CIRCLES - Q&A

EXERCISE 21

1. Multiple Choice Type:
Choose the correct answer from the options given below.

(i) In the given figure, AB is the largest chord of the circle, AB = 18 cm and length of the chord CD is half of length of chord AB, then
(a) OP = 18 cm and CD = 18 cm
(b) OP = 9 cm and CD = 18 cm
(c) OP = 18 cm and CD = 9 cm
(d) OP = 9 cm and CD = 9 cm

Answer: (d)

Steps:
1. The largest chord of a circle is the diameter. Given AB = 18 cm, so the Diameter = 18 cm.
2. The radius of the circle is half of the diameter. Radius (OP) = 18 / 2 = 9 cm.
3. Length of chord CD is half of AB. CD = 18 / 2 = 9 cm.
4. Therefore, OP = 9 cm and CD = 9 cm.

(ii) The largest chord of a circle is 20 cm, then its radius is:
(a) 20 cm
(b) 40 cm
(c) 10 cm
(d) none of these

Answer: (c)

Steps:
1. The largest chord is the diameter.
2. Diameter = 20 cm.
3. Radius = Diameter / 2 = 20 / 2 = 10 cm.

(iii) The relation between the area enclosed by a major segment of a circle and area enclosed by its semi-circle is:
(a) major segment is smaller than the semi-circle.
(b) major segment is bigger than the semi-circle.
(c) major segment is equal to the semi-circle

Answer: (b)

Explanation:
A semi-circle divides the circle into two equal halves. A major segment corresponds to a major arc, which is greater than a semi-circle arc. Therefore, the area of the major segment is greater than the area of the semi-circle.

(iv) The radius of a circle is 5 cm then the length of its longest chord is:
(a) less than 5 cm
(b) equal to 5 cm
(c) equal to 10 cm
(d) less than 10 cm

Answer: (c)

Steps:
1. The longest chord is the diameter.
2. Diameter = 2 × Radius.
3. Diameter = 2 × 5 cm = 10 cm.

(v) The diameter of a circle with centre at point O is 30 cm. A point P lies in the same plane as the circle and OP = 20 cm, then point P lies:
(a) on the circumference of the circle
(b) inside the circle
(c) outside the circle
(d) nothing can be said

Answer: (c)

Steps:
1. Diameter = 30 cm, so Radius = 15 cm.
2. The distance of point P from centre O is OP = 20 cm.
3. Since OP (20 cm) > Radius (15 cm), the point lies outside the circle.

(vi) The angle between a line AB and the diameter of a circle is 90°; line AB is:
(a) a tangent of the circle
(b) not a tangent of the circle
(c) nothing can be said

Answer: (a)

Explanation:
A line perpendicular to the diameter at its endpoint on the circle is a tangent to the circle.

(vii) A sector of a circle can be equal to its segment:
(a) Yes
(b) No
(c) none of these two

Answer: (a)

Explanation:
In the case of a semi-circle, the sector (bounded by two radii which form the diameter, and the semi-circular arc) is exactly the same region as the segment (bounded by the diameter chord and the semi-circular arc).

(viii) The given figure shows a circle with centre O. If length of arc AQB is double of arc APB and angle AOB = 65°; then angle AOC is :
(a) 65°
(b) 2 × 65°
(c) 1/2 × 65°
(d) none of these

Answer: (c)

Steps:
Assuming the question asks for the angle at the circumference (Angle ACB) subtended by the same arc AB:
1. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
2. Angle at circumference = (1/2) × Angle at centre.
3. Angle = (1/2) × 65°.

The following questions are Assertion-Reason based questions. Choose your answer based on the codes given below.
(a) Both A and R are correct, and R is the correct explanation for A.
(b) Both A and R are correct, and R is not the correct explanation for A.
(c) A is true, but R is false.
(d) A is false, but R is true.

(ix) Statement 1: The angle between a radius and a tangent of a circle is 90°
Statement 2: At the point of contact, the angle between radius and tangent is equal to one right angle.
Which of the following options is correct?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (a)

Explanation:
Both statements describe the fundamental property of tangents: the radius meets the tangent at a 90° angle (a right angle) at the point of contact.

(x) Assertion (A): The radius of a circle is 20 cm, then the length of its largest chord is also 20 cm.
Reason (R): Longest chord of a circle is equal to its diameter.

