⭕ Circle theorems practice

Answer key

1. (d) Opposite angles in a cyclic quadrilateral sum to 180° — ABCD is a cyclic quadrilateral, and angle ABC and angle ADC are a pair of opposite angles in it — they don't sit next to each other around the quadrilateral. The theorem that fixes the sum of a cyclic quadrilateral's opposite angles at 180° is exactly the one needed here. The other three theorems don't apply to this figure at all: the angle-in-a-semicircle theorem gives a 90° angle only when one side is a diameter, and no diameter is mentioned; the angle-at-the-centre theorem needs a centre point and an inscribed angle on the same arc, and no centre is given here; the tangent–radius theorem gives a 90° angle only where a tangent line touches the circle, and there is no tangent in this figure. Only the cyclic-quadrilateral theorem fits what is actually drawn.
2. (a) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
3. (c) 90° — A tangent to a circle always meets the radius drawn to the point of contact at a right angle, so the angle between the tangent and the radius at P is 90°. 180° confuses the tangent with the diameter through P, as if the radius continued in a straight line into the tangent. 45° halves the true angle by mistake. 60° comes from confusing this fact with the angle of an equilateral triangle.
4. (a) 30° — Method: the exterior angles of any convex polygon add up to 360°, and in a regular polygon they are all equal, so divide 360° by the number of sides. Working: 360 ÷ 12 = 30. Answer: 30°. The distractors: 150° is the interior angle, 180 − 30, which answers for the wrong angle at the vertex; 15° comes from dividing 180 by 12, using the angles on a straight line instead of the full turn; 36° comes from dividing 360 by 12 − 2 = 10, carrying the subtraction of 2 out of the interior angle sum formula into a calculation that does not need it.
5. (b) 128° — Angle AOC and angle ABC stand on the same arc AC (the minor arc, which does not contain B), so by the angle at the centre theorem angle ABC = 104° ÷ 2 = 52°. ABCD is a cyclic quadrilateral, so its opposite angles ABC and ADC sum to 180°: angle ADC = 180° − 52° = 128°. A candidate who finds angle ABC = 52° correctly but then treats opposite angles as equal, as in a parallelogram, instead of supplementary, writes down 52° and stops there. Skipping the halving step and using 104° as angle ABC gives 180° − 104° = 76°. Reading off the given centre angle itself as the final answer, without applying either theorem, gives 104°. Work through both theorems in order and you land on 128°.
6. (c) 5 cm — Diameter = circumference ÷ π, so 31.4 ÷ 3.14 = 10 cm, and the radius is half the diameter, so 10 ÷ 2 = 5 cm. 10 cm is the diameter itself, given as the radius by forgetting the final halving step. 15.7 cm comes from halving the circumference, 31.4 ÷ 2 = 15.7, and stopping there, treating half the circumference as the radius without ever dividing by π. 2.5 cm comes from halving the correct radius again, effectively dividing by 2 twice instead of once.
7. (d) 13 — Method: OP and PQ meet at a right angle because of the tangent–radius fact, so triangle OPQ is right-angled at P; use Pythagoras' theorem. Working: OQ² = OP² + PQ² = 5² + 12² = 25 + 144 = 169; OQ = √169 = 13. A student who answers 17 has simply added the two given lengths (5 + 12) instead of using Pythagoras' theorem. A student who answers 7 has subtracted the two given lengths (12 − 5) instead of using Pythagoras' theorem. A student who answers 144 has correctly squared 12 but stopped there, forgetting to add 5² and take the square root. Answer: 13 cm.
8. (c) 68° — Since PA and PB are both tangents from the external point P, PA = PB, so triangle PAB is isosceles with equal base angles at A and B. The angles of triangle PAB sum to 180°, so angle PAB + angle PBA = 180° − 44° = 136°, and since the two base angles are equal, angle PAB = 136° ÷ 2 = 68°. Angle PAB is the angle between the tangent at A and the chord AB, so by the alternate segment theorem it equals the angle in the alternate segment, angle ACB = 68°. Using angle APB directly as angle ACB, without using the isosceles triangle to find angle PAB first, gives 44°. Halving angle APB directly, rather than subtracting it from 180° before halving, gives 22°. Finding 180° − 44° = 136° correctly but forgetting to divide by 2 for one base angle leaves 136°.
9. (b) 30 cm — The perpendicular from the centre of a circle to a chord bisects the chord, so this line, half the chord and the radius form a right-angled triangle. Using Pythagoras' theorem, half the chord = √(17² − 8²) = √(289 − 64) = √225 = 15 cm. The full chord AB is twice this length: AB = 2 × 15 = 30 cm. Stopping after finding the half-chord, without doubling it for the whole chord, gives 15 cm. Adding the radius and the perpendicular distance directly, 17 + 8 = 25 cm, ignores that these two lengths are the two shorter sides of a right-angled triangle, not parts of a straight line. Subtracting instead, 17 − 8 = 9 cm, makes the same mistake in the other direction.
