Explore Exams
Login Signup
Sign in Open App

All questions of Circles for Class 10 Exam

Clear all your doubts with EduRev

If PA and PB are tangents to the circle with centre O such that ∠APB = 40∘, then ∠OAB is equal to
  • a)
    30∘
  • b)
    40∘
  • c)
    25∘
  • d)
    20∘
Correct answer is option 'D'. Can you explain this answer?

Parth Basu answered
Let ∠OAB = ∠OBA = x [Opposite angles of opposite equal radii] And ∠AOB =180° -  40° = 140°
Now, in triangle AOB,
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x +140° = 180°
⇒ 2x = 40°
⇒ x = 20° 
∴ ∠OAB = 20°

A circle is inscribed in ΔABC having sides 8 cm, 10 cm and 12 cm as shown in the figure. Then,
 
  • a)
    AD = 7 cm, BE = 5 cm.
  • b)
    AD = 8 cm, BE = 5 cm.
  • c)
    AD = 8 cm, BE = 6 cm.
  • d)
    AD = 5 cm, BE = 7 cm.
Correct answer is option 'A'. Can you explain this answer?

Arun Khatri answered
Let AD = x and BE = y
∴ BD = 12 -  x ⇒  BE = 12 - x [BD = BE = Tangents to a circle from an external point]
⇒ y = 12 - x ⇒ x+y = 12.......(i)
Also, AF = x and CF = 10 - x and CE = 8 - y
Also, AF = x and CF = 10 — x and CE = 8 -  y
∴ 10 - x = 8 - yx - y = 2  (ii)
On solving eq. (i) and (ii), we get x = 7 and y = 5
⇒ AD = 7 cm and BE = 5cm 

At one end of a diameter PQ of a circle of radius 5 cm, tangent XPYis drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is
  • a)
    6 cm.
  • b)
    5 cm.
  • c)
    7 cm.
  • d)
    8 cm
Correct answer is option 'D'. Can you explain this answer?

Prasad Chavan answered
Here, OP = 00 = 5 cm [Radii]
And OR = PR -  OP = 8 - 5 = 3 cm Also, OA = 5 cm [Radius]
Now, in triangle AOQ, OA2 = OR2 + AR2 ⇒ 52 = 32 + AR2
⇒AR2 = 25-9= 160 AR= 4cm
Since, perpendicular from centre of a circle to a chord bisects the chord.
∴ AB = AR+ BR = 4+ 4 = 8cm 

The length of tangent PQ, from an external point P is 24 cm. If the distance of the point P from the centre is 25 cm, then the diameter of the circle is 
  • a)
    7 cm
  • b)
    14 cm.
  • c)
    15 cm.
  • d)
    12 cm
Correct answer is option 'B'. Can you explain this answer?

Swara Chopra answered
Here ∠OPQ = 90°[Angle between tangent and radius through the point of contact]
∴ OQ2 = OP2 + PQ2 (25)2 = OP2 + (24)2 ⇒ OP2 = 625 - 576 ⇒ OP2 = 49
⇒ OP = 7 cm
Therefore, the diameter = 2 x OP = 2 x 7 = 14 cm 

In the given figure, perimeter of quadrilateral ABCD is
  • a)
    48 units
  • b)
    34 units
  • c)
    36 units
  • d)
    28 units
Correct answer is option 'B'. Can you explain this answer?

Anjana Bajaj answered
Here SD = RD = 5 units [Tangents from an external point]
And PB = QB = 4 cm [Tangents from an external point]
∴ QC = 10 - QB = 10 - 4 = 6 units
Also QC = RC = 6 units [Tangents from an extemal point]
∴ CD = RD + RC = 5 + 6 = 11 units
Also AP = AS = 2 units [Tangents from an external point]
∴ AD = AS + DS = 2 + 5 = 7 units Therefore. Perimeter of quadrilateral ABCD = 6 + 10 + 11 + 7 = 34 units 

The diameter of the front and rear wheels of a tractor are 80 cm and 200 cm respectively. What are the number of revolution that a rear wheel makes to cover the distance which the front wheel covers in 800 revolutions?
  • a)
    640
  • b)
    320
  • c)
    240
  • d)
    300
Correct answer is option 'B'. Can you explain this answer?