Answer: (d)

Explanation:
Reason (R) is true: The longest chord is the diameter. However, Assertion (A) is false because if the radius is 20 cm, the diameter (largest chord) should be 2 × 20 = 40 cm, not 20 cm.

(xi) Assertion (A): When the radius of a circle is increased by 3 cm, its diameter will increase by 2 × 3 cm.
Reason (R): Original radius is r cm, original diameter 2r cm. Increase in length of diameter = 2(r + 3) cm - 2r = 6 cm.

Answer: (a)

Explanation:
Since Diameter = 2 × Radius, increasing the radius by 'x' increases the diameter by '2x'. Both the assertion and the calculation in the reason are correct.

(xii) Assertion (A): If a circle is of radius 8 cm and the length of its minor arc is also 8 cm, then the length of major arc = 8(2π - 1) cm
Reason (R): For every circle, sum of the length of its major arc and length of its minor arc is equal to circumference of the circle.

Answer: (a)

Steps:
1. Circumference = 2πr = 2π(8) = 16π.
2. Major Arc = Circumference - Minor Arc.
3. Major Arc = 16π - 8 = 8(2π - 1).
Both statements are correct and R explains A.

(xiii) Assertion (A): A, B and C are three points on the circumference of a circle such that BC is diameter of the circle. If AC = 4 cm and AB = 3 cm, then diameter BC = 4 cm + 3 cm.
Reason (R): Angle BAC = angle of semi-circle = 90° and so BC² = AB² + AC²

Answer: (d)

Steps:
1. Reason (R) is true: Angle in a semi-circle is 90°, so triangle ABC is a right-angled triangle. By Pythagoras theorem, BC² = AB² + AC².
2. Assertion (A) is false: BC = √(3² + 4²) = √(9 + 16) = √25 = 5 cm. It is not equal to 4 + 3 (7 cm).

2. The following figure shows a circle with centre O. Use the figure to fill in the blanks in each of the following:
(i) AB = Diameter
(ii) Radius = OA and OB
(iii) Chords = AC and BC
(iv) Diameter = AB
(v) AB = 2 × Radius

3. M is a fixed point in a plane and a point P moves in the same plane such that PM = 10 cm.
State:
(i) the name of the figure formed
(ii) the length of the radius of the circle
(iii) the length of the diameter of the circle.
Can a chord of length 16 cm be drawn in this circle? Give reason.

Answer:

(i) The figure formed is a Circle.
(ii) The radius is the constant distance PM, so Radius = 10 cm.
(iii) Diameter = 2 × Radius = 2 × 10 = 20 cm.
Yes, a chord of length 16 cm can be drawn because the longest chord (diameter) is 20 cm, and 16 cm is less than 20 cm.

4. The radius of a circle is 6 cm. Find its diameter. If O is the centre of the circle; state, giving reasons, the positions of points A, B and C; if:
(i) OA = 4.8 cm
(ii) OB = 7.5 cm
(iii) OC = 6 cm

Answer:

Diameter = 2 × Radius = 2 × 6 = 12 cm.
(i) Point A is in the interior of the circle because OA (4.8 cm) < Radius (6 cm).
(ii) Point B is in the exterior of the circle because OB (7.5 cm) > Radius (6 cm).
(iii) Point C is on the circumference of the circle because OC (6 cm) = Radius (6 cm).

5. Fill in the blanks:
(i) An arc is the part of the circumference.
(ii) Diameter of a circle bisects the circle (or circumference).
(iii) The part of the circumference greater than the semi-circle is called major arc.
(iv) Sector of a circle is its region bounded by two radii and an arc.
(v) The segment of a circle is the region bounded by a chord and an arc.
(vi) A tangent of a circle meets the circle at one point.
(vii) The number of tangents that can be drawn through a point on its circumference = one.
(viii) The number of tangents that can be drawn through a point outside the circle is two.

6. O is the centre of a circle with diameter 30 cm. P is a point outside the circle and PA is tangent of the circle. Find:
(i) the length of tangent PA; if OP = 39 cm.
(ii) the distance between O and P, if the length of the tangent PA is 20 cm

Answer:

Given Diameter = 30 cm, Radius (OA) = 15 cm. The tangent is perpendicular to the radius at the point of contact (Angle OAP = 90°).

