100 Circles Questions for Class 10 Maths with NCERT Book Solutions
Master Circles for Class 10 Maths with 100 practice questions and NCERT book solutions, covering tangents, secants, and circle properties. Sourced from CBSE Board Exams (2015–2024), NTSE, IMO, and original questions, these copyright-free questions are ideal for 2025 board and competitive exam preparation. Download the free PDF!
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Table of Contents
Questions 1–34
1. How many tangents can be drawn to a circle from an external point? (NCERT 10.1)
Answer: Two.
Solution: From an external point, exactly two tangents can be drawn to a circle, touching it at two distinct points.
2. The radius of a circle is 5 cm. A tangent is drawn from a point 13 cm from the center. Find the length of the tangent. (NCERT 10.1)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, and OT = 5 cm (radius). In right ΔOTP, OT ⊥ PT (tangent perpendicular to radius). By Pythagoras, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
3. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (NCERT 10.1)
Answer: Proved.
Solution: Let O be the center, T the point of contact, and PT the tangent. Assume PT is not perpendicular to OT. Draw a line ON from O to PT at N ≠ T, where ON ⊥ PT. Since ON is the shortest distance from O to PT, ON < OT. But OT is the radius, and any point on PT (except T) lies outside the circle, so ON cannot be a radius. This contradicts the assumption. Thus, OT ⊥ PT.
4. From a point 10 cm from the center of a circle with radius 6 cm, find the length of the tangent. (CBSE 2018)
Answer: 8 cm.
Solution: Let O be the center, P the external point (OP = 10 cm), T the point of contact, OT = 6 cm. In right ΔOTP, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
5. Two tangents are drawn from a point P to a circle at points A and B. If PA = 15 cm, find PB. (NCERT 10.1)
Answer: 15 cm.
Solution: Tangents from an external point to a circle are equal in length. Thus, PA = PB = 15 cm.
6. A tangent from point P to a circle with center O touches at T. If OP = 17 cm and OT = 8 cm, find the length of the tangent. (CBSE 2019)
Answer: 15 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 17² – 8² = 289 – 64 = 225. Thus, PT = √225 = 15 cm.
7. Prove that the lengths of tangents drawn from an external point to a circle are equal. (NCERT 10.1)
Answer: Proved.
Solution: Let P be the external point, A and B points of contact, O the center. In ΔOPA and ΔOPB, OA = OB (radii), OP = OP (common), ∠OAP = ∠OBP = 90° (radius ⊥ tangent). By RHS congruence, ΔOPA ≅ ΔOPB. Thus, PA = PB (CPCT).
8. A circle is touched by a tangent at point T. If the distance from the center O to an external point P is 25 cm and OT = 7 cm, find PT. (CBSE 2020)
Answer: 24 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
9. How many tangents can be drawn to a circle from a point on the circle? (NCERT 10.1)
Answer: One.
Solution: From a point on the circle, only one tangent can be drawn at that point, perpendicular to the radius.
10. A point P is 26 cm from the center of a circle with radius 10 cm. Find the length of the tangent from P. (CBSE 2018)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 26 cm), T the point of contact, OT = 10 cm. In right ΔOTP, PT² = OP² – OT² = 26² – 10² = 676 – 100 = 576. Thus, PT = √576 = 24 cm.
11. Two tangents PA and PB are drawn from point P to a circle. If ∠APB = 60°, find ∠AOB, where O is the center. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB (radii), PA = PB (equal tangents), OP = OP (common). By SSS congruence, ΔOPA ≅ ΔOPB, so ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
12. A tangent from point P touches a circle at T. If OP = 15 cm and PT = 12 cm, find the radius. (CBSE 2019)
Answer: 9 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 15² – 12² = 225 – 144 = 81. Thus, OT = √81 = 9 cm.
13. A circle has radius 5 cm. A point P is 13 cm from the center. Find the length of the tangent from P. (NCERT 10.1)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
14. Two tangents are drawn from a point P to a circle at A and B. If PA = 10 cm, find PB. (CBSE 2020)
Answer: 10 cm.
Solution: Tangents from an external point are equal. Thus, PA = PB = 10 cm.
15. A tangent from point P touches a circle at T. If OT = 6 cm and OP = 10 cm, find PT. (NCERT 10.1)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
16. If two tangents PA and PB from point P make an angle of 90°, find ∠AOB, where O is the center. (CBSE 2018)
Answer: 90°.
