Printable · GCSE Higher · ages 14-16
Circle theorems worksheet — GCSE Higher
Fifteen questions on "circle theorems" — DfE statement G10. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Circle theorems worksheet — GCSE Higher
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- 1.A, B, C and D are points on a circle with centre O, placed in that order around the circle so that ABCD is a cyclic quadrilateral. B lies on the major arc AC and D lies on the minor arc AC. Angle AOC = 104°. Work out the size of angle ADC.
- 2.A student is proving that opposite angles of a cyclic quadrilateral ABCD sum to 180°, letting a = angle DAB and c = angle BCD, and using the fact that the angle at the centre is twice the angle at the circumference. She has already written: 'By the angle at the centre theorem, the reflex angle BOD equals 2a and the non-reflex angle BOD equals 2c.' Which statement must come immediately after this one in a correct proof?
- 3.In cyclic quadrilateral ABCD, with vertices in that order around the circle, the diagonal BD is drawn. In triangle ABD, angle ABD = 35° and angle ADB = 65°. Side DC is extended beyond C to a point E. Work out the size of angle BCE.
- 4.A circle with centre O has radius 17 cm. A chord AB is drawn so that the perpendicular distance from O to AB is 8 cm. Work out the length of the chord AB.
- 5.From an external point P, two tangents PA and PB touch a circle with centre O at A and B. Angle AOB = 112°. Work out the size of angle APB.
- 6.A circle has centre O and radius 13 cm. AB is a chord of the circle, and M is the midpoint of AB. OM is drawn, and OM = 5 cm. Work out the length of AB.
- 7.A, B, C and D are points on a circle with centre O, in that order around the circle. B lies on the major arc AC, and angle AOC = 152°. In triangle ACD, angle ACD = 30°. Work out the size of angle DAC.
- 8.AB is a diameter of a circle with centre O, and C is a point on the circle. A student writes four statements to prove that angle ACB = 90°. Statement 1: OA = OC, since both are radii, so triangle OAC is isosceles with angle OAC = angle OCA. Statement 2: OB = OC, since both are radii, so triangle OBC is isosceles with angle OBC = angle OCB. Statement 3: in triangle ABC, the three angles sum to 360°, so angle OAC + angle OBC + angle ACB = 360°, meaning 2 × angle ACB = 360° and angle ACB = 180°. Statement 4: A, O and B lie on a straight line, since AB is a diameter through the centre O. Which one of these four statements is mathematically incorrect?
- 9.To prove that the angle at the centre is twice the angle at the circumference, a student draws radii OA and OC, where A and C are points on the circle and O is the centre. The student's first step is: 'Since OA = OC, both being radii of the circle, triangle OAC is isosceles, so angle OAC = angle OCA.' Is this first step correct?
- 10.AT is a diameter of a circle with centre O. PT is a tangent to the circle at the point T, and P is a point on this tangent such that A, T and P form a triangle. Angle PAT = 28°. Work out the size of angle APT.
- 11.O is the centre of a circle, and PT is a tangent to the circle at the point T. A is a point on the circle, with OA and OT both radii and angle AOT = 122°. At the point T, the chord TA lies between the radius TO and the tangent TP. Work out the size of angle ATP, the angle between the chord and the tangent.
- 12.At a cycling event, a circular track has centre O. Two flags, F and M, are fixed on the track. A course marker P is also on the track, positioned so that angle FOP = 55° and angle POM = 43°, with P between F and M as seen from O. A spectator S stands on the major arc FM, on the opposite side of the track from P. The event programme needs angle FSM. Work out angle FSM.
- 13.From an external point P, two tangents PA and PB touch a circle with centre O at points A and B. Angle APB = 44°. C is a point on the major arc AB. Using the fact that PA = PB, and the alternate segment theorem, work out the size of angle ACB.
- 14.From an external point P, two tangents PA and PB touch a circle with centre O at the points A and B. Angle APB = 50°. Work out the size of angle OAB.
- 15.A, B and C are points on a circle with centre O. B is on the major arc AC. Angle AOC = (4x + 20)°. Work out the size of angle ABC in terms of x, in its simplest form.
Answer key
- (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°.
- (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.
