = Radius of the circle.

Also, AO = BO = Radius

∴ ΔOAB is an equilateral triangle.

Thus, ∠AOB = ∠OBA = ∠OAB = 60°

Also, angle subtended by an arc at the center of the circle is twice the angle subtended by it at any other point in the remaining part of the circle.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = 1/2 (∠AOB)

⇒ ∠ACB = 1/2 (60°) = 30°

Topic in NCERT: Angle Subtended by a Chord at a Point

Line in NCERT: "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."

Test: Circles- 1 - Question 6
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The given figure shows two intersecting circles. If ∠ABC = 75o, then the measure of ∠PAD is

Detailed Solution: Question 6

In the given figure, the points A, B, C, and D form a cyclic quadrilateral because they lie on the circumferences of the two intersecting circles.

The opposite angles of a cyclic quadrilateral are supplementary, i.e.,

It is given that ∠ABC=75∘

Using the property of cyclic quadrilaterals:

Test: Circles- 1 - Question 7
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In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75o, ∠ABD = 58o and ∠ADC = 77o , AC and BD intersect at P. the measure of ∠DPC is

Detailed Solution: Question 7

∠DBA = ∠DCA = 58° …(1)

[Angles in same segment]  ABCD is a cyclic quadrilateral : 

Sum of opposite angles = 180 degrees 

∠A +∠C = 180° 

75° + ∠C = 180° 

∠C = 105° 

Again, 

∠ACB + ∠ACD = 105° 

∠ACB + 58° = 105° 

or ∠ACB = 47° …(2) 

Now,  ∠ACB = ∠ADB = 47° [Angles in same segment] 

Also, ∠D = 77° (Given) 

Again From figure, ∠BDC + ∠ADB = 77° 

∠BDC + 47° = 77° 

∠BDC = 30° 

In triangle DPC 

∠PDC + ∠DCP + ∠DPC = 180° 

30° + 58° + ∠DPC = 180° 

or ∠DPC = 92° 

Test: Circles- 1 - Question 8
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For what value of x in the figure, points A, B, C and D are concyclic ?

Detailed Solution: Question 8

In a cyclic quadrilateral, opposite angles are supplementary.

Here Given the opposite interior angles as ∠A = 89o and ∠C = 81o + x

Using the property ∠A + ∠C = 180o

89o + (81o + x) = 180o
So, 89o + 81o + x = 180o 
170o + x = 180o
Therefore x = 180o - 170o = 10o

Hence x = 10o, which corresponds to option D

Test: Circles- 1 - Question 9
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Greatest chord of a circle is called its

Detailed Solution: Question 9

As we know that the greatest chord of a circle is called its diameter.
Hence, option (d) is correct.

Topic in NCERT: Equal Chords and their Distances from the Centre

Line in NCERT: "The distance of the diameter, which is the longest chord from the centre? Since the centre lies on it, the distance is zero."

Test: Circles- 1 - Question 10
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In the given figure PQ and RS are two equal chords of a circle with centre O. OA and OB are perpendiculars on chords PQ and RS, respectively. If ∠AOB = 140o, then ∠PAB is equal to

Detailed Solution: Question 10

We are given:

  • ∠AOB=140∘
  • PQ and RS are equal chords.
  • OA and OB are perpendiculars on the chords PQ and RS, respectively.
  • We need to find ∠PAB.

The angle subtended by a chord at the center is twice the angle subtended on the remaining part of the circle.

  • The chord AB subtends ∠AOB=140∘ at the center.
  • Therefore, the angle subtended by AB at any point on the circle (e.g., ∠PAB) is:

Substitute the given value of ∠AOB:

Test: Circles- 1 - Question 11
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AD is diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then the distance of AB from the centre of the circle is

Detailed Solution: Question 11

Test: Circles- 1 - Question 12
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In the given figure, O is the centre of the circle. If ∠CAB = 40o and ∠CBA = 110o, the value of x is :

Detailed Solution: Question 12

Property of Angles in a Triangle: In △ABC, the sum of all angles is 180∘

Using the Property of Angles at the Center and the Circumference: The angle subtended by an arc at the center is twice the angle subtended at any point on the circumference.

Here, the angle subtended by arc AC at the circumference is ∠ACB=30∘. Therefore, the angle subtended by the same arc at the center (∠AOC) is:

Test: Circles- 1 - Question 13
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Angle formed in minor segment of a circle is

Detailed Solution: Question 13

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment.
The angle formed by the chord in the minor segment will always be obtuse.

Test: Circles- 1 - Question 14
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In the given figure PQ = QR = RS and ∠PTS = 75o then the measure of ∠QOR is

Detailed Solution: Question 14

Test: Circles- 1 - Question 15
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In the given figure, O is the centre of the circle ABE is a straight line,. If ∠DBE = 95o then ∠AOD is equal to

Detailed Solution: Question 15

AE is a straight line so,

angle ABD + angle EBD = 180°

angle ABD + 95° = 180°

angle ABD = 180° -  95°

angle ABD = 85°

so, it is interior angle of circle

So, angle AOD is double of angle suspended on the circle

angle AOD = 2ABD

angle AOD = 2(85°)

angle AOD = 170°

Test: Circles- 1 - Question 16
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The given figure shows two congruent circles with centre O and O’ intersecting at A and B. If ∠AO′B = 50o, then the measure of ∠APB is

Detailed Solution: Question 16

As we are given that, both the triangle are congruent which means their corresponding angles are equal.  Therefore, ∠AOB = AO’B = 50° Now, by degree measure theorem, we have  ∠APB = ∠AOB/2 = 25° 

Test: Circles- 1 - Question 17
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Angle inscribed in a semicircle is :

Detailed Solution: Question 17

Topic in NCERT: Angle in a semicircle is a right angle.