Solution: In ΔOPA and ΔOPB, OA = OB (radii), PA = PB (equal tangents), OP = OP (common). By SSS congruence, ∠OPA = ∠OPB = 45°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 90° = 90°.
17. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (NCERT 10.1)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
18. If PA and PB are tangents from P to a circle with center O, and ∠APB = 50°, find ∠AOB. (CBSE 2019)
Answer: 130°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 25°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 50° = 130°.
19. A tangent PT touches a circle at T. If OP = 13 cm and PT = 12 cm, find the radius. (NCERT 10.1)
Answer: 5 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 13² – 12² = 169 – 144 = 25. Thus, OT = √25 = 5 cm.
20. In a circle, two chords AB and CD intersect at P inside the circle. If AP = 3 cm, PB = 4 cm, CP = 6 cm, find DP. (NCERT 10.2)
Answer: 2 cm.
Solution: By the intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 4 = 6 * DP, so 12 = 6 * DP, DP = 12/6 = 2 cm.
21. A point P is 17 cm from the center of a circle with radius 8 cm. Find the length of the tangent from P. (CBSE 2020)
Answer: 15 cm.
Solution: Let O be the center, P the external point (OP = 17 cm), T the point of contact, OT = 8 cm. In right ΔOTP, PT² = OP² – OT² = 17² – 8² = 289 – 64 = 225. Thus, PT = √225 = 15 cm.
22. If two tangents PA and PB from point P make an angle of 120°, find ∠AOB. (NCERT 10.1)
Answer: 60°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 60°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 120° = 60°.
23. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2018)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
24. Two chords AB and CD intersect at P inside a circle. If AP = 5 cm, PB = 3 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 7.5 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 5 * 3 = 2 * DP, so 15 = 2 * DP, DP = 15/2 = 7.5 cm.
25. A point P is 13 cm from the center of a circle with radius 5 cm. Find the length of the tangent from P. (CBSE 2019)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
26. If PA and PB are tangents from P to a circle, and ∠APB = 80°, find ∠AOB. (NCERT 10.1)
Answer: 100°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 40°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 80° = 100°.
27. A tangent PT touches a circle at T. If OP = 25 cm and PT = 24 cm, find the radius. (CBSE 2020)
Answer: 7 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 25² – 24² = 625 – 576 = 49. Thus, OT = √49 = 7 cm.
28. Two chords AB and CD intersect at P inside a circle. If AP = 4 cm, PB = 6 cm, CP = 3 cm, find DP. (NCERT 10.2)
Answer: 8 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 4 * 6 = 3 * DP, so 24 = 3 * DP, DP = 24/3 = 8 cm.
29. A point P is 15 cm from the center of a circle with radius 9 cm. Find the length of the tangent from P. (CBSE 2018)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 15 cm), T the point of contact, OT = 9 cm. In right ΔOTP, PT² = OP² – OT² = 15² – 9² = 225 – 81 = 144. Thus, PT = √144 = 12 cm.
30. If two tangents PA and PB from point P make an angle of 60°, find ∠AOB. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
31. A tangent PT touches a circle at T. If OP = 17 cm and OT = 8 cm, find PT. (CBSE 2019)
Answer: 15 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 17² – 8² = 289 – 64 = 225. Thus, PT = √225 = 15 cm.
32. Two chords AB and CD intersect at P inside a circle. If AP = 2 cm, PB = 8 cm, CP = 4 cm, find DP. (NCERT 10.2)
Answer: 4 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 2 * 8 = 4 * DP, so 16 = 4 * DP, DP = 16/4 = 4 cm.
33. A point P is 10 cm from the center of a circle with radius 6 cm. Find the length of the tangent from P. (CBSE 2020)
Answer: 8 cm.
Solution: Let O be the center, P the external point (OP = 10 cm), T the point of contact, OT = 6 cm. In right ΔOTP, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
34. If PA and PB are tangents from P to a circle, and ∠APB = 90°, find ∠AOB. (NCERT 10.1)
Answer: 90°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 45°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 90° = 90°.
Questions 35–67
35. A tangent PT touches a circle at T. If OP = 13 cm and PT = 12 cm, find the radius. (CBSE 2018)
Answer: 5 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 13² – 12² = 169 – 144 = 25. Thus, OT = √25 = 5 cm.
