Circle MCQ Quiz - Objective Question with Answer for Circle - Download Free PDF

Given:

A triangle ABC is circumscribing a circle.

AN = 7 cm

BN = 8 cm

AC = 18 cm

Formula Used:

Tangents drawn from an external point to a circle are equal in length.

Calculation:

Since the tangents from an external point to a circle are equal in length:

AN = AM = 7 cm (Tangents from point A)

BN = BL = 8 cm (Tangents from point B)

CM = CL (Tangents from point C)

We are given AC = 18 cm.

AC = AM + CM

18 = 7 + CM

CM = 18 - 7

CM = 11 cm

Since CM = CL, we have CL = 11 cm.

Now, we can find the length of BC.

BC = BL + CL

BC = 8 + 11

BC = 19 cm

∴ The length of BC is 19 cm.

Given:

Two circles touch each other externally at P.

AB is a direct common tangent to the two circles.

Points of contact are A and B, respectively.

∠PAB = 55º

Formula used:

In a triangle, the sum of all angles is 180º.

∠PAB + ∠ABP + ∠APB = 180º

Calculation:

Since the circles touch externally, ∠APB = 90º (property of tangents).

⇒ ∠PAB + ∠ABP + 90º = 180º

⇒ 55º + ∠ABP + 90º = 180º

⇒ ∠ABP = 180º - 145º

⇒ ∠ABP = 35º

∴ The correct answer is option (3).

Given:

O is the center of the circle.

∠AOB = 140°

∠OAC = 50°

Formula Used:

Angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle.

In a triangle, angles opposite to equal sides are equal.

Sum of angles in a triangle is 180°.

Calculation:

Consider triangle OAC. Since OA and OC are radii of the same circle, OA = OC.

Therefore, ∠OCA = ∠OAC = 50° (angles opposite to equal sides are equal).

In triangle OAC, ∠AOC + ∠OAC + ∠OCA = 180° (sum of angles in a triangle).

⇒ ∠AOC + 50° + 50° = 180°

⇒ ∠AOC + 100° = 180°

⇒ ∠AOC = 180° - 100°

⇒ ∠AOC = 80°

The reflex angle ∠AOC at the center is 360° - ∠AOC = 360° - 80° = 280°.

Angle subtended by arc ABC at the center is reflex ∠AOC = 280°.

Angle subtended by arc ABC at the circumference is ∠ABC.

∠ABC = (1/2) × reflex ∠AOC = (1/2) × 280° = 140°.

Consider triangle OAB. Since OA and OB are radii of the same circle, OA = OB.

Therefore, ∠OBA = ∠OAB.

In triangle OAB, ∠AOB + ∠OAB + ∠OBA = 180°.

⇒ 140° + ∠OAB + ∠OAB = 180°

⇒ 140° + 2∠OAB = 180°

⇒ 2∠OAB = 180° - 140°

⇒ 2∠OAB = 40°

⇒ ∠OAB = 20°

∠BAC = ∠OAC - ∠OAB = 50° - 20° = 30°.

∴ The value of angle ∠BAC is 30°.

Given:

Parallel chords AB = 36 cm

Parallel chords CD = 48 cm

Distance between the chords = 42 cm

Formula Used:

The perpendicular from the center of a circle to a chord bisects the chord.

Pythagoras theorem: (radius)² = (half of chord length)² + (distance from center)²

Diameter = 2 × radius

Calculation:

Let the radius of the circle be r.

Let the distance of chord AB from the center be x cm.

Then the distance of chord CD from the center will be (42 - x) cm.

For chord AB:

r² = (36/2)² + x²

⇒ r² = 18² + x²

⇒ r² = 324 + x² (Equation 1)

For chord CD:

r² = (48/2)² + (42 - x)²

⇒ r² = 24² + (42 - x)²

⇒ r² = 576 + (1764 - 84x + x²)

⇒ r² = 576 + 1764 - 84x + x²

⇒ r² = 2340 - 84x + x² (Equation 2)

Equating Equation 1 and Equation 2:

324 + x² = 2340 - 84x + x²

⇒ 324 = 2340 - 84x

⇒ 84x = 2340 - 324

⇒ 84x = 2016

⇒ x = 2016 / 84

⇒ x = 24 cm

Substitute the value of x in Equation 1:

r² = 324 + (24)²

⇒ r² = 324 + 576

⇒ r² = 900

⇒ r = \(\sqrt{900}\)

⇒ r = 30 cm

Diameter = 2 × radius = 2 × 30 = 60 cm

∴ The diameter of the circle is 60 cm.