Aishwarya gupta answered
To solve this problem, we can use the concept of the circumference of a circle and the relationship between the circumference and diameter.

Given:
Diameter of the front wheel = 80 cm
Diameter of the rear wheel = 200 cm

Step 1: Calculate the circumference of each wheel
Circumference of a circle = π * diameter

Circumference of the front wheel = π * 80 cm
Circumference of the rear wheel = π * 200 cm

Step 2: Find the ratio of the circumferences
Since the front and rear wheels are connected to the same axle, they will cover the same distance in one revolution. Therefore, the ratio of the circumferences is equal to the ratio of the number of revolutions.

Ratio of the circumferences = (Circumference of the rear wheel) / (Circumference of the front wheel)

Ratio of the circumferences = (π * 200 cm) / (π * 80 cm) = 200 cm / 80 cm = 5/2

Step 3: Find the number of revolutions the rear wheel makes to cover the distance covered by the front wheel in 800 revolutions
Since the ratio of the circumferences is equal to the ratio of the number of revolutions, we can set up the following proportion:

(Revolutions of the rear wheel) / 800 = 5/2

Cross-multiplying, we get:

2 * (Revolutions of the rear wheel) = 800 * 5
2 * (Revolutions of the rear wheel) = 4000
Revolutions of the rear wheel = 4000 / 2
Revolutions of the rear wheel = 2000

Therefore, the rear wheel makes 2000 revolutions to cover the distance covered by the front wheel in 800 revolutions.

Hence, the correct answer is option B) 320.

In the given figure, If TP and TQ are two tangents to a circle with centre O, so that ∠POQ = 110o then ∠PTQ is equal to :
  • a)
    80∘
  • b)
    70∘
  • c)
    60∘
  • d)
    90∘
Correct answer is option 'B'. Can you explain this answer?

Prasad Chavan answered
Since the angle between the two tangents drawn from an external point to a circle in supplementary of the angle between the radii of the circle through the point of contact.
∴ ∠PTQ = 180∘−110∘ = 70∘

In the given figure, AB and PQ intersect at M. If A and B are centres of circles then _______.
  • a)
    PM = MQ
  • b)
    PQ ⊥ AB
  • c)
     Both (A) and (B)
  • d)
    PQ = AB
Correct answer is option 'C'. Can you explain this answer?

Priyanka Kapoor answered
In the given figure, PQ is the common chord of the two circles.
⇒ AB bisects the common chord PQ at M.
∴ PM = MQ
Moreover, PQ is perpendicular to AB.
∴ Option (C) is correct.

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the the bigger circle. BD is a tangent to the  smaller circle touching it at D. Find the length AD.
  • a)
    19 cm
  • b)
    20 cm
  • c)
    16 cm
  • d)
    √105
Correct answer is option 'A'. Can you explain this answer?

Seema joshi answered
To find the length AD, we can use the Pythagorean theorem.

Since AB is a diameter of the bigger circle, its length is equal to the diameter of the bigger circle, which is twice the radius. So, AB = 2 * 13 cm = 26 cm.

Since BD is tangent to the smaller circle, it is perpendicular to AD. Therefore, triangle ABD is a right triangle.

Using the Pythagorean theorem, we have:

(AD)^2 + (BD)^2 = (AB)^2
(AD)^2 + (8 cm)^2 = (26 cm)^2
(AD)^2 + 64 cm^2 = 676 cm^2
(AD)^2 = 676 cm^2 - 64 cm^2
(AD)^2 = 612 cm^2
AD = √(612 cm^2)
AD ≈ 24.7 cm

Therefore, the length AD is approximately 24.7 cm.