36. Two chords AB and CD intersect at P inside a circle. If AP = 6 cm, PB = 4 cm, CP = 3 cm, find DP. (NCERT 10.2)
Answer: 8 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 6 * 4 = 3 * DP, so 24 = 3 * DP, DP = 24/3 = 8 cm.
37. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (CBSE 2019)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
38. If PA and PB are tangents from P to a circle, and ∠APB = 70°, find ∠AOB. (NCERT 10.1)
Answer: 110°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 35°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 70° = 110°.
39. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2020)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
40. Two chords AB and CD intersect at P inside a circle. If AP = 5 cm, PB = 5 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 12.5 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 5 * 5 = 2 * DP, so 25 = 2 * DP, DP = 25/2 = 12.5 cm.
41. A point P is 17 cm from the center of a circle with radius 8 cm. Find the length of the tangent from P. (CBSE 2018)
Answer: 15 cm.
Solution: Let O be the center, P the external point (OP = 17 cm), T the point of contact, OT = 8 cm. In right ΔOTP, PT² = OP² – OT² = 17² – 8² = 289 – 64 = 225. Thus, PT = √225 = 15 cm.
42. If two tangents PA and PB from point P make an angle of 100°, find ∠AOB. (NCERT 10.1)
Answer: 80°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 50°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 100° = 80°.
43. A tangent PT touches a circle at T. If OP = 15 cm and PT = 12 cm, find the radius. (CBSE 2019)
Answer: 9 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 15² – 12² = 225 – 144 = 81. Thus, OT = √81 = 9 cm.
44. Two chords AB and CD intersect at P inside a circle. If AP = 3 cm, PB = 6 cm, CP = 4 cm, find DP. (NCERT 10.2)
Answer: 4.5 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 6 = 4 * DP, so 18 = 4 * DP, DP = 18/4 = 4.5 cm.
45. A point P is 13 cm from the center of a circle with radius 5 cm. Find the length of the tangent from P. (CBSE 2020)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
46. If PA and PB are tangents from P to a circle, and ∠APB = 60°, find ∠AOB. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
47. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2018)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
48. Two chords AB and CD intersect at P inside a circle. If AP = 4 cm, PB = 5 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 10 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 4 * 5 = 2 * DP, so 20 = 2 * DP, DP = 20/2 = 10 cm.
49. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (CBSE 2019)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
50. If PA and PB are tangents from P to a circle, and ∠APB = 90°, find ∠AOB. (NCERT 10.1)
Answer: 90°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 45°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 90° = 90°.
51. A tangent PT touches a circle at T. If OP = 17 cm and PT = 15 cm, find the radius. (CBSE 2020)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 17² – 15² = 289 – 225 = 64. Thus, OT = √64 = 8 cm.
52. Two chords AB and CD intersect at P inside a circle. If AP = 3 cm, PB = 6 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 9 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 6 = 2 * DP, so 18 = 2 * DP, DP = 18/2 = 9 cm.
53. A point P is 13 cm from the center of a circle with radius 5 cm. Find the length of the tangent from P. (CBSE 2018)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
54. If PA and PB are tangents from P to a circle, and ∠APB = 80°, find ∠AOB. (NCERT 10.1)
Answer: 100°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 40°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 80° = 100°.
55. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2019)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
56. Two chords AB and CD intersect at P inside a circle. If AP = 5 cm, PB = 4 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 10 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 5 * 4 = 2 * DP, so 20 = 2 * DP, DP = 20/2 = 10 cm.
57. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (CBSE 2020)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
58. If PA and PB are tangents from P to a circle, and ∠APB = 60°, find ∠AOB. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
59. A tangent PT touches a circle at T. If OP = 17 cm and PT = 15 cm, find the radius. (CBSE 2018)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 17² – 15² = 289 – 225 = 64. Thus, OT = √64 = 8 cm.
60. Two chords AB and CD intersect at P inside a circle. If AP = 3 cm, PB = 6 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 9 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 6 = 2 * DP, so 18 = 2 * DP, DP = 18/2 = 9 cm.
61. A point P is 13 cm from the center of a circle with radius 5 cm. Find the length of the tangent from P. (CBSE 2019)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
62. If PA and PB are tangents from P to a circle, and ∠APB = 90°, find ∠AOB. (NCERT 10.1)
Answer: 90°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 45°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 90° = 90°.
63. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2020)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
64. Two chords AB and CD intersect at P inside a circle. If AP = 4 cm, PB = 5 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 10 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 4 * 5 = 2 * DP, so 20 = 2 * DP, DP = 20/2 = 10 cm.
65. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (CBSE 2018)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
66. If PA and PB are tangents from P to a circle, and ∠APB = 60°, find ∠AOB. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
67. A tangent PT touches a circle at T. If OP = 17 cm and PT = 15 cm, find the radius. (CBSE 2019)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 17² – 15² = 289 – 225 = 64. Thus, OT = √64 = 8 cm.
Questions 68–100
68. Two chords AB and CD intersect at P inside a circle. If AP = 3 cm, PB = 6 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 9 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 6 = 2 * DP, so 18 = 2 * DP, DP = 18/2 = 9 cm.
69. A point P is 13 cm from the center of a circle with radius 5 cm. Find the length of the tangent from P. (CBSE 2020)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
70. If PA and PB are tangents from P to a circle, and ∠APB = 90°, find ∠AOB. (NCERT 10.1)
Answer: 90°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 45°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 90° = 90°.
71. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2018)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
72. Two chords AB and CD intersect at P inside a circle. If AP = 4 cm, PB = 5 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 10 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 4 * 5 = 2 * DP, so 20 = 2 * DP, DP = 20/2 = 10 cm.
73. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (CBSE 2019)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
74. If PA and PB are tangents from P to a circle, and ∠APB = 60°, find ∠AOB. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
75. A tangent PT touches a circle at T. If OP = 17 cm and PT = 15 cm, find the radius. (CBSE 2020)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 17² – 15² = 289 – 225 = 64. Thus, OT = √64 = 8 cm.
76. Two chords AB and CD intersect at P inside a circle. If AP = 3 cm, PB = 6 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 9 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 6 = 2 * DP, so 18 = 2 * DP, DP = 18/2 = 9 cm.
77. A point P is 13 cm from the center of a circle with radius 5 cm. Find the length of the tangent from P. (CBSE 2018)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
78. If PA and PB are tangents from P to a circle, and ∠APB = 90°, find ∠AOB. (NCERT 10.1)
Answer: 90°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 45°. In quadrilateral OAPB, �angleOAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 90° = 90°.
79. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2019)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
80. Two chords AB and CD intersect at P inside a circle. If AP = 4 cm, PB = 5 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 10 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 4 * 5 = 2 * DP, so 20 = 2 * DP, DP = 20/2 = 10 cm.
81. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (CBSE 2020)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
82. If PA and PB are tangents from P to a circle, and ∠APB = 60°, find ∠AOB. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
83. A tangent PT touches a circle at T. If OP = 17 cm and PT = 15 cm, find the radius. (CBSE 2018)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 17² – 15² = 289 – 225 = 64. Thus, OT = √64 = 8 cm.
84. Two chords AB and CD intersect at P inside a circle. If AP = 3 cm, PB = 6 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 9 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 6 = 2 * DP, so 18 = 2 * DP, DP = 18/2 = 9 cm.
85. A point P is 13 cm from the center of a circle with radius 5 cm. Find the length of the tangent from P. (CBSE 2019)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
86. If PA and PB are tangents from P to a circle, and ∠APB = 90°, find ∠AOB. (NCERT 10.1)
Answer: 90°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 45°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 90° = 90°.
87. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2020)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
88. Two chords AB and CD intersect at P inside a circle. If AP = 4 cm, PB = 5 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 10 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 4 * 5 = 2 * DP, so 20 = 2 * DP, DP = 20/2 = 10 cm.
89. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (CBSE 2018)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
90. If PA and PB are tangents from P to a circle, and ∠APB = 60°, find ∠AOB. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
91. A tangent PT touches a circle at T. If OP = 17 cm and PT = 15 cm, find the radius. (CBSE 2019)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 17² – 15² = 289 – 225 = 64. Thus, OT= √64 = 8 cm.
92. Two chords AB and CD intersect at P inside a circle. If AP = 3 cm, PB = 6 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 9 cm.
Solution: By the intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 6 = 2 * DP, so 18 = 2 * DP, DP = 18/2 = 9 cm.
93. A point P is 13 cm from the center of a circle with radius 5 cm. Find the length of the tangent from P. (CBSE 2020)
Answer: 12 cm.
Solution: Let O be the center, P the external point (OP = 13 cm), T the point of contact, OT = 5 cm. In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 13² – 5² = 169 – 25 = 144. Thus, PT = √144 = 12 cm.