Line in NCERT: "Angle in a semicircle is a right angle."

Test: Circles- 1 - Question 18
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Number of circles that can be drawn through three non-collinear points is

Detailed Solution: Question 18

1. Understanding Non-Collinear Points:
- Non-collinear points are points that do not all lie on the same straight line. For example, if we have three points A, B, and C, they form a triangle if they are non-collinear.

2. Circle through Two Points:
- If we take any two points, say A and B, an infinite number of circles can be drawn through these two points. This is because circles can be drawn with different radii and centers that still pass through points A and B.

3. Adding the Third Point:
- When we add a third point C, which is not on the line formed by A and B, we can only draw one unique circle that passes through all three points A, B, and C. This is because a circle is uniquely defined by three non-collinear points.

4. Conclusion:
- Therefore, the number of circles that can be drawn through three non-collinear points is exactly one.

Final Answer: The answer is (a) 1.

Test: Circles- 1 - Question 19
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In the given figure, if ∠ABC = 50o and ∠BDC = 40o, then ∠BCA is equal to

Detailed Solution: Question 19

Understand the Geometry:

  • ∠ABC and ∠BCA are angles of triangle △ABC.
  • ∠BDC is an exterior angle of △ABC, subtending the same arc BC.

Angle Subtended by the Same Arc: In a cyclic quadrilateral, the angles subtended by the same arc on the opposite sides of the circle are related. Here:

  • ∠ABC and ∠ADC are subtended by the arc AC.
  • ∠ABC+∠ADC=180∘ (cyclic quadrilateral property).

However, we focus on triangle ABC for simplicity.

Exterior Angle Property: The exterior angle ∠BDC of a triangle is equal to the sum of the two non-adjacent interior angles. Therefore, in △ABC:

∠BDC=∠CAB+∠BCA

Substituting the given values:

40∘=50∘+∠BCA

Solve for ∠BCA: Rearrange to find ∠BCA:

∠BCA=90∘

Test: Circles- 1 - Question 20
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In the given figure, chords AB and CD intersect each other at right angles. Then, ∠x+∠y is equal to

Detailed Solution: Question 20

In the circle, AB and CD are two chords which intersect each other at P at right angle i.e. ∠CPB=90o.

∠CAB and ∠CDB are in the same segment.

∴ ∠CDB=∠CAB=x

Now, in ΔPDB, Ext. ∠CPB=∠D+∠DBP

⇒90o = x+y (∵CD⊥AB)

Hence, x+y = 90o

Test: Circles- 1 - Question 21
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In the given figure if ∠CAB = 49o and ∠ADC = 43o, then the measure of ∠ACB is

Detailed Solution: Question 21

Angle ADC = Angle CBA(: AC is the same point that makes the angle ADC and angle CBA.)
43° = Angle CBA
In triangle ACB
Angle BAC + Angle ACB + Angle ABC = 180°
49° + Angle ACB +43° = 180°
180° - 92° = ∠ACB
88° = Angle ACB

Test: Circles- 1 - Question 22
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In the given figure, O is the centre of the circle. If ∠QPR is 50o, then ∠QOR is :

Detailed Solution: Question 22

The angle subtended by a chord (or arc) at the center of the circle is twice the angle subtended at the circumference of the circle.

Mathematically:

∠QOR = 2 × ∠QPR

Substituting ∠QPR=50∘ into the formula:

∠QOR = 2 × 50∘ = 100∘

Test: Circles- 1 - Question 23
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 If TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to

Detailed Solution: Question 23

As per the given question:

We can see, OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents TQ.

So, OP ⊥ PT and TQ ⊥ OQ

∴ ∠OPT = ∠OQT = 90°

Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360°

So, ∠PTQ + ∠POQ + ∠OPT + ∠OQT = 360°

Now, by putting the respective values, we get,

⇒ ∠PTQ + 90° + 110° + 90° = 360°

⇒ ∠PTQ = 70°

Test: Circles- 1 - Question 24
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In the given figure, O is the centre of a circle. If ∠OAB=40∘ and C is a point on the circle then ∠ACB=?

Detailed Solution: Question 24

(b) 50°
OA = OB
⇒ ∠OBA = ∠OAB = 40°
Now, ∠AOB = 180° - (40° + 40°) = 100°
∴ ∠ACB=1/2∠AOB=1/2×100°=50°

Test: Circles- 1 - Question 25
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In the given figure, O is the centre of the circle and AB is a chord. The perpendicular from O to AB meets it at point M. If OM = 9 cm and OA = 15 cm, then the length of chord AB is:

Detailed Solution: Question 25


Since OM ⟂ AB,
Δ OAM is a right angle triangle.
M is the midpoint of chord AB.
So,
Using Pythagores theorem in Δ OAM

Substitute values:

Thus,
AB = 2 × AM = 2 × 12 = 24cm [As M is mid point ]
So, AB = 24 cm.

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This test contains 25 MCQs on Circles- 1 with detailed solutions for each question. It is part of the Mathematics (Maths) Class 9 course on EduRev, designed to align with the current Class 9 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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