CHAPTER 16: CIRCLE - Questions & Answers

EXERCISE 16(A)

1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) A chord of length 6 cm is drawn in a circle of diameter 10 cm, its distance from the centre of the circle is : (i) 6 cm (ii) 8 cm (iii) 4 cm (iv) 10 cm
Answer:
Step 1: Radius of the circle = Diameter / 2 = 10 / 2 = 5 cm.
Step 2: A perpendicular from the centre to a chord bisects the chord. So, half-chord = 6 / 2 = 3 cm.
Step 3: Using Pythagoras theorem: (Radius)² = (Distance)² + (Half-chord)².
Step 4: 5² = Distance² + 3² => 25 = Distance² + 9 => Distance² = 16.
Step 5: Distance = 4 cm.
Correct Option: (iii) 4 cm

(b) The given figure shows two concentric circles and AD is a chord. The relation between AB and CD is : (i) AB = CD (ii) AB > CD (iii) AB < CD (iv) AB ≠ CD
Answer:
Step 1: Draw a perpendicular from the common centre O to the chord AD, meeting it at M.
Step 2: For the larger circle, the perpendicular bisects the chord AD, so AM = MD.
Step 3: For the smaller circle, the perpendicular bisects the chord BC, so BM = MC.
Step 4: Subtracting the two equations: AM - BM = MD - MC.
Step 5: This leaves AB = CD.
Correct Option: (i) AB = CD

(c) In the given figure, chord AB is larger than chord CD. The relation between OM and ON is : (i) OM = ON (ii) OM < ON (iii) OM > ON (iv) OM + ON = AB
Answer:
Step 1: Recall the theorem: The larger chord of a circle is always closer to the centre.
Step 2: Since chord AB > chord CD, chord AB is closer to the centre O.
Step 3: Therefore, the perpendicular distance OM must be less than the distance ON.
Step 4: Hence, OM < ON.
Correct Option: (ii) OM < ON

(d) The line joining the mid-points of two chords passes through its centre, then the chords are : (i) not parallel to each other (ii) equal to each other (iii) parallel to each other (iv) not equal to each other
Answer:
Step 1: The line segment from the centre of a circle to the mid-point of a chord is perpendicular to the chord.
Step 2: If a single straight line passes through the centre and the mid-points of both chords, it is perpendicular to both.
Step 3: Since both chords are perpendicular to the same straight line, they must be parallel.
Correct Option: (iii) parallel to each other

(e) In the given figure, O and O' are centres of two circles, AB // CD // OO', then which of the following is not true : (i) AB = 2 × OO' (ii) CD = 2 × OO' (iii) AB = CD (iv) AB ≠ CD
Answer:
Step 1: By the theorem for intersecting equal circles, a line passing through an intersection point parallel to the line of centres is bisected by the circles, such that its length is twice the distance between the centres.
Step 2: Therefore, AB = 2 × OO' and CD = 2 × OO'.
Step 3: Since both are equal to 2 × OO', it means AB = CD.
Step 4: The only incorrect statement is that they are not equal.
Correct Option: (iv) AB ≠ CD

2. A chord of length 8 cm is drawn at a distance of 3 cm from the centre of a circle. Calculate the radius of the circle.
Answer:
Step 1: The perpendicular from the centre bisects the chord. Half-chord = 8 / 2 = 4 cm.
Step 2: Let the radius be r. The radius, half-chord, and distance form a right-angled triangle.
Step 3: Applying Pythagoras theorem: r² = 3² + 4².
Step 4: r² = 9 + 16 = 25.
Step 5: r = √25 = 5 cm.
The radius of the circle is 5 cm.

3. The radius of a circle is 17·0 cm and the length of perpendicular drawn from its centre to a chord is 8·0 cm. Calculate the length of the chord.
Answer:
Step 1: Let the half-length of the chord be x. Using Pythagoras theorem: r² = d² + x².
Step 2: 17² = 8² + x².
Step 3: 289 = 64 + x² => x² = 289 - 64 = 225.
Step 4: x = √225 = 15 cm.
Step 5: Length of the total chord = 2 × 15 = 30 cm.
The length of the chord is 30 cm.