(i) In right-angled triangle OAP:
OP² = OA² + PA²
39² = 15² + PA²
1521 = 225 + PA²
PA² = 1521 - 225 = 1296
PA = √1296 = 36 cm.

(ii) In right-angled triangle OAP:
OP² = OA² + PA²
OP² = 15² + 20²
OP² = 225 + 400 = 625
OP = √625 = 25 cm.

7. The following figure shows a circle with centre O and a diameter AB.
(i) Name the angle APB.
(ii) State the measure of angle APB.
(iii) If AP = 12 cm and OA = 10 cm; find the lengths of AB and BP.
(iv) If 4AP = 3PB = 12 cm; find the radius of the circle.

Answer:

(i) Angle APB is an angle in a semi-circle.
(ii) Angle APB = 90°.
(iii) OA = 10 cm, so Diameter AB = 2 × 10 = 20 cm.
In right triangle APB (since ∠APB=90°):
AB² = AP² + BP²
20² = 12² + BP²
400 = 144 + BP²
BP² = 256
BP = 16 cm.
(iv) Given 4AP = 12 ⇒ AP = 3 cm. Given 3PB = 12 ⇒ PB = 4 cm.
In right triangle APB:
AB² = AP² + PB²
AB² = 3² + 4² = 9 + 16 = 25
AB = 5 cm.
Radius = AB / 2 = 5 / 2 = 2.5 cm.

8. Find the length of the tangent to a circle with radius 5 cm, from a point at a distance of 13 cm from its centre.

Answer: 12 cm

Steps:
Let radius r = 5 cm, distance d = 13 cm, and length of tangent be L.
Using Pythagoras theorem (since tangent is perpendicular to radius):
d² = r² + L²
13² = 5² + L²
169 = 25 + L²
L² = 144
L = 12 cm.

9. The following figure shows a circle with diameter AB and centre at point O. If angle CAB = 35°; find the measure of angle ABC.

Answer: 55°

Steps:
1. Since AB is the diameter, angle ACB is the angle in a semi-circle.
2. Therefore, ∠ACB = 90°.
3. In triangle ABC, the sum of angles is 180°.
∠ABC + ∠ACB + ∠CAB = 180°
∠ABC + 90° + 35° = 180°
∠ABC = 180° - 125° = 55°.

10. Draw a circle with radius 3 cm and centre at point O. Draw two radii OA and OB of the circle drawn such that ∠AOB = 60°. Join A and B and measure the length of AB.

Answer: Length of AB = 3 cm

Explanation:
1. Draw circle with radius 3 cm.
2. Draw radii OA and OB with angle 60° between them.
3. Since OA = OB (radii) and angle AOB = 60°, triangle AOB is an isosceles triangle with a 60° vertex angle, which makes it an Equilateral Triangle.
4. Therefore, side AB = OA = OB = 3 cm.

11. Draw a circle with radius 4 cm. Draw a chord AB of length 6 cm. Shade its minor segment by horizontal lines and major segment by vertical lines.

Answer:

Steps to Construct:
1. Draw a circle with a radius of 4 cm using a compass.
2. Mark a point A on the circumference.
3. Set the compass width to 6 cm. Place the needle on A and cut an arc on the circle to mark point B.
4. Join A and B to form the chord AB.
5. The smaller region bounded by chord AB and the arc is the minor segment (shade with horizontal lines).
6. The larger region is the major segment (shade with vertical lines).

12. Draw a circle with centre at point P. Draw its radii PA and PB such that angle APB = 90°. Shade the minor sector of the circle by horizontal lines and its major sector by vertical lines.

Answer:

Steps to Construct:
1. Draw a circle with centre P.
2. Draw a radius PA.
3. Construct a radius PB perpendicular to PA (90° angle) at P.
4. The region inside the 90° angle (resembling a slice of pizza) is the minor sector (shade horizontally).
5. The remaining region (270° angle) is the major sector (shade vertically).

13. In a circle with centre O and diameter AB; angle APB is the angle of semi-circle. If PA = PB find the measure of each angle of the triangle APB.

