⭕ Circle theorems practice

Answer key

1. (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.
2. (c) Tangent-chord angle = angle in the alternate segment. — This diagram has a diameter AC, a centre O, and a cyclic quadrilateral ABCD, but no tangent anywhere in it. 'Angle in a semicircle = 90°' applies directly, because AC is a diameter. 'Opposite angles of a cyclic quadrilateral sum to 180°' applies directly, because ABCD is a cyclic quadrilateral. 'Angle at the centre = twice angle at circumference' applies directly, because O is given as the centre of the circle. The tangent-chord fact relates the angle between a tangent and a chord to an angle elsewhere in the circle, and since this diagram has no tangent, there is nothing in it for that fact to describe — so it is the one theorem that does not apply here.
3. (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.
4. (a) Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°. — A correct proof must build in a strictly logical order: the reflex and non-reflex angles at O, which together make a complete turn about the centre, must be added to 360° before 2a and 2c can be combined and divided. The statement 'Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°' is the only one that follows directly from having 2a and 2c already established, and it is exactly what is needed before the final division step. Jumping straight to 'Divide by 2 throughout to get a + c = 180°' skips the step that justifies why 2a + 2c equals 360° in the first place — there is nothing yet to divide. Repeating 'Angle BOD is 2a and 2c, by the centre theorem' does not move the proof forward at all, since that fact has already been established in the sentence given. Introducing 'OB = OD are radii, so triangle OBD is isosceles' brings in an unrelated triangle and an unrelated method that plays no part in this particular proof.
5. (d) tangent — A line that touches a circle at exactly one point, without crossing into the circle, is called a tangent. A chord is a straight line joining two points ON the circle, so it touches at two points, not one. A radius runs from the centre to the circle's edge, not along the outside of it. A diameter is a chord that passes through the centre, also touching the circle at two points.
6. (d) Minor arc — Method: compare the lengths of the two arcs formed by the two points, and recall the term for the shorter one. Working: the two points split the circumference into two arcs; the shorter one is the minor arc and the longer one is the major arc. A student who answers major arc has picked the longer arc by mistake. A student who answers minor segment has confused the curved boundary with the enclosed two-dimensional region. A student who answers chord has named the straight line joining the two points instead of the curved arc. Answer: minor arc.
7. (a) 71° — Method: first use the straight line through T to find the tangent-chord angle RTC, then apply the alternate segment theorem, which states that this angle equals the angle subtended by the chord in the alternate segment. Working: the two rails form a straight line through T, so angle RTC = 180 − 109 = 71 degrees. D lies in the alternate segment of the chord TC, so angle TDC = angle RTC = 71°. Answer: 71°. Find angle RTC FIRST from the straight line before applying the theorem: using the given 109° directly, doubling the tangent-chord angle, or taking its complement from 90° all give the wrong angle at D.
8. (a) 7.2 cm — The distance around the mat is the circumference, π × diameter = 3.14 × 20 = 62.8 cm. Yasmin started with 70 cm, so the ribbon left over is 70 − 62.8 = 7.2 cm. 62.8 cm is the circumference itself, the amount of ribbon used, not what is left over. 50 cm comes from subtracting the diameter instead of the circumference: 70 − 20 = 50. 20 cm is simply the diameter of the mat and involves no calculation with the ribbon length at all.
9. (b) 62° — Because AT is a diameter, the radius OT lies along AT, and the tangent PT meets every radius at 90°, by the tangent–radius theorem. So PT meets AT at T at a right angle: angle ATP = 90°. The angles of triangle APT sum to 180°, so angle APT = 180° − 90° − 28° = 62°. Copying the given angle PAT straight across, as though the triangle were isosceles, gives 28°. Adding the two known angles instead of subtracting them from 180° gives 90° + 28° = 118°. Naming the right angle itself, angle ATP, instead of the angle that was actually asked for gives 90°. Subtract both known angles from 180°, and 62° is angle APT.
10. (d) The angle between a tangent and a radius is 90° — OP is the radius drawn to the point of contact P, and a circle theorem states that a tangent always meets that radius at a right angle, so angle OPQ = 90°. A tangent is not parallel to the radius it touches — at the point of contact it is perpendicular to that radius, not parallel to it. A tangent does not pass through the centre — a straight line through the centre that also touches the circle at one point would have to be a diameter, which is a different line entirely. The angle-in-a-semicircle theorem needs a triangle drawn inside the circle with a diameter as its longest side; there is no such triangle here, just a tangent and a radius.
11. (a) 65° — OT and OC are both radii, so triangle OTC is isosceles with OT = OC, and its base angles are equal: angle OTC = angle OCT = (180° − 130°) ÷ 2 = 25°. A tangent is perpendicular to the radius at the point of contact, so angle OTP = 90°. Since C lies inside angle OTP, the chord TC splits this right angle into angle OTC and angle PTC, so angle PTC = angle OTP − angle OTC = 90° − 25° = 65°. Stopping after finding the base angle of the isosceles triangle, without subtracting it from the right angle at T, leaves 25° instead of the angle actually asked for. Adding the base angle to the right angle instead of subtracting it, 90° + 25° = 115°, reverses the direction the two angles combine in. Finding the sum of the two base angles of the isosceles triangle, 180° − 130° = 50°, and stopping there, gives another wrong value entirely.
12. (a) 31.4 — Method: first find the circumference of one turn using π × diameter, then multiply by the number of turns. Working: circumference = 3.14 × 0.5 = 1.57 m; total distance = 1.57 × 20 = 31.4 m. A student who answers 15.7 has mistakenly halved the diameter again before multiplying, using 0.25 m instead of 0.5 m. A student who answers 3.14 has simply written down π itself, without completing the circumference or multiplying by the number of turns. A student who answers 62.8 has mistakenly doubled the diameter to 1 m before multiplying, treating the given length as a radius. Answer: 31.4 m.
13. (a) 68° — Method: a tangent meets a radius at 90°, so quadrilateral OAPB has two right angles at A and B; its four angles sum to 360°, so angle APB = 360° − 90° − 90° − angle AOB. Working: angle APB = 360° − 90° − 90° − 112° = 68°. Halving angle AOB, confusing this with the angle-at-centre theorem for a chord, gives 56°; assuming the angle between two tangents is always 90° (true only for the angles at A and B, not at P) gives 90°; and assuming angle APB simply equals angle AOB by symmetry gives 112°. The two right angles at A and B, from the tangent-radius property, are what link angle AOB and angle APB.
14. (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°.
15. (a) 105° — The interior angles of a pentagon add up to (5 − 2) × 180° = 540°. Subtracting the three known angles, 540 − 100 − 110 − 120 = 210°, and this 210° is shared equally between the two angles marked x°, so each one is 210 ÷ 2 = 105°. 210° stops one step early, giving the total of the two unknown angles instead of one of them. 108° is the interior angle of a regular pentagon, which does not apply here since this pentagon's angles are not all equal. 55° comes from halving one of the given angles, 110°, instead of halving the remaining total.
16. (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.
17. (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.
18. (b) 58° — Method: the angle at the centre is TWICE the angle at the circumference when both stand on the same arc. Working: angle AOC is the angle at the centre standing on arc AC, and angle ABC is the angle at the circumference standing on the same arc AC, with B on the major arc, so angle AOC = 2 × angle ABC. Rearranging, angle ABC = 116 ÷ 2 = 58 degrees. Answer: 58°. Halve the centre angle, do not double it, and use the angle AOC exactly as given, 116°: you do not need its reflex angle here, because B sits on the major arc, which is precisely the arrangement the theorem is stated for.