Let s denote the semiperimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, find BD.
  • a)
    s – b
  • b)
    2s + h
  • c)
    b + s
  • d)
    3b – s
Correct answer is option 'A'. Can you explain this answer?

Niharika dasgupta answered
By the property of a circle tangent to a line, we know that the line segment from the center of the circle to the point of tangency is perpendicular to the line. Therefore, let O be the center of the circle and let M, N, and P be the midpoints of BC, CA, and AB respectively.

Since O is the incenter of triangle ABC, O is equidistant from the sides of the triangle. Therefore, OM = ON = OP.

We can draw radii from O to the points of tangency D, E, and F. Let r be the length of these radii. Then, OD = OE = OF = r.

Since OM = ON = OP, we have that MD = NE = PF = s - a, where s is the semiperimeter of triangle ABC.

By the Pythagorean Theorem, we have that BD^2 = BM^2 + MD^2. Since BM = a/2 and MD = s - a, we have that BD^2 = (a/2)^2 + (s - a)^2.

Expanding and simplifying, we have that BD^2 = a^2/4 + s^2 - 2as + a^2.

Rearranging terms, we have that BD^2 = s^2 - 2as + a^2/4 + a^2.

Factoring, we have that BD^2 = (s - a/2)^2 + a^2/4.

Taking the square root of both sides, we have that BD = sqrt((s - a/2)^2 + a^2/4).

Therefore, the length of BD is sqrt((s - a/2)^2 + a^2/4).

Answer: a) sqrt((s - a/2)^2 + a^2/4)

AB is a chord of length 24 cm of a circle of radius 13 cm. The tangents at A and B intersect at a point C. Find the length AC.
  • a)
    31.2 cm
  • b)
    12 cm
  • c)
    28.8 cm
  • d)
    25 cm
Correct answer is option 'A'. Can you explain this answer?

Ritu Saxena answered
Given, Chord AB = 24 cm, Radius OB = OA = 13 cm.
Draw OP ⊥ AB
In D OPB, OP ⊥ AB ⇒ AP = PB
[Perpendicular from centre on chord bisect the chord] =(1/2)AB= 12
Also, OB2 = OP2 + PB2
⇒ (13)2 = OP2 + PB2 ⇒ 169 = OP2 + (12)2
⇒ OP2 = 169 - 144 = 25 ⇒ OP = 5 cm
In Δ BPC, BC2 = x2 + BP2 [By pythagoras theorem]
BC2 = x2 + 144 ...(i)
In ΔOBC, OC2 = OB2 + BC2
⇒ (x + 5)2 = (13)2 + BC2 ⇒ x = 288/10=28.8cm

Put the value of x in (i), we get
BC2 = x2 + 144 = ((144)2/25) + 144⇒ BC = 31.2
⇒ AC = BC = 31.2 cm

The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm, then what is area of sector?
  • a)
    14.6 cm2
  • b)
    15.6 cm2
  • c)
    19.6 cm2
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Priyanka Kapoor answered
Let O be the centre of a circle of radius 5.2 cm. Let OABO be the sector with perimeter 16.4 cm.

OA + O B + arc AB = 27.2
5.2 + 5 .2 + arc AB = 16.4
⇒ arc AB = 16.4 - 16.4 = 6 cm
Area of sector OACBO

The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.
  • a)
    252 cm2
  • b)
    264 cm2
  • c)
    184 cm2
  • d)
    1284 cm2
Correct answer is option 'B'. Can you explain this answer?

Smita das answered
To find the area of the face of the clock described by the minute hand in 35 minutes, we need to find the length of the arc covered by the minute hand and then use it to calculate the area.

Length of Arc Covered by the Minute Hand:
The minute hand of the clock is 12 cm long. In 60 minutes, it covers a complete circle, which is equivalent to the circumference of the clock face. The formula for the circumference of a circle is given by:

C = 2πr

where C is the circumference and r is the radius.