4. A chord of length 24 cm is at a distance of 5 cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the centre.
Answer:
Step 1: For the first chord, half-length = 24 / 2 = 12 cm.
Step 2: Find the radius of the circle (r): r² = 5² + 12² = 25 + 144 = 169.
Step 3: r = √169 = 13 cm.
Step 4: For the second chord at 12 cm distance, let its half-length be y.
Step 5: 13² = 12² + y² => 169 = 144 + y² => y² = 25 => y = 5 cm.
Step 6: Total length of the second chord = 2 × y = 2 × 5 = 10 cm.
The length of the chord is 10 cm.

5. In the following figure, AD is a straight line. OP ⊥ AD and O is the centre of both the circles. If OA = 34 cm, OB = 20 cm and OP = 16 cm; find the length of AB.
Answer:
Step 1: In the right-angled triangle OPA (larger circle), AP² = OA² - OP².
Step 2: AP² = 34² - 16² = 1156 - 256 = 900 => AP = 30 cm.
Step 3: In the right-angled triangle OPB (smaller circle), BP² = OB² - OP².
Step 4: BP² = 20² - 16² = 400 - 256 = 144 => BP = 12 cm.
Step 5: From the figure, length of AB = AP - BP.
Step 6: AB = 30 - 12 = 18 cm.
The length of AB is 18 cm.

6. In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn. Find the distance between the chords, if both the chords are : (i) on the opposite sides of the centre, (ii) on the same side of the centre.
Answer:
Step 1: Half-length of the 30 cm chord = 15 cm. Its distance from centre (d1) = √(17² - 15²) = √(289 - 225) = √64 = 8 cm.
Step 2: Half-length of the 16 cm chord = 8 cm. Its distance from centre (d2) = √(17² - 8²) = √(289 - 64) = √225 = 15 cm.
Step 3: (i) If chords are on opposite sides, distance = d1 + d2 = 8 + 15 = 23 cm.
Step 4: (ii) If chords are on the same side, distance = d2 - d1 = 15 - 8 = 7 cm.
Distance on opposite sides is 23 cm and on the same side is 7 cm.

7. Two parallel chords are drawn in a circle of diameter 30·0 cm. The length of one chord is 24·0 cm and the distance between the two chords is 21·0 cm; find the length of the other chord.
Answer:
Step 1: Radius = 30 / 2 = 15 cm.
Step 2: Half-length of the first chord = 24 / 2 = 12 cm.
Step 3: Distance of first chord from centre (d1) = √(15² - 12²) = √(225 - 144) = √81 = 9 cm.
Step 4: Since distance between chords is 21 cm (> 15 cm), they must be on opposite sides.
Step 5: Distance of second chord from centre (d2) = 21 - 9 = 12 cm.
Step 6: Half-length of second chord = √(15² - 12²) = √(225 - 144) = √81 = 9 cm.
Step 7: Full length of the second chord = 9 × 2 = 18 cm.
The length of the other chord is 18 cm.

8. A chord CD of a circle, whose centre is O, is bisected at P by a diameter AB. Given OA = OB = 15 cm and OP = 9 cm. Calculate the lengths of : (i) CD (ii) AD (iii) CB.
Answer:
Step 1: Diameter AB bisects chord CD at P, meaning OP ⊥ CD. Radius OC = 15 cm.
Step 2: (i) In right ΔOPC, CP² = OC² - OP² = 15² - 9² = 225 - 81 = 144 => CP = 12 cm. CD = 2 × 12 = 24 cm.
Step 3: (ii) In right ΔAPD, AP = OA + OP = 15 + 9 = 24 cm. DP = CP = 12 cm. AD = √(AP² + DP²) = √(24² + 12²) = √720 = 12√5 cm.
Step 4: (iii) In right ΔCPB, BP = OB - OP = 15 - 9 = 6 cm. CP = 12 cm. CB = √(BP² + CP²) = √(6² + 12²) = √180 = 6√5 cm.
(i) CD = 24 cm, (ii) AD = 12√5 cm, (iii) CB = 6√5 cm.

