⭕ Circle theorems practice

Answer key

1. (c) 1/4 — A full turn at the centre of a circle is 360°, so a sector's fraction of the circle is its angle divided by 360°: 90 ÷ 360 = 1/4. 1/2 would be the fraction for a sector with an angle of 180°, not 90°. 3/4 is the fraction of the rest of the circle, the major sector left over from the 270° that is not part of this sector. 1/8 would be the fraction for a sector with an angle of 45°, half of 90°.
2. (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.
3. (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.
4. (b) 13 — Using the sum of interior angles formula, (n − 2) × 180° = 1980°, so n − 2 = 1980 ÷ 180 = 11, and n = 11 + 2 = 13. 11 stops after the division, forgetting to add 2 back to find n. 15 adds 2 twice by mistake, giving 11 + 2 + 2. 22 divides 1980 by 90 instead of 180.
5. (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.
6. (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.
7. (a) 24 cm — The line from the centre to the midpoint of a chord is perpendicular to the chord, so triangle OMA has a right angle at M. By Pythagoras' theorem, AM² = OA² − OM² = 169 − 25 = 144, so AM = 12 cm. AB is twice AM, since M is the midpoint: AB = 2 × 12 = 24 cm. Finding AM = 12 cm correctly but forgetting to double it for the full chord gives 12 cm. Working out 169 − 25 = 144 and forgetting to take the square root gives 144 cm. Subtracting first and then doubling the wrong way, (13 − 5) × 2, gives 16 cm. Doubling AM after finding it correctly is the step that's missing from all three — do it, and you get 24 cm.
8. (b) 2x + 2y = 180° — Method: use the fact that OA, OB and OC are all radii, so triangles OAC and OBC are each isosceles, and set up the angle sum of triangle ABC as a whole. Working: since OA = OC, triangle OAC is isosceles, so angle OAC = angle OCA = x. Since OB = OC, triangle OBC is isosceles, so angle OBC = angle OCB = y. The angle at A in triangle ABC is x, the angle at B is y, and the angle at C is angle OCA plus angle OCB, which is x + y. The three angles of triangle ABC sum to 180°, giving x + y + (x + y) = 180°, which is 2x + 2y = 180°. Answer: 2x + 2y = 180°. Dividing this by 2 gives x + y = 90°, and since angle ACB = x + y, this proves angle ACB = 90°: the point C was never restricted to any particular position on the circle, so the proof holds for every choice of C. Include ALL THREE angles of the triangle in the sum, do not use a quadrilateral's 360° total, and remember angle ACB is the SUM x + y, not x alone.
9. (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.
10. (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.
11. (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.
12. (a) 1260° — The sum of the interior angles of a polygon with n sides is (n − 2) × 180°. For a nonagon, n = 9, so the sum is (9 − 2) × 180° = 7 × 180° = 1260°. 1620° uses 9 × 180° without subtracting 2 from n first. 140° is the size of a single interior angle of a regular nonagon (1260° ÷ 9), not the sum of all nine. 1440° uses (n − 1) × 180° = 8 × 180° instead of (n − 2) × 180°.
13. (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.
14. (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.
15. (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.
16. (b) (2x + 10)° — By the angle at the centre theorem, angle ABC is half of angle AOC, because both stand on the same arc AC: angle ABC = (4x + 20)° ÷ 2 = (2x + 10)°. Writing down the centre angle itself, without halving at all, gives (4x + 20)°. Halving only the constant term and leaving the x-term unchanged gives (4x + 10)°. Doubling the centre angle instead of halving it gives (8x + 40)°. Halve every term in the expression, and (2x + 10)° is what you get.
17. (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.
18. (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.