In this case, the radius is equal to the length of the minute hand, which is 12 cm. So, the circumference of the clock face is:

C = 2π(12) = 24π cm

In 60 minutes, the minute hand covers the entire circumference. Therefore, in 35 minutes, it covers:

35/60 * 24π = 7/12 * 24π = 14π cm

Area of the Face of the Clock:
To find the area of the face of the clock described by the minute hand in 35 minutes, we need to calculate the area of the sector formed by the minute hand.

The formula for the area of a sector of a circle is given by:

A = (θ/360) * πr^2

where A is the area, θ is the central angle in degrees, and r is the radius.

In this case, the central angle is 35/60 * 360 = 210 degrees (since the minute hand covers 35 minutes out of 60 minutes, and each minute corresponds to 6 degrees on the clock face).

Substituting the values into the formula, we get:

A = (210/360) * π(12)^2
= (7/12) * π(144)
= 7π * 12
= 84π cm^2

Approximating π to 3.14, we have:

Area = 84 * 3.14
≈ 264 cm^2

Therefore, the area of the face of the clock described by the minute hand in 35 minutes is approximately 264 cm^2.

Hence, option B is the correct answer.

A copper wire when bent in the form of a square encloses an area of 484 cm2. The same wire is now bent in the form of a circle. Find the area enclosed by the circle.
  • a)
    616 cm2
  • b)
    456 cm2
  • c)
    216 cm2
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Anoushka Nambiar answered
To solve this problem, we need to use the formulas for the area of a square and the area of a circle.

1. Area of a square:
The area of a square can be found by multiplying the length of one side by itself. Therefore, if the copper wire is bent in the form of a square and encloses an area of 484 cm^2, we can find the length of one side using the formula:
Area of square = side^2
484 = side^2
Taking the square root of both sides, we get:
side = √484
side = 22 cm

2. Perimeter of a square:
The perimeter of a square can be found by multiplying the length of one side by 4. In this case, the perimeter of the square formed by the copper wire is:
Perimeter of square = 4 * side
Perimeter of square = 4 * 22
Perimeter of square = 88 cm

3. Circumference of a circle:
The circumference of a circle can be found using the formula:
Circumference = π * diameter
Since the copper wire is bent in the form of a square, the length of the wire is equal to the perimeter of the square. Therefore, the circumference of the circle formed by the copper wire is:
Circumference of circle = 88 cm

4. Radius of the circle:
To find the radius of the circle, we can use the formula:
Radius = Circumference / (2 * π)
Radius = 88 / (2 * 3.14)
Radius ≈ 14 cm

5. Area of a circle:
Finally, we can find the area of the circle using the formula:
Area of circle = π * radius^2
Area of circle = 3.14 * 14^2
Area of circle ≈ 3.14 * 196
Area of circle ≈ 615.44 cm^2

Therefore, the area enclosed by the circle formed by the copper wire is approximately 616 cm^2, which corresponds to option A.

The side of a square is 10 cm. What is the area of circumscribed circle?
  • a)
    78.5 cm2
  • b)
    157 cm2
  • c)
    135 cm2
  • d)
    314 cm2
Correct answer is option 'B'. Can you explain this answer?

Ashish Choudhary answered
Given:
Side of the square = 10 cm

To find:
The area of the circumscribed circle.

Solution:
To find the area of the circumscribed circle, we need to find the radius of the circle first.

Finding the radius:
The diagonal of a square divides it into two congruent right-angled triangles.
Let's consider one of these triangles.

In a right-angled triangle, the hypotenuse (diagonal of the square) is equal to the diameter of the circle and the length of one side of the square is equal to the radius of the circle.

