⭕ 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) 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°.
3. (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.
4. (c) VW — Method: within one circle a chord's distance from the centre is fixed by its length, because a chord passing nearer the centre cuts further across the circle; so the chord that lies closest to the centre is simply the longest one listed. Working: the four lengths are 6 cm, 10 cm, 14 cm and 15 cm. Placing them in order, the greatest is 15 cm, and that length belongs to VW, so VW lies closest to the centre. Answer: VW. The distractors: PQ comes from reversing the rule and taking the shortest chord to be the one tucked nearest the centre; TU comes from knowing that the very longest chord is a diameter, deciding that such a chord passes through the centre rather than lying close to it, ruling the 15 cm chord out on that ground and taking the next longest; RS comes from reading 'closest to the centre' as 'nearest the middle of the list of lengths' and picking a middling value.
5. (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.
6. (b) segment — The region enclosed between a chord and the arc it cuts off is called a segment. A sector is the different region already named in the question, enclosed by the two radii and the arc, not by the chord. A tangent is a straight line that touches the circle at one point, not an enclosed region at all. The circumference is the total distance around the whole circle, a length, not a region.
7. (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.
8. (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.
9. (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.
10. (b) 61° — Triangle OAT is isosceles because OA = OT, both radii, so its base angles are equal: angle OTA = angle OAT = (180° − 122°) ÷ 2 = 29°. The tangent PT meets the radius OT at 90°, by the tangent–radius theorem, so angle OTP = 90°. Since TA lies between TO and TP, angle OTP splits into angle OTA and angle ATP: angle ATP = 90° − 29° = 61°. Stopping after the isosceles step and reporting the base angle itself gives 29°. Finding the base angles as 180° − 122° without halving, then subtracting from 90°, gives 90° − 58° = 32°. Adding the base angle to 90° instead of subtracting it gives 90° + 29° = 119°. Subtract the base angle from the right angle, and 61° is what's left.
11. (a) 90° — Method: the angle in a semicircle theorem states that the angle subtended at the circumference by a diameter is always a right angle, whatever the circle's size. Working: since AC is a diameter, the angle ABC subtended at the circumference by AC is 90 degrees, regardless of the circle's radius. Answer: 90°. This result holds for every position of B on the circle and for every size of circle, so no length or radius is ever needed: do not confuse angle ABC with the straight angle along the diameter itself, which is 180°, and do not halve the right angle again.
12. (d) Concentric circles — Method: focus on what the two circles have in common — their centre, not their size. Working: both circles share exactly the same centre point but have different radii, which is the defining feature of this pair of circles. A student who answers congruent circles has confused 'same centre' with 'same size', but congruent circles simply have equal radii and need not share a centre. A student who answers tangential circles has confused circles that touch each other at one point with ones that share a centre. A student who answers similar circles has used the general term for the same shape at different sizes, missing the specific 'same centre' fact. Answer: concentric circles.
13. (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.
14. (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.
15. (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.
16. (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.
17. (c) 140° — Method: OAPB is a quadrilateral whose angles sum to 360°; the tangent-radius angles at A and B are each 90°, and angle APB is given, so angle AOB is whatever is left. Working: angle OAP = angle OBP = 90°, so angle AOB = 360 − 90 − 90 − 40 = 140 degrees. Answer: 140°. Both tangent-radius angles are 90° each and must both be subtracted, not just one, and the quadrilateral's angles sum to 360°, not 180°: do not assume angle AOB simply matches angle APB, which is a different angle in a different part of the figure.
18. (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.