9. A straight line is drawn cutting two equal circles and passing through the mid-point M of the line joining their centres O and O'. Prove that the chords AB and CD, which are intercepted by the two circles, are equal.
Answer:
Step 1: Draw perpendiculars OP ⊥ AB and O'Q ⊥ CD.
Step 2: In right-angled ΔOPM and ΔO'QM, ∠OPM = ∠O'QM = 90°.
Step 3: ∠OMP = ∠O'MQ (Vertically opposite angles).
Step 4: OM = O'M (Given, M is the mid-point of OO').
Step 5: By AAS congruence, ΔOPM ≅ ΔO'QM.
Step 6: Therefore, OP = O'Q (by CPCTC).
Step 7: In equal circles, chords equidistant from the centres are equal. Hence, AB = CD.

10. M and N are the mid-points of two equal chords AB and CD respectively of a circle with centre O. Prove that : (i) ∠BMN = ∠DNM, (ii) ∠AMN = ∠CNM.
Answer:
Step 1: Join OM and ON. Since equal chords are equidistant from the centre, OM = ON.
Step 2: Therefore, ΔOMN is isosceles, meaning ∠OMN = ∠ONM.
Step 3: Since M and N are mid-points, OM ⊥ AB and ON ⊥ CD. Thus, ∠OMB = ∠OND = 90°.
Step 4: (i) ∠BMN = ∠OMB - ∠OMN = 90° - ∠OMN. And ∠DNM = ∠OND - ∠ONM = 90° - ∠ONM. Hence, ∠BMN = ∠DNM.
Step 5: (ii) ∠AMN = ∠OMA + ∠OMN = 90° + ∠OMN. And ∠CNM = ∠ONC + ∠ONM = 90° + ∠ONM. Hence, ∠AMN = ∠CNM.

11. Two equal chords AB and CD of a circle with centre O, intersect each other at point P inside the circle. Prove that : (i) AP = CP, (ii) BP = DP
Answer:
Step 1: Draw OM ⊥ AB and ON ⊥ CD. Join OP.
Step 2: Equal chords are equidistant from the centre, so OM = ON.
Step 3: In right ΔOMP and ΔONP, hypotenuse OP is common, and OM = ON.
Step 4: By RHS congruence, ΔOMP ≅ ΔONP. Thus, MP = NP (by CPCTC).
Step 5: Since OM and ON bisect the chords, AM = ½AB and CN = ½CD. Since AB = CD, AM = CN.
Step 6: (i) AP = AM + MP and CP = CN + NP. Therefore, AP = CP.
Step 7: (ii) Since AB = CD and AP = CP, subtracting them gives AB - AP = CD - CP => BP = DP.

12. In the following figure, OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q. Prove that : (HOTS) (i) Δ OPA ≅ Δ OQC, (ii) Δ BPC ≅ Δ BQA.
Answer:
Step 1: For the square OABC, side OA = OC and ∠AOC = 90°.
Step 2: OP and OQ are radii of the same circle, so OP = OQ.
Step 3: (i) In ΔOPA and ΔOQC: OA = OC, OP = OQ, and ∠POA = ∠QOC = 90°. By SAS congruence, ΔOPA ≅ ΔOQC.
Step 4: (ii) From the first proof, AP = CQ (by CPCTC).
Step 5: In ΔBPC and ΔBQA: side BC = BA (square sides), PC = OC - OP and QA = OA - OQ. Since OC=OA and OP=OQ, PC = QA.
Step 6: Also, ∠BCP = ∠BAQ = 90°. By SAS congruence, ΔBPC ≅ ΔBQA.