Using the Pythagorean theorem, we can find the length of the diagonal of the square:
(diagonal)^2 = (side)^2 + (side)^2
(diagonal)^2 = 10^2 + 10^2
(diagonal)^2 = 100 + 100
(diagonal)^2 = 200
diagonal = √200 = 10√2 cm

Since the diagonal of the square is equal to the diameter of the circle, the diameter of the circle is 10√2 cm.

So, the radius of the circle = (10√2)/2 = 5√2 cm.

Finding the area of the circle:
The area of a circle is given by the formula: A = πr^2, where r is the radius of the circle.

Substituting the value of the radius, we get:
A = π(5√2)^2
A = π(25*2)
A = 50π cm^2

Now, let's approximate the value of π to 3.14.

A ≈ 50 * 3.14
A ≈ 157 cm^2

Therefore, the area of the circumscribed circle is approximately 157 cm^2.

Answer:
The correct answer is option b) 157 cm^2.

Two circles of radii 10 cm and 8 cm intersect each other and the length of common chord is 12 cm. The distance between their centres is.
  • a)
    √7 cm
  • b)
    3√7 cm
  • c)
    4√7 cm
  • d)
    (8 + 2√7) cm
Correct answer is option 'D'. Can you explain this answer?

Akanksha Deshpande answered
To find the distance between the centers of the two circles, we can draw a line connecting the centers and draw a perpendicular line from the center of one circle to the common chord.

Let O1 be the center of the larger circle with radius 10 cm, and O2 be the center of the smaller circle with radius 8 cm. Let AB be the common chord, with length 12 cm.

Since the common chord is perpendicular to the line connecting the centers, we can draw a right triangle O1O2B, where O1O2 is the distance between the centers and O1B and O2B are the radii of the circles.

Using the Pythagorean theorem, we can find the length of O1O2:

O1O2^2 = O1B^2 + O2B^2
O1O2^2 = 10^2 + 8^2
O1O2^2 = 164

Taking the square root of both sides, we find:

O1O2 = √164
O1O2 = 2√41

Therefore, the distance between the centers of the two circles is 2√41 cm.

From a point Q, the length of tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm, radius of circle is : 
  • a)
    10 cm
  • b)
    8 cm
  • c)
    7 cm
  • d)
    6 cm
Correct answer is option 'C'. Can you explain this answer?

Anjana Bajaj answered
Here ∠OPQ = 90° [Tangent makes right angle with the radius at the point of contact]
∴ OQ2 = OP2 + PQ2 (25)2 = OP2 + (24)2
⇒ OP2 = 676 - 576 = 49
⇒ OP = 7 cm Therefore, the radius of the circle is 7 cm 

If PQR is a tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 60∘, then ∠AQB is equal to
 
  • a)
    60∘
  • b)
    45∘
  • c)
    90∘
  • d)
    30∘
Correct answer is option 'A'. Can you explain this answer?

Anmol Chatterjee answered
Since AB II PA and BQ is intersectin.
∴ ∠BQR = LQBA = 60° [Alternate angles]
And ∠BQR = ∠QAB = 60° [Alternate segment theorem] Now, in triangle AQB,
∠AQB + ∠QBA + ∠BAQ = 180°
⇒ ∠AQ,B + 60° + 60° = 180°
⇒ ∠LAQB = 60° 

How the tangent at any point of a circle and radius through the point is related?
  • a)
    perpendicular to each other
  • b)
    parallel to each other
  • c)
    having same length
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Rohit Sharma answered
The tangent at any point of a circle and the radius through this point are perpendicular to each other as shown below:

In the figure above,
∠OPA +∠OPQ = 180∘ ........(1) [Linear pair angle]
Also, ∠OPA = ∠OPQ ......(2)
∴ ∠OPA +∠OPQ = 180∘
2 ×∠OPQ = 180∘
∠OPQ = 90∘
Hence, OP is perpendicular to PQ ⇒ OP is perpendicular to AB.