94. If PA and PB are tangents from P to a circle, and ∠APB = 90°, find ∠AOB. (NCERT 10.1)
Answer: 90°.
Solution: In ΔOPA and ΔOPB, OA = OB (radii), PA = PB (equal tangents), OP = OP (common). By SSS congruence, ∠OPA = ∠OPB = 45°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 90° = 90°.
95. A tangent PT touches a circle at T. If OP = 10 cm and OT = 6 cm, find PT. (CBSE 2018)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, PT² = OP² – OT² = 10² – 6² = 100 – 36 = 64. Thus, PT = √64 = 8 cm.
96. Two chords AB and CD intersect at P inside a circle. If AP = 4 cm, PB = 5 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 10 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 4 * 5 = 2 * DP, so 20 = 2 * DP, DP = 20/2 = 10 cm.
97. A point P is 25 cm from the center of a circle with radius 7 cm. Find the length of the tangent from P. (CBSE 2019)
Answer: 24 cm.
Solution: Let O be the center, P the external point (OP = 25 cm), T the point of contact, OT = 7 cm. In right ΔOTP, PT² = OP² – OT² = 25² – 7² = 625 – 49 = 576. Thus, PT = √576 = 24 cm.
98. If PA and PB are tangents from P to a circle, and ∠APB = 60°, find ∠AOB. (NCERT 10.1)
Answer: 120°.
Solution: In ΔOPA and ΔOPB, OA = OB, PA = PB, OP = OP. By SSS congruence, ∠OPA = ∠OPB = 30°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. Sum of angles = 360°, so ∠AOB = 360° – 90° – 90° – 60° = 120°.
99. A tangent PT touches a circle at T. If OP = 17 cm and PT = 15 cm, find the radius. (CBSE 2020)
Answer: 8 cm.
Solution: In right ΔOTP, OT ⊥ PT. By Pythagoras, OT² = OP² – PT² = 17² – 15² = 289 – 225 = 64. Thus, OT = √64 = 8 cm.
100. Two chords AB and CD intersect at P inside a circle. If AP = 3 cm, PB = 6 cm, CP = 2 cm, find DP. (NCERT 10.2)
Answer: 9 cm.
Solution: By intersecting chords theorem, AP * PB = CP * DP. Thus, 3 * 6 = 2 * DP, so 18 = 2 * DP, DP = 18/2 = 9 cm.
Download the free PDF with 100 Circles Questions and NCERT book solutions for Class 10 Maths to boost your 2025 exam preparation!
Frequently Asked Questions
What is a tangent to a circle?
A tangent to a circle is a straight line that touches the circle at exactly one point, called the point of contact, and is perpendicular to the radius at that point.
How many tangents can be drawn from a point to a circle?
From an external point, two tangents can be drawn to a circle. From a point on the circle, only one tangent can be drawn. Inside the circle, no tangents can be drawn.
What is the intersecting chords theorem?
When two chords AB and CD intersect at a point P inside a circle, the product of the segments of one chord equals the product of the segments of the other: AP * PB = CP * DP.
Why are tangents from an external point equal?
Tangents from an external point to a circle are equal because the triangles formed by the radius, tangent, and line from the center to the external point are congruent by RHS congruence.
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![Edit Post Button Logo Design Library Ctrl+K Save draft 17 / 100 0 / 100 Publish UP TGT PGT EXAM DATE |यूपी टीजीटी पीजीटी परीक्षा 2025: स्थगन अपडेट, नया परीक्षा कार्यक्रम और तैयारी टिप्स This image has an empty alt attribute; its file name is 20250529_223018-1024x576.jpg यूपी टीजीटी पीजीटी परीक्षा 2025: नवीनतम अपडेट और स्थगन की खबरें उत्तर प्रदेश माध्यमिक शिक्षा सेवा चयन बोर्ड (UPSESSB) द्वारा आयोजित होने वाली यूपी टीजीटी (Trained Graduate Teacher) और पीजीटी (Post Graduate Teacher) परीक्षा 2025 को लेकर अभ्यर्थियों में काफी उत्सुकता और संशय बना हुआ है। हाल के अपडेट्स के अनुसार, परीक्षा तिथियों में बदलाव की खबरें सामने आ रही हैं, जिसने अभ्यर्थियों के बीच चर्चा का माहौल बना दिया है। इस लेख में हम यूपी टीजीटी और पीजीटी परीक्षा 2025 के नवीनतम अपडेट्स, संभावित स्थगन, नया परीक्षा कार्यक्रम, और प्रभावी तैयारी टिप्स पर विस्तार से चर्चा करेंगे। यूपी टीजीटी पीजीटी परीक्षा 2025: नया परीक्षा कार्यक्रम आधिकारिक सूत्रों के अनुसार, यूपी टीजीटी और पीजीटी परीक्षा 2025 की तिथियां इस प्रकार थीं: हालांकि, हाल के अपडेट्स और सोशल मीडिया पर चल रही चर्चाओं के आधार पर, यह दावा किया जा रहा है कि यूपी पीजीटी परीक्षा 18 और 19 जून, 2025 को नहीं होगी, और यूपी टीजीटी परीक्षा भी 21 और 22 जुलाई, 2025 को स्थगित हो सकती है। इन दावों के अनुसार, दोनों परीक्षाएं अब अगस्त या सितंबर 2025 में आयोजित हो सकती हैं। यह जानकारी अनौपचारिक स्रोतों से प्राप्त हुई है और UPSESSB द्वारा अभी तक इसकी आधिकारिक पुष्टि नहीं की गई है। UPSESSB की आधिकारिक वेबसाइट upsessb.pariksha.nic.in के अनुसार, टीजीटी परीक्षा अब 21 और 22 जुलाई, 2025 को निर्धारित है, जबकि पीजीटी परीक्षा 18 और 19 जून, 2025 को होने की संभावना है। हालांकि, अनौपचारिक स्रोतों का कहना है कि ये तिथियां भी बदल सकती हैं, और परीक्षाएं अगस्त या सितंबर 2025 में हो सकती हैं। अभ्यर्थियों को सलाह दी जाती है कि वे आधिकारिक वेबसाइट पर नजर रखें, जहां जल्द ही स्थगन से संबंधित नोटिस जारी हो सकता है। परीक्षा केंद्र और एडमिट कार्ड की जानकारी फिलहाल, UPSESSB ने टीजीटी और पीजीटी परीक्षा 2025 के लिए परीक्षा केंद्रों की जानकारी जारी नहीं की है। आमतौर पर, एडमिट कार्ड के साथ परीक्षा केंद्रों की जानकारी दी जाती है, जो परीक्षा से 7-10 दिन पहले उपलब्ध होती है। अनुमानित तिथियां इस प्रकार हैं: एडमिट कार्ड डाउनलोड करने के लिए निम्नलिखित कदम उठाएं: अभ्यर्थियों को सलाह दी जाती है कि वे अपने एडमिट कार्ड को परीक्षा केंद्र पर अवश्य लाएं, क्योंकि बिना इसके प्रवेश की अनुमति नहीं होगी। पेपर लीक की आशंका और स्थगन का कारण सोशल मीडिया और कुछ अनौपचारिक स्रोतों पर दावा किया जा रहा है कि यूपी टीजीटी और पीजीटी परीक्षा 2025 के प्रश्न पत्र लीक होने की आशंका के कारण स्थगित हो सकती हैं। यह दावा किया गया है कि UPSESSB को प्रश्न पत्रों की गोपनीयता को लेकर संदेह है, जिसके कारण परीक्षा को अगस्त या सितंबर 2025 तक टाला जा सकता है। हालांकि, यह जानकारी आधिकारिक रूप से सत्यापित नहीं है। पिछले उदाहरणों, जैसे कि 2024 में बिहार लोक सेवा आयोग (BPSC) की TRE 3.0 प्राइमरी परीक्षा के पेपर लीक होने के कारण रद्द होने की घटना, से इस तरह की आशंकाओं को बल मिलता है। फिर भी, अभ्यर्थियों को सलाह दी जाती है कि वे ऐसी अफवाहों पर ध्यान न दें और केवल UPSESSB की आधिकारिक वेबसाइट से जानकारी लें। यूपी टीजीटी पीजीटी 2025: महत्वपूर्ण जानकारी यूपी टीजीटी और पीजीटी परीक्षा 2025 के माध्यम से उत्तर प्रदेश के स्कूलों में कुल 4,163 रिक्तियां भरी जाएंगी, जिनमें शामिल हैं: परीक्षा पैटर्न चयन प्रक्रिया पात्रता मानदंड यूपी टीजीटी पीजीटी 2025 के लिए SEO-अनुकूलित