13. The length of common chord of two intersecting circles is 30 cm. If the diameters of these two circles be 50 cm and 34 cm, calculate the distance between their centres.
Answer:
Step 1: Let the centres be O and O'. The common chord AB = 30 cm. The line joining centres bisects the common chord at M. So AM = 15 cm.
Step 2: Radii are r1 = 50 / 2 = 25 cm, and r2 = 34 / 2 = 17 cm.
Step 3: In right ΔOMA, OM = √(25² - 15²) = √(625 - 225) = √400 = 20 cm.
Step 4: In right ΔO'MA, O'M = √(17² - 15²) = √(289 - 225) = √64 = 8 cm.
Step 5: Total distance between centres = OM + O'M = 20 + 8 = 28 cm.
The distance between their centres is 28 cm.

14. The line joining the mid-points of two chords of a circle passes through its centre. Prove that the chords are parallel.
Answer:
Step 1: Let the chords be AB and CD, and their mid-points M and N. The line passing through centre O is MN.
Step 2: The line segment from the centre to a mid-point is perpendicular to the chord. Thus, OM ⊥ AB (∠OMA = 90°).
Step 3: Similarly, ON ⊥ CD (∠ONC = 90°).
Step 4: Since MN is a straight line transversal, the sum of consecutive interior angles is 90° + 90° = 180°.
Step 5: Hence, the chords AB and CD are parallel.

15. In the following figure, the line ABCD is perpendicular to PQ; where P and Q are the centres of the circles. Show that : (i) AB = CD, (ii) AC = BD.
Answer:
Step 1: Let PQ intersect ABCD at M. Since PQ ⊥ ABCD, PM ⊥ AD and QM ⊥ BC.
Step 2: For the larger circle, perpendicular from centre P bisects chord AD. Thus, AM = MD.
Step 3: For the smaller circle, perpendicular from centre Q bisects chord BC. Thus, BM = MC.
Step 4: (i) Subtract the second from the first: AM - BM = MD - MC => AB = CD.
Step 5: (ii) Adding BC to both sides: AB + BC = CD + BC => AC = BD.

EXERCISE 16(B)

1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) In the given figure, arc APB = arc CQD, then : (i) AB = CD (ii) AB > CD (iii) AB < CD (iv) none of the above
Answer:
Step 1: Equal arcs of a circle subtend equal chords.
Step 2: Since arc APB = arc CQD, their corresponding chords are equal.
Step 3: Therefore, AB = CD.
Correct Option: (i) AB = CD

(b) In the given figure, O is centre of the circle and ∠COD is greater than ∠AOB, then : (i) AB > CD (ii) AB < CD (iii) AB = CD (iv) AB + CD = AD
Answer:
Step 1: In a circle, a greater angle subtended at the centre means a longer corresponding chord.
Step 2: Since ∠COD > ∠AOB, chord CD must be greater than chord AB.
Step 3: This implies AB < CD.
Correct Option: (ii) AB < CD

(c) In a circle, O is its centre and AB, CD are its two chords. If AB : CD = 3 : 2, then ratio between ∠AOB and ∠COD is : (i) 1 : 1 (ii) 3 : 2 (iii) 2 : 5 (iv) 3 : 5
Answer:
Step 1: Assuming the proportional ratio applies to the arcs (a standard class 9 textbook convention), arc AB : arc CD = 3 : 2.
Step 2: The angles subtended at the centre are directly proportional to the arcs.
Step 3: Therefore, ∠AOB : ∠COD = 3 : 2.
Correct Option: (ii) 3 : 2

(d) In the given figure, O is centre of the circle and ABC is an equilateral triangle, then ∠AOB is equal to : (i) 105° (ii) 90° (iii) 60° (iv) 120°
Answer:
Step 1: An equilateral triangle has 3 equal sides, which act as 3 equal chords.
Step 2: Equal chords subtend equal angles at the centre of the circle.
Step 3: The total central angle is 360°, so ∠AOB = 360° / 3 = 120°.
Correct Option: (iv) 120°

(e) In the given figure, O is centre of the circle and chord AB : chord CD = 5 : 3. If angle DOC = 60°; then ∠AOB is : (i) 120° (ii) 75° (iii) 100° (iv) 80°
Answer:
Step 1: Using arc proportionality convention: arc AB : arc CD = 5 : 3, which implies ∠AOB : ∠COD = 5 : 3.
Step 2: Given ∠DOC = 60°, we write ∠AOB / 60° = 5 / 3.
Step 3: ∠AOB = (5 / 3) × 60° = 100°.
Correct Option: (iii) 100°

2. In the given figure, a square is inscribed in a circle with centre O. Find : (i) ∠BOC (ii) ∠OCB (iii) ∠COD (iv) ∠BOD. Is BD a diameter of the circle?
Answer:
Step 1: A square has 4 equal sides. Each side subtends an angle of 360° / 4 = 90° at the centre.
Step 2: (i) ∠BOC = 90°.
Step 3: (ii) In ΔOBC, OB=OC, making it isosceles. ∠OCB = (180° - 90°) / 2 = 45°.
Step 4: (iii) ∠COD = 90°.
Step 5: (iv) ∠BOD = ∠BOC + ∠COD = 90° + 90° = 180°.
Step 6: Since ∠BOD is exactly 180°, it forms a straight line. Yes, BD is a diameter.

3. In the given figure, AB is a side of a regular pentagon and BC is a side of a regular hexagon. Find: (i) ∠AOB (ii) ∠BOC (iii) ∠AOC (iv) ∠OBA (v) ∠OBC (vi) ∠ABC
Answer:
Step 1: (i) Pentagon has 5 sides. ∠AOB = 360° / 5 = 72°.
Step 2: (ii) Hexagon has 6 sides. ∠BOC = 360° / 6 = 60°.
Step 3: (iii) ∠AOC = ∠AOB + ∠BOC = 72° + 60° = 132°.
Step 4: (iv) In isosceles ΔAOB, ∠OBA = (180° - 72°) / 2 = 108° / 2 = 54°.
Step 5: (v) In isosceles ΔOBC, ∠OBC = (180° - 60°) / 2 = 120° / 2 = 60°.
Step 6: (vi) ∠ABC = ∠OBA + ∠OBC = 54° + 60° = 114°.

4. In the given figure, arc AB and arc BC are equal in length. If ∠AOB = 48°, find: (i) ∠BOC (ii) ∠OBC (iii) ∠AOC (iv) ∠OAC
Answer:
Step 1: (i) Equal arcs subtend equal central angles. ∠BOC = ∠AOB = 48°.
Step 2: (ii) In isosceles ΔOBC, ∠OBC = (180° - 48°) / 2 = 132° / 2 = 66°.
Step 3: (iii) ∠AOC = ∠AOB + ∠BOC = 48° + 48° = 96°.
Step 4: (iv) In isosceles ΔOAC, ∠OAC = (180° - 96°) / 2 = 84° / 2 = 42°.

5. In the given figure, the lengths of arcs AB and BC are in the ratio 3 : 2. If ∠AOB = 96°, find: (i) ∠BOC (ii) ∠ABC
Answer:
Step 1: (i) Central angles are proportional to arc lengths: ∠AOB / ∠BOC = 3 / 2. So, 96° / ∠BOC = 3 / 2 => ∠BOC = (96° × 2) / 3 = 64°.
Step 2: (ii) In ΔAOB, ∠OBA = (180° - 96°) / 2 = 42°. In ΔBOC, ∠OBC = (180° - 64°) / 2 = 58°.
Step 3: ∠ABC = ∠OBA + ∠OBC = 42° + 58° = 100°.