What is the value of y and x respectively in the given figure?
  • a)
    x = 55°, y = 45°
  • b)
    x = 45°, y = 55°
  • c)
    x = 65°, y = 55°
  • d)
    x = 55°, y = 65°
Correct answer is option 'B'. Can you explain this answer?

Priyanka Kapoor answered
In ΔABO,
∠y + 25° + 100° = 180°
⇒ y = 55°
∠C + ∠A = 180° [∵ ABCD is cycle]
⇒ ∠ = 80°
In ΔPBC,
∠x + 55° + 80° = 180°
⇒ ∠x = 45°

In the given figure, O is the centre of the circle, then ∠OZ is______.
  • a)
    2 ∠XZY
  • b)
    2 ∠Y
  • c)
    2 ∠Z
  • d)
    2 (∠XZY + ∠YXZ)
Correct answer is option 'D'. Can you explain this answer?

Ritu Saxena answered
OX= OY = OZ = r    (radius of circle)
Taking ∠X = θ and ∠Z = α
Then in ΔYXO, OX = OY ⇒ ∠X = ∠OYX = θ
Now, in ΔOZY OZ= OY⇒ ∠Z= ∠OYZ = α
Now, XOZY is a quadrilateral,
⇒ ∠X+∠Y+ ∠Z+∠O = 360°
⇒ θ + θ + α + α +∠O = 360°
⇒ ∠O = 360° - 2(α + θ)
⇒ ∠O = 360° - 2 (∠OZX + ∠XZY + ∠OXZ + ∠ZXY)
⇒ ∠O = 360° - 2 (∠OZX + ∠OXZ) - 2(∠XZY + ∠ZXY)
⇒ ∠O = 360° - 2(180° - ∠O) - 2(∠XZY + ∠ZXY)
⇒ ∠O = 360° - 360° + 2∠O - 2(∠XZY + ∠ZXY)

Match the columns.
  • a)
    (a) → (p), (b) → (r), (c) → (q)
  • b)
    (a) → (r), (b) → (q), (c) → (p)
  • c)
    (a) → (q), (b) → (r), (c) → (p)
  • d)
    (a) → (r), (b) → (p), (c) → (q)
Correct answer is option 'B'. Can you explain this answer?

Ritu Saxena answered
(a) Draw OC perpendicular on AB,
In Δ OBC & Δ OAC
∠OCB = ∠OCA (each 90°)
OB = OA (radii of same circle)

OC is common side in both triangles
∴ ΔOBC ≅ ΔOAC (RHS congruence)

Now, in ΔOCA (OA)2 = (OC)2 + (CA)2

Hence, radius of circle = 5 cm
(b) : In ΔOPT,
(OP)2 = (OT)2 + (PT)2 
⇒ (x + 4)2 = x2 + (8)2
⇒ x2 + 16 + 8x = x2 + 64
⇒ 8x = 48 ⇒ x = 6 cm
(c) : PQ = 10 cm
We know, length of tangents drawn from an external point to a circle are equal.
⇒ PQ = PR ...(i)
Also, SQ = SU ...(ii)
and TU = TR ...(iii)
Now, perimeter of DPST
= PS + ST + PT = PS + SU + UT + PT
= PS + SQ + TR + PT (Using (ii) & (iii))
= PQ + PR = PQ + PQ (Using (i))
= 2 PQ = 2 × 10 = 20 cm.

In the given figure, a circle touches all four sides of a quadrilateral PQRS, whose sides are PQ = 6.5 cm, QR = 7.3 cm, and PS = 4.2 cm, then RS is equal to
  • a)
    4.7 cm.
  • b)
    7.3 cm.
  • c)
    5.3 cm.
  • d)
    5 cm.
Correct answer is option 'D'. Can you explain this answer?

Mahesh Basu answered
Let P,Q,R,S be the points where the circle touches the quadrilateral .From a point out side of the circle  the two tangents to the circle are equal.
AP = AS,BP = BQ , CR = CQ , and DR = DS
Adding AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AD + BC
6 + 4 = AD + 7
AD = 3 cm .

