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.
Higher only
Circle theorems worksheet — GCSE Higher
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- 1.A, B and C are points on a circle with centre O, and AC is a diameter of the circle. Work out the size of angle ABC.
- 2.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.
- 3.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.
- 4.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.
- 5.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.
- 6.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.
- 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.ABCD is a cyclic quadrilateral, with its vertices in that order around the circle. Angle DAB = 108°. Work out the size of angle BCD.
- 9.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.
- 10.A skateboard ramp is designed so that a straight rail RT is tangent to a circular curve at the point T, and TC is a chord of the curve from T to another point C on the curve. A second rail continues straight through T on the other side, so that the two rails together form a straight line, and the angle between that second rail and the chord TC is 109°. A support point D is placed on the curve, on the major arc TC. Work out the size of angle TDC, the angle subtended by the chord TC at D.
- 11.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.
- 12.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?
- 13.From an external point P, two tangents PA and PB touch a circle with centre O at the points A and B. Angle APB = 46°. C is a point on the major arc AB, the arc on the opposite side of AB from P. Work out the size of angle ACB.
- 14.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?
- 15.PT is a tangent to a circle with centre O, touching the circle at T. OT is a radius. Angle OPT = 27°, where P is a point outside the circle. Work out the size of angle POT.
Answer key
- (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.
- (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.
- (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) 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.
- (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.
- (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.
- (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.
- (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.
- (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.
- (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.
- (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) 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.
- (a) 67° — PA and PB are tangents from the same point, so OAPB is a kite with right angles at A and B, by the tangent–radius theorem. The kite's angles sum to 360°, so angle AOB = 360° − 90° − 90° − 46° = 134°. C is on the major arc AB, so by the angle at the centre theorem, angle ACB = 134° ÷ 2 = 67°. Doubling angle APB instead of first finding angle AOB from the kite gives 46° + 46° = 92°. Using angle APB directly as if it were the centre angle, then halving it, gives 46° ÷ 2 = 23°. Finding angle AOB = 134° correctly but forgetting to halve it for the angle at the circumference gives 134°. Halve angle AOB, once you've found it properly, and you get 67°.
- (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.
- (b) 63° — Method: a tangent meets the radius drawn to the point of contact at a right angle, so triangle OPT has a 90° angle at T; the three angles of the triangle then sum to 180°. Working: angle OTP = 90°, angle OPT = 27°, so angle POT = 180 − 90 − 27 = 63 degrees. Answer: 63°. The tangent-radius angle is a fixed 90°, not something to assume equal to another angle in the triangle, and the three angles of ANY triangle sum to 180°, never 360°: that total belongs to a quadrilateral, not a triangle.
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