⭕ Circle theorems practice

Answer key

1. (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°.
2. (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.
3. (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
4. (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.
5. (b) 80° — Method: opposite angles of a cyclic quadrilateral sum to 180°, so form and solve an equation for x, then substitute back to find angle DAB. Working: (2x + 10) + (3x − 5) = 180, so 5x + 5 = 180, x = 35, and angle DAB = 2(35) + 10 = 80°. Using 360° instead of 180° as the total (as if the two angles were the whole circle rather than a pair of opposite angles) gives x = 71 and angle DAB = 152°; solving correctly for x but reporting x itself instead of substituting it back into 2x + 10 gives 35°; and mishandling the signs when combining the two expressions, using (2x + 10) + (3x + 5) = 180 instead of (2x + 10) + (3x − 5) = 180, gives x = 33 and angle DAB = 76°. Opposite angles of a cyclic quadrilateral — not adjacent ones — are the pair that sums to 180°.
6. (c) 72° — Method: opposite angles in a cyclic quadrilateral always sum to 180°. Working: angle DAB and angle BCD are opposite angles of the cyclic quadrilateral ABCD, so angle BCD = 180 − 108 = 72 degrees. Answer: 72°. Opposite angles in a cyclic quadrilateral are SUPPLEMENTARY, not equal, so do not simply copy the given angle; and do not subtract from 360°, which is the rule for angles round a point, or confuse this with the angle-in-a-semicircle theorem, which gives a fixed 90° that has nothing to do with this quadrilateral.
7. (d) 20 — Alternate angles between parallel lines are equal, so 3x + 10 = 5x − 30. Rearranging, 10 + 30 = 5x − 3x, so 40 = 2x, and x = 20. −10 comes from a sign error when rearranging, moving a term to the wrong side and getting −20 = 2x instead. 25 comes from wrongly treating the two angles as co-interior and adding them to 180°: (3x + 10) + (5x − 30) = 180 gives 8x − 20 = 180, so x = 25. 47.5 makes the same co-interior mistake but sets the sum equal to 360° instead of 180°, giving 8x − 20 = 360 and x = 47.5.
8. (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.
9. (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.
10. (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.
11. (c) 32 — Method: alternate angles between parallel lines are equal, so 2x + 10 = 74. Working: subtracting 10 from both sides gives 2x = 64; dividing by 2 gives x = 32. Answer: x = 32. A candidate who forgets to subtract 10 first and divides 74 by 2 directly gets 37. A candidate who treats the angles as co-interior instead of alternate, so that the two expressions add to 180° rather than being equal, gets 48 after solving. A candidate who makes a sign error and treats the equation as 2x equalling 10 minus 74 instead of 74 minus 10 gets −32.
12. (c) 15 — The exterior angle is 180° − 156° = 24°, and the number of sides of a regular polygon is 360° divided by the exterior angle, so 360 ÷ 24 = 15. 24° is the exterior angle itself, stopping one step before the final division. 17 comes from finding 15 correctly and then adding 2, muddling the exterior angle rule with the (n − 2) that appears in the interior angle sum formula. 7.5 comes from dividing 180 by the exterior angle instead of 360, using the angles on a straight line rather than the total of the exterior angles of a polygon.
13. (a) 25° — PA and PB are tangents from the same external point, so PA = PB and OAPB is a kite with right angles at A and B, by the tangent–radius theorem: angle OAP = angle OBP = 90°. The four angles of the kite sum to 360°, so angle AOB = 360° − 90° − 90° − 50° = 130°. Triangle OAB is isosceles because OA = OB, both radii, so its base angles are equal: angle OAB = (180° − 130°) ÷ 2 = 25°. Using angle APB itself as the apex angle of triangle OAB instead of angle AOB gives (180° − 50°) ÷ 2 = 65°. Using a triangle's angle sum on the four-sided kite, and so counting only one of its two right angles, treats angle AOB as 180° − 90° − 50° = 40°, and halving what is then left of triangle OAB gives (180° − 40°) ÷ 2 = 70°. Finding angle AOB = 130° correctly but forgetting to halve for the isosceles base angle gives 180° − 130° = 50°. Halve that remaining angle, and 25° is what's left for angle OAB.
14. (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.
15. (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.
16. (a) 60° — Method: the angles of a triangle add up to 180°, so add the three expressions, solve for x and then substitute back into the expression for angle B. Working: 2x + 3x + 4x = 9x, so 9x = 180 and x = 20. Angle B is 3x, so angle B = 3 × 20 = 60. Answer: 60°. The distractors: 20° is the value of x, from stopping as soon as the equation is solved instead of substituting back; 120° comes from using 360° as the angle sum, which gives x = 40 and 3x = 120; 80° is 4x, the angle at C, from substituting into the wrong expression.
17. (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.
18. (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.