In the adjoining figure, the measure Of PR is 
  • a)
    18 cm.
  • b)
    12 cm.
  • c)
    10 cm.
  • d)
    16 cm.
Correct answer is option 'D'. Can you explain this answer?

Niharika Mehta answered
Here ∠Q = 90° [Angle between tangent and radius through the point of contact] Now, in triangle OPQ, OP2 = QO2 + PQ2
⇒ OP2 = (6)2 + (8)2 = 36 + 64 = 100
⇒ OP = 10 cm
∴ PR = OP + OR = OP + OQ [OR = OQ = Radii]
⇒ PR = 10+6 = 16cm

In figure, AB is a chord of a circle and AT is a tangent at A such that ∠BAT = 60o, measure of ∠ACB is :
  • a)
    90∘
  • b)
    150∘
  • c)
    110∘
  • d)
    120∘
Correct answer is option 'D'. Can you explain this answer?

Debanshi Sen answered
Since OA is perpendicular to AT. then ∠OAT = 90°
⇒ ∠OAB + ∠BAT = 90°
⇒ ∠OAB + 60° = 90° ⇒ ∠OAB = 30°
∴ ∠OAB = ∠O8A = 30" [Angles opposite to radii]
∴ ∠OAB = 180° - (30° +30° ) = 120° [Angle sum property of a triangle]
∴ Reflex ∠AOB = 360° - 120° = 240°
Now, since the degree measure of an arc of a circle is twice the angle subtended by it any point of the alternate segment of the circle with respect to the arc.
∴ Reflex ∠AOB = 2∠ACB ⇒ 240° = 2∠ACB ⇒ ∠ACB = 120°

In the given figure, PT touches the circle at R whose centre is O. Diameter SQ when produced meets PT at P. Given ∠SPR = x° and ∠QRP = y°. Then,
  • a)
    x° + 2y° = 90°
  • b)
    2x° + y° = 90°
  • c)
    x° + y° = 120°
  • d)
    3x°+2y° = 120°
Correct answer is option 'A'. Can you explain this answer?

Vivek Bansal answered
Consider chord QR
∴∠QRP = ∠QSR
⇒ ∠QSR = y° [∴ QRP = y°]
⇒ ∠PSR = y°
Since angle in a semicircle is a right angle. 
∴∠QRS = 90°
Now, ∠PRS = ∠QRP + ∠QRS
⇒ ∠PRS = y° + 90°
In ΔPRS, we have
∠SPR + ∠PRS + ∠PSR = 180°
⇒ x° + y°+ 90° + y° = 180° 
⇒ x° + 2y° = 90°

If O is the centre of a circle, AOC is its diameter and B is a point on the circle such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT =
  • a)
    40°
  • b)
    50°
  • c)
    60°
  • d)
    65°
Correct answer is option 'B'. Can you explain this answer?

Ritu Saxena answered
∠ABC = 90° [Angle in a semicircle ]
In ∆ ABC, we have
∠ACB + ∠CAB + ∠ABC = 180°

⇒ 50° + ∠CAB + 90° = 180°
⇒ ∠CAB = 40°
Now, ∠CAT = 90° ⇒ ∠CAB + ∠BAT = 90°
⇒ 40° + ∠BAT = 90°⇒ ∠BAT = 50°

In the figure given alongside the length of PR is
  • a)
    24 cm.
  • b)
    28 cm.
  • c)
    26 cm.
  • d)
    20 cm.
Correct answer is option 'C'. Can you explain this answer?