Given:

OP = 17 cm

PA = 12 cm

PB = 22.5 cm

Formula Used:

In such a case = PA × PB = PC × PD

Calculation:

Let radius = x

⇒ PC = 17 - x

and PD = 17 + x

According to Question

PA × PB = PC × PD

⇒ 12 × 22.5 = (17 - x)(17 + x)

⇒ 270 = 289 - x² 

⇒ x²  = 19 

⇒ x = √19 

⇒ r = √(19) cm

The radius of the circle is √(19) cm.

Given:

Two circles touch each other externally at P.

AB is a direct common tangent to the two circles, A and B are points of contact, and ∠PAB = 40°.

Concept used:

If two circles touch each other externally at some point and a direct common tangent is drawn to both circles, the angle subtended by the direct common tangent at the point where two circles touch each other is 90°.

Calculation:

According to the concept, ∠APB = 90°

Considering ΔAPB,

∠ABP

⇒ 90° - ∠PAB

⇒ 90° - 40° = 50°

∴ The measure of ∠ABP is 50°.

Given:

The area of the sector of a circle of radius 28 cm is 112 cm2.

Calculation:

Area of sector = πr²θ/360°

112 = π(28)²θ/360°

θ = 360/22

Length of the arc = θ/360°(2πr)

⇒ 360/360 × 22/22 × 28/7 × 2

⇒  8 cm

Therefore, the length of the arc is 8 cm.

Alternate Method Area of sector = 1/2 × r × l

r = radius of the circle

l = length of arc

As per question,

⇒ 112 = 1/2 × 28 × l

⇒ l = 8 cm

Therefore, the length of the arc is 8 cm.

Given:

Radius of large circle (R) = 16 cm

Radius of small circle (r) = 8 cm

Distance between centre (D) = 26 cm

Formula used:

Direct common tangent = √{D² - (R - r)²}

Calculation:

Direct common tangent = √{D2 - (R - r)2}

⇒ AB = √{262 - (16 - 8)2}

⇒ AB = √{676 - 64} = √612 = 2 × √153

∴ The correct answer is 2√153. 

Given:

ΔPQR is an equilateral triangle inscribed in a circle.

S is any point on the arc QR.

Concept used:

Each of the angles of an equilateral triangle is equal and 60° 

The angles subtended by an arc in the same segment of a circle are equal.

Calculation:

Both ∠PRQ and ∠PSQ are in the subtended by the arc PQ in the same segment, so

∠PRQ = ∠PSQ = 60°

 ∴ The correct option is 2

Given:

The radius of a circle is 10 cm. The angle made by chord AB at the center of this circle is 60 degrees.

Concept used:

1. In an Isosceles triangle, the two angles opposite to the equal sides are equal to each other.

2. In an equilateral triangle, since the three sides are equal therefore the three angles, opposite to the equal sides, are equal in measure.

Calculation:

OA = OB = 10 cm

ΔOAB is an isosceles triangle.

So, ∠OAB = ∠OBA = \(\frac {(180 - 60)^\circ}{2}\) = 60° 

Since ∠OAB = ∠OBA = ∠AOB = 60°, ΔOAB is an equilateral triangle.

So, OA = OB = AB = 10 cm

∴ The length of this chord is 10 cm.

Given:

The radius of the circles is 8 cm

Calculation:

According to the diagram,

AD = DB

O1O2 = 8

Again O1A = O2A = 8 [Radius of the circle]

∠ADO1 = 90°

O1D = O2D = 4

AD = √(64 - 16)

⇒ √48 = 4√3

AB = 2 × 4√3 = 8√3

∴ The length of the common chord is 8√3 cm

Formula used:

Direct common tangent = √ (distance between the two centers² - ( r₁  - r₂)²

Calculation: 

Distance between the two centers is = d cm

As per the formula,

12 = √ (d2 - ( 8  - 3)².

⇒ 144 = d2 - 25

⇒ d² = 169

⇒ d = 13

∴ The correct option is 3

Given:

Radius of 1st circle (r₁) = 10 cm

Radius of 2nd circle (r₂) = 5 cm

Formula used:

Direct common tangent = 2 × √(r₁ × r₂)

Calculation:

Direct common tangent = 2 × √(r1 × r2)

⇒  2 × √(10 × 5)

⇒ 10√2 cm

∴ The correct answer is 10√2 cm.

Concept Used : 

The angle at the point of contact of a tangent to a circle is a right angle. 

Calculation :

According to question,

⇒ ∠PAB = 65° 

Now In ΔAPB,

⇒ ∠A  + ∠B + ∠P = 180° 

⇒ 65° + ∠B + 90° = 180° 

⇒ ∠B = 180° - 155° = 25° 

∴ The correct answer is 25°.

Given:

PT = 8 cm

PA = 6 cm

Formula used

PT² = PA × PB

Here, PT is tangent

Calculation:

Let AB be "x"

PT² = PA × PB

8² = 6(6 + x)

32/3 = 6 + x

x = 14/3

The value of AB is 14/3 cm.