⭕ Circle theorems practice

Answer key

1. (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.
2. (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.
3. (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.
4. (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.
5. (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.
6. (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.
7. (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°.
8. (c) minor segment — A chord splits a circle into two segments; the smaller of the two is called the minor segment and the larger one the major segment. "major segment" names the larger region, the opposite of what is asked for. "sector" is a different region altogether, enclosed by two radii and an arc, not by a chord. "arc" is a curved length along the circumference, not an enclosed region at all.
9. (b) 35° — Method: angle CBE and angle ABC lie on a straight line at B, so they are supplementary; angle ABC then equals angle ADC because B and D are both on the major arc AC (angles in the same segment). Working: angle ABC = 180 − 145 = 35 degrees, using angles on a straight line. Since B and D are both on the major arc AC, angle ADC = angle ABC = 35°. Answer: 35°. Find angle ABC FIRST from the straight line at B before applying the circle theorem: using the given 145° directly, doubling or halving it, or subtracting it from 180° a second time all give the wrong angle.
10. (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.
11. (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.
12. (b) Minor segment — Method: compare the sizes of the two regions cut off by the chord, and recall the term used for the smaller one. Working: the chord creates two segments; the smaller region is called the minor segment and the larger one the major segment. A student who answers major segment has picked the larger region by mistake instead of the smaller one. A student who answers minor arc has named the curved boundary rather than the two-dimensional region it encloses. A student who answers semicircle has wrongly assumed the chord must pass through the centre. Answer: minor segment.
13. (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.
14. (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°.
15. (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.
16. (b) 62 — Method: co-interior (allied) angles between parallel lines add up to 180°. Working: 180 − 118 = 62. Answer: 62°. A candidate who treats co-interior angles as equal, like corresponding angles, gives 118. A candidate who uses 360° instead of 180°, working out 360 − 118, gets 242. A candidate who subtracts as if the angles were complementary, working out 118 − 90, gets 28.
17. (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.
18. (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.