Saumya Chakraborty answered
Here ∠Q = 90° and LS = 90° [Angle between tangent and radius through the point of contact]
Now, in triangle OPQ, OP2 = OQ2 + QP2
⇒ OP2 = (3)2 + (4)2
⇒ OP2 = 16 + 9 = 25
⇒ OP = 5 cm Again in triangle RSO', O'R2 = O'S2 + FS2
⇒ O'R2 = (5)2 + (12)2
⇒ O'R2 = 25 +144 = 169
⇒ O'R = 13 cm
∴ PR = OP+OQ+O's+OR = 5+ 3+ 5+ 13 = 26 m
 

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q, such that OQ = 15 cm. Length of PQ is 
 
  • a)
    12 cm
  • b)
    13m
  • c)
    10√2 cm
  • d)
    none
Correct answer is option 'C'. Can you explain this answer?

Abhiram Gupta answered
Since OP is perpendicular to PQ, then ∠OPQ = 90°
Now, in triangle OPQ.
OQ2 = OP2 + PQ2 ⇒ (15)2 = (5)2 + PQ2 ⇒ PQ2 = 225 -  25
⇒ PQ2 = 200 ⇒ PQ = 10√2cm

In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and RQ = 4 cm, then OR is equal to
  • a)
    2.5 cm.
  • b)
    12 cm.
  • c)
    5 cm.
  • d)
    10 cm.
Correct answer is option 'C'. Can you explain this answer?

Kds Coaching answered
Given: SQ = 6 cm and RQ = 4 cm.
To find the length of OR, we can use the properties of the right triangle OQR, where:
  • OQ is the radius of the circle.
  • Since SQ = 6 cm, then OQ = OS = 6/2 = 3 cm.
  • Angle Q is a right angle (90°).
Using the Pythagorean theorem in triangle OQR:
  • OR2 = QR2 + OQ2
  • Substituting the known values: OR2 = (4 cm)2 + (3 cm)2
  • Calculating: OR2 = 16 + 9 = 25
  • Thus, OR = √25 = 5 cm.
The final result is that OR is equal to 5 cm.

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, then BA : AT is
  • a)
    3 : 1
  • b)
    4 : 1
  • c)
    2 : 1
  • d)
    3 : 2
Correct answer is option 'C'. Can you explain this answer?

Ritu Saxena answered
∠BPA = 90° (Angle in semicircle)
In ∆ BPA, ∠ABP + ∠BPA + ∠PAB = 180°
⇒ 30° + 90° + ∠PAB = 180°
⇒ ∠PAB = 60°
Also, ∠POA = 2∠PBA
⇒ ∠POA = 2 × 30° = 60°
⇒ OP = AP ...(i)
(side opposite to equal angles)

In ∆OPT, ∠OPT = 90°
∠POT = 60° and ∠PTO = 30° [angle sum property of a D]
Also ∠APT + ∠ATP = ∠PAO [exterior angle property]
∴ ∠APT + 30°= 60° ⇒ ∠APT = 30°
∴ AP = AT ...(ii) (side opposite to equal angles)
From (i) and (ii), AT = OP = radius of the circle; and AB = 2r
⇒ AB = 2AT ⇒ AB/AT = 2 ⇒ AB : AT = 2 :1

Two concentric circles of radii a and b, where a > b, are given. The length of a chord of the larger circle which touches the other circle is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Ritu Saxena answered
In figure, AB is a chord of circle C1 which is a tangent to C2.
Since, tangent is perpendicular to radius through point of contact
∴ ∠OCA = 90° ⇒ OA = a, OC = b
In ∆OCA, (OA)2 = (OC)2 + (AC)2
⇒ a2 = b2 + (AC)2 ⇒ AC = 
∴ Length of chord AB = 2AC = 2 

Chapter doubts & questions for Circles - International Mathematics Olympiad (IMO) for Class 10 2025 is part of Class 10 exam preparation. The chapters have been prepared according to the Class 10 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Class 10 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Circles - International Mathematics Olympiad (IMO) for Class 10 in English & Hindi are available as part of Class 10 exam. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free.

Top Courses Class 10

Related Class 10 Content

© EduRev
Education Revolution
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily