Circle Theorems II (Copy)
1.
In a circle, chord AB = chord CD.
State and explain the relationship between their distances from the centre O.
2.
In a circle, chord PQ = chord RS. OP ⟂ PQ and OT ⟂ RS.
Show that OP = OT.
3.
AB is a chord of a circle, centre O. A perpendicular from O meets AB at M.
Prove that AM = MB.
4.
AB is a chord of length 12 cm. Distance from centre O to AB is 5 cm.
Radius of circle = 13 cm.
Find OM, where M is midpoint of AB.
5.
A chord of length 10 cm lies 12 cm from the centre of the circle.
Find the radius of the circle.
6.
The radius of a circle is 25 cm. A chord 48 cm long is drawn.
Find its distance from the centre.
7.
In a circle of radius 10 cm, a chord is 12 cm long.
Find its distance from the centre.
8.
In a circle of radius 17 cm, a chord is 30 cm long.
Find its perpendicular distance from the centre.
9.
Two chords of equal length subtend equal angles at the centre.
Prove this using circle properties.
10.
In a circle, chords AB and CD are parallel and equal.
Show that their midpoints are collinear with the centre.
11.
O is the centre of a circle. AB is a chord.
If ∠AOB = 90°, find distance from O to chord AB, given OA = 10 cm.
12.
Two tangents are drawn from external point P to a circle at points A and B.
Prove that PA = PB.
13.
Tangents PA and PB are drawn from external point P to a circle.
If ∠APB = 40°, find ∠AOB, where O is centre.
14.
From an external point, two tangents are drawn to a circle of radius 5 cm.
The distance between the tangents at points of contact is 8 cm.
Find distance of external point from centre.
15.
Tangents are drawn from point P to a circle.
Explain why line OP bisects ∠APB (O is centre, A and B are contact points).
16.
Two tangents PA and PB are drawn from P to a circle of radius 13 cm.
OP = 15 cm.
Find length of tangent PA.
17.
From an external point P, two tangents are drawn to a circle of radius 12 cm.
If OP = 20 cm, find length of tangent PA.
18.
Tangents from external point P to circle meet at A and B.
Prove ΔOAP ≅ ΔOBP.
19.
In a circle, two equal chords AB and CD are drawn equidistant from the centre.
Show that AB = CD and AM = CN, where M and N are midpoints.
20.
A circle has radius 10 cm. A chord 16 cm long is drawn.
Find distance of chord from centre.
21.
In a circle of radius 13 cm, a chord of length 10 cm is drawn.
Find its distance from the centre.
22.
Two equal tangents PA and PB are drawn from P to a circle with centre O.
Explain why triangle OAP is right-angled and isosceles.
23.
From an external point P, two tangents PA and PB are drawn.
Prove that line OP bisects angle ∠APB.
24.
Tangents from point P to a circle have lengths 24 cm each.
Distance from P to centre O is 25 cm.
Find radius of the circle.
25.
In a circle, chords AB and CD are equal. OM and ON are perpendicular distances from centre to AB and CD.
Show that OM = ON.
26.
AB is a chord, O is centre. Radius = 17 cm, AB = 30 cm.
Find perpendicular distance from O to AB.
27.
Equal tangents PA and PB are drawn to a circle.
If AB = 14 cm and OP = 13 cm, find radius.
28.
A tangent is drawn at point A on circle with centre O.
If radius OA = 7 cm, prove tangent ⟂ OA.
29.
From external point P, tangents PA and PB are drawn.
If PA = 15 cm, PB = 15 cm, and OP = 17 cm, find radius of circle.
30.
Tangents from point P to a circle make angle 120° at P.
Find angle subtended at centre by chord AB (where A and B are tangent points).
1. Equal chords are equidistant from centre.
Reason: Circle symmetry — distance depends only on chord length.
2. OP = OT.
Reason: Equal chords are equidistant from centre.
3. AM = MB.
Reason: Perpendicular from centre to chord bisects chord.
4. Half chord AM = √(13² − 5²) = √(169 − 25) = √144 = 12 ÷ 2 = 6.
OM = 5.
Reason: Right-angled triangle OMA, Pythagoras.
5. Half chord = 10 ÷ 2 = 5.
Radius = √(12² + 5²) = √(144 + 25) = √169 = 13 cm.
6. Half chord = 48 ÷ 2 = 24.
Distance = √(25² − 24²) = √(625 − 576) = √49 = 7 cm.
7. Half chord = 12 ÷ 2 = 6.
Distance = √(10² − 6²) = √(100 − 36) = √64 = 8 cm.
8. Half chord = 30 ÷ 2 = 15.
Distance = √(17² − 15²) = √(289 − 225) = √64 = 8 cm.
9. Equal chords subtend equal angles at centre.
Reason: Symmetry of circle.
10. Midpoints and centre collinear.
Reason: Perpendicular bisectors of parallel equal chords pass through centre.
11. ∠AOB = 90°, OA = 10.
Half chord = OA × sin(45°) = 10/√2 ≈ 7.07.
Distance = √(10² − 7.07²) = 7.07 cm.
12. PA = PB.
Reason: Tangents from external point are equal.
13. ∠AOB = 180° − ∠APB = 140°.
Reason: Quadrilateral OAPB cyclic.
14. Radius = 5, distance between tangent points = 8.
This forms isosceles trapezium. Use right triangle: OP² = r² + (½ × 8)².
OP² = 25 + 16 = 41.
OP = √41 ≈ 6.4 cm.
15. OP bisects ∠APB.
Reason: ΔOAP ≅ ΔOBP (radii equal, tangents equal).
16. OP² = OA² + PA².
15² = 13² + PA².
225 − 169 = 56.
PA = √56 ≈ 7.48 cm.
17. OP² = OA² + PA².
20² = 12² + PA².
400 − 144 = 256.
PA = 16 cm.
18. ΔOAP ≅ ΔOBP (OA=OB radius, PA=PB tangents, OP common).
Hence congruent triangles.
19. Equal chords → equal distance from centre → equal halves.
20. Half chord = 8.
Distance = √(10² − 8²) = √(100 − 64) = √36 = 6 cm.
21. Half chord = 10 ÷ 2 = 5.
Distance = √(13² − 5²) = √(169 − 25) = √144 = 12 cm.
22. OA ⟂ tangent.
OA = OB.
So ΔOAP right-angled isosceles.
23. By symmetry, OP bisects ∠APB.
Reason: Congruent triangles OAP, OBP.
24. OP² = OA² + PA².
25² = r² + 24².
625 = r² + 576.
r² = 49 → r = 7 cm.
25. OM = ON.
Reason: Equal chords equidistant from centre.
26. Half chord = 15.
Distance = √(17² − 15²) = √(289 − 225) = √64 = 8 cm.
27. Use Pythagoras in ΔOAP.
OP² = OA² + PA².
13² = r² + 7².
169 − 49 = r² → r²=120 → r=√120≈10.95 cm.
28. Tangent ⟂ radius at point of contact.
So ∠OAT = 90°.
29. OP² = OA² + PA².
17² = r² + 15².
289 − 225 = 64.
r²=64 → r=8 cm.
30. ∠APB = 120°.
Quadrilateral OAPB cyclic → ∠AOB = 180 − 120 = 60°.
So chord AB subtends 60° at centre.