10. (b) 31.4 cm — Circumference = π × diameter, so 3.14 × 10 = 31.4 cm. 15.7 cm comes from using the radius, 5 cm, in place of the diameter: 3.14 × 5 = 15.7, which is only half the circumference. 78.5 cm comes from using the area formula π × radius² instead of the circumference formula: 3.14 × 5² = 3.14 × 25 = 78.5. 62.8 cm comes from keeping the 2 from the radius form of the formula, C = 2 × π × radius, but putting the full diameter into it: 2 × 3.14 × 10 = 62.8.
11. (c) 115 — Method: use corresponding angles to carry the 65° angle from the lower rafter up to the upper rafter, then use angles on a straight line to move to the other side of the strut. Working: the angle above the upper rafter and to the left of the strut corresponds to the given angle, so it is 65°; the angle above the upper rafter and to the right of the strut lies on a straight line with it, so it is 180 − 65 = 115. Answer: 115°. A candidate who assumes the angle stays 65° without allowing for the move from the left of the strut to the right of it gives 65. A candidate who uses 90° instead of 180°, working out 90 − 65, gets 25. A candidate who adds instead of subtracting, working out 180 + 65, gets 245.
12. (b) It is the longest of the three chords — Method: in any circle the length of a chord is decided by how far the chord lies from the centre, because a chord passing nearer the centre cuts further across the circle. Working: the chord through O lies at a distance of zero from the centre, and no chord can lie closer than that, so no chord of the circle can be longer than it; a chord through the centre is a diameter. The other two chords lie at some distance greater than zero, so each of them falls short of that maximum. Answer: It is the longest of the three chords. The distractors: It is the shortest of the three chords comes from reversing the rule and picturing a chord near the centre as a short line tucked inside; It is the same length as the other two chords comes from carrying the fact that all radii of a circle are equal across to chords, which are not all equal; It is half the length of each of the other two chords comes from confusing a chord through the centre with a radius, which really is half a diameter.
13. (c) chord — A diameter is a straight line joining two points on the circle that happens to pass through the centre. The general term for any straight line joining two points on a circle is a chord — every diameter is a chord, but not every chord is a diameter. Calling it a radius is wrong because a radius runs from the centre to just ONE point on the circle, not between two points on the circle. A tangent touches the circle at only one point, so it cannot be a diameter at all. An arc is a curved part of the circle's edge, not a straight line.
14. (b) 160° — B is on the minor arc AC, so the angle at the centre theorem applies to the reflex angle AOC: reflex angle AOC = 2 × 100° = 200°. The non-reflex angle AOC is the rest of the full turn: 360° − 200° = 160°. Reporting the reflex angle itself, without subtracting it from 360°, gives 200°. Subtracting angle ABC from 180° instead, as if this were a cyclic quadrilateral, gives 80°. Halving angle ABC instead of doubling it gives 50°. Double first, then take the angle away from a full turn, and 160° is what's left.
15. (a) 104° — Method: find each inscribed angle separately using the angle-at-the-centre theorem: for a point on the arc NOT cut off by the given central angle, halve that central angle; for a point on the OTHER arc, halve the REFLEX central angle instead; then subtract the smaller from the larger. Working: for C on the major arc, angle ACB = 76 ÷ 2 = 38 degrees. For D on the minor arc, D sees the reflex angle at the centre, 360 − 76 = 284 degrees, so angle ADB = 284 ÷ 2 = 142 degrees. The difference is 142 − 38 = 104 degrees. Answer: 104°. Both inscribed angles need the theorem applied separately, using the correct arc's central angle each time (the reflex angle for D), and the question asks for the DIFFERENCE between the two, not either angle on its own and not their sum, 180°, which is simply the opposite-angle total for the cyclic quadrilateral ACBD.
16. (d) 60° — Method: the six angles at the centre together make one complete turn of 360°, and because the hexagon is regular they are all equal, so divide 360° by 6. Working: 360 ÷ 6 = 60. Answer: 60°. The distractors: 120° is the interior angle of a regular hexagon, 720 ÷ 6, which is the angle at a vertex and not the angle at the centre; 45° comes from dividing 360 by 8, treating the hexagon as though it had eight sides; 30° comes from halving the angle at the centre, as though each of the six triangles were split again by a line of symmetry.
17. (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
18. (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.