- (a) 80° — In triangle ABD, the three angles sum to 180°: angle BAD = 180° − 35° − 65° = 80°. ABCD is a cyclic quadrilateral, so the exterior angle at C, angle BCE, is equal to the interior angle at the opposite vertex, angle BAD: angle BCE = 80°. Adding the two given angles in triangle ABD instead of subtracting them from 180°, 35° + 65° = 100°, and then applying the exterior-angle rule correctly still gives the wrong angle BAD and so the wrong exterior angle, 100°. Halving the correct angle BAD, 80° ÷ 2 = 40°, confuses the exterior-angle rule with a different circle theorem in which one angle is half of another. Doubling it instead, 80° × 2 = 160°, makes the same kind of confusion in the other direction.
- (b) 30 cm — The perpendicular from the centre of a circle to a chord bisects the chord, so this line, half the chord and the radius form a right-angled triangle. Using Pythagoras' theorem, half the chord = √(17² − 8²) = √(289 − 64) = √225 = 15 cm. The full chord AB is twice this length: AB = 2 × 15 = 30 cm. Stopping after finding the half-chord, without doubling it for the whole chord, gives 15 cm. Adding the radius and the perpendicular distance directly, 17 + 8 = 25 cm, ignores that these two lengths are the two shorter sides of a right-angled triangle, not parts of a straight line. Subtracting instead, 17 − 8 = 9 cm, makes the same mistake in the other direction.
- (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.
- (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.
- (a) 46° — Method: use the angle at the centre, then the cyclic quadrilateral, then the angle sum of a triangle. Working: angle ABC = 152° ÷ 2 = 76° (angle at the centre is twice the angle at the circumference); angle ADC = 180° − 76° = 104° (opposite angles of a cyclic quadrilateral sum to 180°); angle DAC = 180° − 104° − 30° = 46° (angle sum of triangle ACD). Skipping the halving step and taking angle ABC = 152° gives angle ADC = 180° − 152° = 28° and then angle DAC = 122°; using angle ABC = 76° directly as angle ADC, skipping the cyclic quadrilateral step, gives 74°; and finding angle DAC as 180° − 104° without subtracting the given 30° gives 76°. Each of the three steps must be carried through in order — halve, then subtract from 180°, then subtract from 180° again with the extra angle.
- (b) Statement 3 — a triangle's angles are said to sum to 360° — The angles of any triangle sum to 180°, not 360° — Statement 3 uses the wrong total, and that error is what sends its final line to the impossible claim that angle ACB = 180°. The correct working is angle OAC + angle OBC + angle ACB = 180°, and since angle ACB = angle OCA + angle OCB, this gives 2 × angle ACB = 180°, so angle ACB = 90°, which is the actual theorem. Statement 1 correctly identifies OA and OC as equal radii, making triangle OAC isosceles — nothing wrong there. Statement 2 correctly does the same for triangle OBC. Statement 4 correctly states that A, O and B are collinear, since a diameter passes through the centre — also nothing wrong there. Statement 3 is the one to flag: it is the angle sum it quotes that is wrong, not the diagram or the radii.
- (b) Yes — OA = OC (radii), so the base angles are equal. — Method: check the definition of a radius and the isosceles triangle property the step relies on. Working: every radius of a circle has the same length, so OA = OC regardless of where A and C sit on the circle — this alone makes triangle OAC isosceles, and the base angles opposite the two equal sides, angle OAC and angle OCA, must be equal. The verdict is Yes, for exactly that reason. Claiming OA and OC are only equal if they are drawn to 'the very same point' misunderstands what a radius is — A and C can be any two points on the circle and OA still equals OC. Claiming the triangle is isosceles because angle AOC is 90° reverses the logic: nothing in the step has fixed angle AOC at 90°, and even if it had, that alone would not explain why OA = OC. Claiming the triangle would need to be equilateral confuses isosceles (two equal sides) with equilateral (three equal sides) — only two sides, OA and OC, are being compared here.
- (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.
- (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.
- (d) 49° — Method: first combine the two given angles at the centre to find the whole central angle FOM, then apply the angle-at-the-centre theorem, which halves the central angle to give the angle at the circumference on the major arc. Working: angle FOM = angle FOP + angle POM = 55 + 43 = 98 degrees. Since S is on the major arc FM, angle FSM = 98 ÷ 2 = 49 degrees. Answer: 49°. Add BOTH given angles to find the whole angle FOM before you halve anything: halving only one of them, doubling the total instead of halving it, or subtracting from 180° all give the wrong angle for the programme.
- (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°.
- (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.
- (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.
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