6. In the given figure, AB = BC = DC and ∠AOB = 50°. Find : (i) ∠AOC (ii) ∠AOD (iii) ∠BOD (iv) ∠OAC (v) ∠ODA
Answer:
Step 1: Equal chords subtend equal angles. ∠BOC = 50° and ∠COD = 50°.
Step 2: (i) ∠AOC = ∠AOB + ∠BOC = 50° + 50° = 100°.
Step 3: (ii) ∠AOD = ∠AOB + ∠BOC + ∠COD = 150°.
Step 4: (iii) ∠BOD = ∠BOC + ∠COD = 50° + 50° = 100°.
Step 5: (iv) In isosceles ΔOAC, ∠OAC = (180° - 100°) / 2 = 40°.
Step 6: (v) In isosceles ΔOAD, ∠ODA = (180° - 150°) / 2 = 15°.

7. In the given figure, AB is a side of a regular hexagon and AC is a side of a regular eight sided polygon. Find : (i) ∠AOB (ii) ∠AOC (iii) ∠BOC (iv) ∠OBC
Answer:
Step 1: (i) AB is hexagon side. ∠AOB = 360° / 6 = 60°.
Step 2: (ii) AC is octagon side. ∠AOC = 360° / 8 = 45°.
Step 3: (iii) From figure, ∠BOC = ∠AOB + ∠AOC = 60° + 45° = 105°.
Step 4: (iv) In isosceles ΔOBC, ∠OBC = (180° - 105°) / 2 = 75° / 2 = 37.5°.

8. In the given figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If ∠AOB = 100°, find: (i) ∠BOC (ii) ∠OAC
Answer:
Step 1: (i) Central angles are proportional to arcs. ∠AOB = 2 × ∠BOC. 100° = 2 × ∠BOC => ∠BOC = 50°.
Step 2: (ii) ∠AOC = ∠AOB + ∠BOC = 100° + 50° = 150°.
Step 3: In isosceles ΔOAC, ∠OAC = (180° - 150°) / 2 = 15°.

TEST YOURSELF

1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) In a circle with centre at point O, chord AB is a side of a square and chord BC is a side of regular hexagon. Then angle AOC is equal to : (i) 120° (ii) 150° (iii) 90° (iv) none of these
Answer:
Step 1: ∠AOB for square = 360°/4 = 90°.
Step 2: ∠BOC for hexagon = 360°/6 = 60°.
Step 3: ∠AOC = 90° + 60° = 150°.
Correct Option: (ii) 150°

(b) AB (= 20 cm) is diameter of the given circle and AP (= 16 cm). The distance of chord AP from centre O is : (i) 12 cm (ii) 18 cm (iii) 9 cm (iv) 6 cm
Answer:
Step 1: Radius = 20 / 2 = 10 cm.
Step 2: Half-chord = 16 / 2 = 8 cm.
Step 3: Distance = √(10² - 8²) = √(100 - 64) = √36 = 6 cm.
Correct Option: (iv) 6 cm

(c) Given O is centre of the circle with chord AB = 8 cm, OA = 5 cm and OD ⊥ AB. The length of CD is : (i) 3 cm (ii) 5 cm (iii) 2 cm (iv) none of these
Answer:
Step 1: Half-chord AD = 8 / 2 = 4 cm.
Step 2: Distance OD = √(OA² - AD²) = √(5² - 4²) = 3 cm.
Step 3: Radius OC = 5 cm. CD = OC - OD = 5 - 3 = 2 cm.
Correct Option: (iii) 2 cm

(d) AB and CD are chords of a circle with centre O. ∠AOB = 60° and angle ∠COD = 45°; the ratio between the lengths of chords AB and CD is : (i) 3 : 4 (ii) 4 : 3 (iii) 7 : 4 (iv) 7 : 3
Answer:
Step 1: Following the text's proportionality convention for arc-angles, ratio = 60° : 45° = 4 : 3.
Correct Option: (ii) 4 : 3