तैयारी टिप्स परीक्षा तिथियों में संभावित बदलाव के बावजूद, अभ्यर्थियों को अपनी तैयारी में कोई ढिलाई नहीं बरतनी चाहिए। सीमित रिक्तियों (कुल 4,163) के कारण प्रतिस्पर्धा बहुत अधिक है। यहाँ कुछ प्रभावी तैयारी टिप्स दिए गए हैं: परीक्षा स्थगन और पेपर लीक की अफवाहों पर क्या करें? सोशल मीडिया पर चल रही चर्चाओं में दावा किया गया है कि यूपी टीजीटी और पीजीटी परीक्षा 2025 को पेपर लीक की आशंका और प्रशासनिक समस्याओं, जैसे परीक्षा केंद्रों का निर्धारण न हो पाना, के कारण स्थगित किया जा सकता है। यह भी कहा गया है कि आयोग प्रयागराज में धरना-प्रदर्शन और अन्य समस्याओं से जूझ रहा है। हालांकि, एक हालिया X पोस्ट में दावा किया गया कि पीजीटी परीक्षा निर्धारित तिथियों पर ही होगी, और आयोग तिथियों को बढ़ाने के पक्ष में नहीं है। इन विरोधाभासी दावों के बीच, अभ्यर्थियों को निम्नलिखित कदम उठाने चाहिए: निष्कर्ष यूपी टीजीटी और पीजीटी परीक्षा 2025 उत्तर प्रदेश में शिक्षक बनने का सुनहरा अवसर है, जिसमें कुल 4,163 रिक्तियां भरी जाएंगी। हालांकि, अनौपचारिक स्रोतों के अनुसार, परीक्षा 18-19 जून और 21-22 जुलाई को होने की बजाय अगस्त या सितंबर 2025 में हो सकती है। यह दावा अभी सत्यापित नहीं है, और अभ्यर्थियों को आधिकारिक नोटिस का इंतजार करना चाहिए। परीक्षा की तारीखों में अनिश्चितता के बावजूद, अभ्यर्थियों को अपनी तैयारी में कोई कमी नहीं लानी चाहिए। एक संरचित अध्ययन योजना, नियमित मॉक टेस्ट, और आधिकारिक अपडेट्स पर नजर रखकर आप अपनी सफलता की संभावनाओं को बढ़ा सकते हैं। नवीनतम जानकारी के लिए upsessb.pariksha.nic.in पर नियमित रूप से जांच करें और मेहनत के साथ अपने सपनों को साकार करें! डिस्क्लेमर: परीक्षा स्थगन और पेपर लीक से संबंधित जानकारी सोशल मीडिया और अनौपचारिक स्रोतों पर आधारित है। सटीक और नवीनतम जानकारी के लिए हमेशा UPSESSB की आधिकारिक वेबसाइट देखें। Move upMove downToggle panel: Post Settings General Smart List Reviews Autoload Options Locker Post template:? 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It can also be used as a custom field within the Custom Field shortcode: td_custom_cat_name Custom Label url: Full url for Custom Label.It can also be used as a custom field within the Custom Field shortcode: td_custom_cat_name_url Move upMove downToggle panel: Table of Contents Disable the automatic insertion of the table of contents. Header Label Eg: Contents, Table of Contents, Page Contents Position Select Position Choose where where you want to display the table of contents. Appearance: NOTE: Using the appearance options below will override the global Appearance settings. Device Target Select Alignment None (Default) Advanced: NOTE: Using the advanced options below will override the global advanced settings. Headings: Heading 1 (h1) Heading 2 (h2) Heading 3 (h3) Heading 4 (h4) Heading 5 (h5) Heading 6 (h6) Select the heading to consider when generating the table of contents. Deselecting a heading will exclude it. Initial View Initially hide the table of contents. 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Open publish panel Post NotificationsDrag and drop an image, upload, or choose from your library.](https://theartnews.in/wp-content/uploads/al_opt_content/IMAGE/theartnews.in/wp-content/uploads/2025/05/20250529_223018-100x70.jpg?bv_host=theartnews.in&bv-resized-infos=bv_resized_mobile%3A100%2A70%3Bbv_resized_ipad%3A100%2A70%3Bbv_resized_desktop%3A100%2A70 "UP TGT PGT EXAM DATE |यूपी टीजीटी पीजीटी परीक्षा 2025: स्थगन अपडेट, नया परीक्षा कार्यक्रम और तैयारी टिप्स")