Answer: ∠APB = 90°, ∠PAB = 45°, ∠PBA = 45°

Steps:
1. Since APB is an angle in a semi-circle, ∠APB = 90°.
2. Given PA = PB, triangle APB is an isosceles right-angled triangle.
3. The base angles of an isosceles triangle are equal. ∠PAB = ∠PBA.
4. Sum of angles: 90° + x + x = 180° ⇒ 2x = 90° ⇒ x = 45°.
5. Thus, the angles are 90°, 45°, and 45°.

14. In the following figure, O is the centre of the circle and AB is a diameter. C and D are points on the circumference of the circle such that ∠CAB = 35° and ∠ABD = 55°. Find the measures of angles CAD and CBD.

Answer: ∠CAD = 70° and ∠CBD = 110°

Steps:
1. O is the centre and AB is diameter. Triangles ADB and ACB are in semi-circles, so ∠ADB = 90° and ∠ACB = 90°.
2. In triangle ABD: ∠DAB = 180° - (90° + 55°) = 35°.
3. In triangle ABC: ∠CBA = 180° - (90° + 35°) = 55°.
4. Assuming C and D are on opposite sides of AB (forming cyclic quadrilateral ACBD):
∠CAD = ∠CAB + ∠DAB = 35° + 35° = 70°.
∠CBD = ∠CBA + ∠DBA = 55° + 55° = 110°.

15. The following figure shows a circle with centre O and diameter PQ. Point R lies on the circumference of the circle such that PR = QR and PR = 4 cm. Calculate the radius of the given circle and state its value correct to two decimal places.

Answer: 2.83 cm

Steps:
1. Since PQ is diameter, ∠PRQ = 90°.
2. Triangle PQR is an isosceles right-angled triangle because PR = QR = 4 cm.
3. By Pythagoras theorem: PQ² = PR² + QR² = 4² + 4² = 16 + 16 = 32.
4. PQ = √32 = 4√2 cm.
5. Radius = PQ / 2 = 2√2 cm.
6. Value = 2 × 1.414 = 2.828... ≈ 2.83 cm.

16. In the following figure, PQ and PR are tangents to a circle with centre O. Line segment AC touches the circle at point B.
(i) State the relation between tangents PQ and PR.
(ii) Also, show that: PQ = PA + AB
(iii) PR = PC + CB
(iv) PQ + PR = Perimeter of ΔPAC.

Answer:

(i) PQ = PR (Tangents drawn from an external point to a circle are equal in length).
(ii) A is a point outside the circle. Tangents from A are AQ and AB. So, AQ = AB.
Since A lies on PQ, PQ = PA + AQ. Substituting AQ with AB, we get PQ = PA + AB.
(iii) C is a point outside the circle. Tangents from C are CR and CB. So, CR = CB.
Since C lies on PR, PR = PC + CR. Substituting CR with CB, we get PR = PC + CB.
(iv) Perimeter of ΔPAC = PA + AC + PC.
AC = AB + CB (since B lies on AC).
Perimeter = PA + (AB + CB) + PC = (PA + AB) + (PC + CB).
From (ii) and (iii), this equals PQ + PR.

17. From an exterior point P, the tangent PA is drawn to a circle with centre O.
(i) If OP = 20 cm and tangent PA = 16 cm, find the diameter of the circle.
(ii) If diameter of the circle is 20 cm and tangent PA = 24 cm; find the length of OP.

Answer:

(i) Find Diameter:
In right triangle OAP (radius OA ⊥ tangent PA):
OP² = OA² + PA²
20² = OA² + 16²
400 = OA² + 256
OA² = 144 ⇒ Radius OA = 12 cm.
Diameter = 2 × 12 = 24 cm.

(ii) Find OP:
Diameter = 20 cm ⇒ Radius OA = 10 cm. PA = 24 cm.
OP² = OA² + PA²
OP² = 10² + 24² = 100 + 576 = 676
OP = √676 = 26 cm.

18. In each figure, given below, O is centre of the circle. Use the given informations to find the value of x :
(Due to image clarity issues, only fully readable parts are solved below)

(v) Figure showing radius 8 cm and hypotenuse 10 cm, find x (tangent).

Answer: 6 cm

Steps:
Assuming a standard right triangle formed by radius, tangent, and line to centre:
x² + 8² = 10²
x² + 64 = 100
x² = 36 ⇒ x = 6 cm.