(e) Statement (1) : O and O' are centres of two equal circles and ABCD is a straight line.
Statement (2) : If OP ⊥ AB, O'Q ⊥ CD and O'Q is greater than OP, then CD > AB
(i) Both the statements are true. (ii) Both the statements are false. (iii) Statement 1 is true, and statement 2 is false. (iv) Statement 1 is false, and statement 2 is true.
Answer:
Step 1: From a straight line through intersection of equal circles, AB = CD implies statement 1 is contextually treated as valid/true in the figure context.
Step 2: Statement 2 states if O'Q > OP, then CD > AB. This is false, because larger distance means a smaller chord.
Correct Option: (iii) Statement 1 is true, and statement 2 is false.

(f) Statement (1) : In a circle with centre O, chord AB : chord BC = 1 : 3. If angle AOC is 160° => angle BOC = 120°.
Statement (2) : AB : BC = 1 : 3 => ∠AOC = 3 × ∠AOB
(i) Both the statements are true. (ii) Both the statements are false. (iii) Statement 1 is true, and statement 2 is false. (iv) Statement 1 is false, and statement 2 is true.
Answer:
Step 1: If ratio is 1:3, ∠AOB=x, ∠BOC=3x. ∠AOC = 4x = 160° => x=40°. ∠BOC = 120°. Statement 1 is True.
Step 2: ∠AOC = 4x, which is 4 × ∠AOB, not 3. Statement 2 is False.
Correct Option: (iii) Statement 1 is true, and statement 2 is false.

(g) Assertion (A) : In the given figure, chord AB = 8 cm, diameter CD = 20 cm, then length of OP = 10 cm.
Reason (R) : OP = √(OA² - AP²) and CP = OC + OP
(i) A is true, R is false. (ii) A is false, R is true. (iii) Both A and R are true and R is the correct reason for A. (iv) Both A and R are true and R is the incorrect reason for A.
Answer:
Step 1: Radius = 10 cm. AP = 4 cm. OP = √(10² - 4²) = √84 ≠ 10 cm. A is false.
Step 2: The formulas given in R are mathematically correct definitions. R is true.
Correct Option: (ii) A is false, R is true.

2. The figure, given below, shows a circle with centre O in which diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4 cm, find the radius of the circle.
Answer:
Step 1: Let the radius be r. Distance OE = OB - EB = r - 4.
Step 2: In right ΔOEC, OC² = OE² + CE² => r² = (r - 4)² + 8².
Step 3: r² = r² - 8r + 16 + 64 => 8r = 80.
Step 4: r = 10 cm.
The radius is 10 cm.

3. In the given figure, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the : (i) radius of the circle. (ii) length of chord CD.
Answer:
Step 1: (i) Half-chord AM = 12 cm. Radius = √(OM² + AM²) = √(5² + 12²) = 13 cm.
Step 2: (ii) For chord CD, half-chord CN = √(r² - ON²) = √(13² - 12²) = 5 cm.
Step 3: Length of CD = 2 × 5 = 10 cm.
Radius is 13 cm and CD is 10 cm.

4. AB and CD are two equal chords of a circle with centre O which intersect each other at right angle at point P. If OM ⊥ AB and ON ⊥ CD; show that OMPN is a square.
Answer:
Step 1: Since AB = CD, they are equidistant from the centre. Hence, OM = ON.
Step 2: In quadrilateral OMPN, ∠OMP = 90°, ∠ONP = 90°, and ∠MPN = 90° (given).
Step 3: The fourth angle ∠MON is also 360° - 270° = 90°.
Step 4: A rectangle with equal adjacent sides (OM = ON) is a square.

5. The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centres.
Answer:
Step 1: Half-chord = 24 / 2 = 12 cm.
Step 2: Distance = √(Radius² - half-chord²) = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.
Distance is 5 cm.

6. Prove that equal chords of congruent circles subtend equal angles at their centres.
Answer:
Step 1: Let the chords be AB and C'D' with centres O and O'. AB = C'D'.
Step 2: OA = OB = O'C' = O'D' (radii of congruent circles).
Step 3: By SSS congruence criterion, ΔAOB ≅ ΔC'O'D'.
Step 4: By CPCTC, ∠AOB = ∠C'O'D'.

7. Draw two circles of different radii. How many points these circles can have in common? What is the maximum number of common points ?
Answer:
Step 1: Two non-overlapping circles have 0 common points.
Step 2: Touching circles have 1 common point.
Step 3: Intersecting circles have exactly 2 common points.
Step 4: The maximum number of common points is 2.

8. Suppose you are given a circle. Describe a method by which you can find the centre of this circle.
Answer:
Step 1: Draw any two non-parallel chords on the given circle.
Step 2: Draw the perpendicular bisectors of both these chords.
Step 3: The point where these two perpendicular bisectors intersect is the centre of the circle.

9. Given two equal chords AB and CD of a circle, with centre O, intersecting each other at point P. Prove that : (i) AP = CP (ii) BP = DP
Answer:
Step 1: Draw OM ⊥ AB and ON ⊥ CD. Since AB = CD, OM = ON.
Step 2: By RHS congruence, ΔOMP ≅ ΔONP, leading to MP = NP.
Step 3: AM = CN (half of equal chords).
Step 4: (i) AP = AM + MP = CN + NP = CP.
Step 5: (ii) BP = AB - AP = CD - CP = DP.

10. In a cricle of radius 10 cm, AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively. Calculate the distance between the chords, if they are on : (i) the same side of the centre. (ii) the opposite sides of the centre.
Answer:
Step 1: Distances from centre = √(10² - 8²) = 6 cm and √(10² - 6²) = 8 cm.
Step 2: (i) Same side distance = 8 - 6 = 2 cm.
Step 3: (ii) Opposite side distance = 8 + 6 = 14 cm.
Distances are 2 cm and 14 cm.

11. In the given figure, O is the centre of the circle with radius 20 cm and OD is perpendicular to AB. If AB = 32 cm, find the length of CD.
Answer:
Step 1: Radius OC = 20 cm. Half-chord AC = 32 / 2 = 16 cm.
Step 2: Distance OD = √(20² - 16²) = √(400 - 256) = √144 = 12 cm.
Step 3: CD = Radius (OC) - OD = 20 - 12 = 8 cm.
The length of CD is 8 cm.

12. In the given figure, AB and CD are two equal chords of a circle, with centre O. If P is the mid-point of chord AB, Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.
Answer:
Step 1: OP ⊥ AB and OQ ⊥ CD. Since AB = CD, OP = OQ.
Step 2: ΔOPQ is isosceles. ∠OPQ = (180° - 150°) / 2 = 15°.
Step 3: ∠APO = 90°.
Step 4: ∠APQ = ∠APO - ∠OPQ = 90° - 15° = 75°.
The angle is 75°.

13. In the given figure, AOC is the diameter of the circle, with centre O. If arc AXB is half of arc BYC, find ∠BOC.
Answer:
Step 1: The ratio of arc AXB to arc BYC is 1 : 2. So, ∠AOB : ∠BOC = 1 : 2.
Step 2: Let ∠AOB = x and ∠BOC = 2x.
Step 3: AOC is a straight line, so x + 2x = 180° => 3x = 180° => x = 60°.
Step 4: ∠BOC = 2x = 2 × 60° = 120°.
The angle is 120°.

14. The circumference of a circle, with centre O, is divided into three arcs APB, BQC and CRA such that : arc APB / 2 = arc BQC / 3 = arc CRA / 4. Find ∠BOC. (HOTS)
Answer:
Step 1: The arcs are in the ratio 2 : 3 : 4.
Step 2: Total parts = 2 + 3 + 4 = 9 parts. These correspond to 360°.
Step 3: ∠BOC is subtended by arc BQC, which represents 3 parts.
Step 4: ∠BOC = (3 / 9) × 360° = 120°.
The angle is 120°.