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.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.
- 2.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.
- 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, B, C and D are points on a circle, with B and D both on the same major arc AC. E lies on the straight line through A and B, beyond B, so that angle CBE = 145°. Work out the size of angle ADC.
- 5.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.
- 6.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?
- 7.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.
- 8.Points A, B, C and D lie on a circle with centre O, in that order around the circle, and AC is a diameter. Which of the following circle facts does NOT apply to this diagram?
- 9.A, B, C and D are points on a circle, and ABCD is a cyclic quadrilateral with the vertices in that order around the circle. Angle ABC = 108° and angle ADC = 72°. Which circle theorem is the reason that angle ABC + angle ADC = 180°?
- 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.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.
- 12.A, B and C are points on a circle with centre O. B is on the minor arc AC. Angle ABC = 100°. Work out the size of the non-reflex angle AOC.
- 13.A, B and C are points on a circle with centre O, and AB is a diameter. Prove that angle ACB = 90°, using the fact that triangle OAC and triangle OBC are both isosceles. In the proof, angle OAC = angle OCA = x and angle OBC = angle OCB = y. Which equation correctly expresses the angle sum of triangle ABC in terms of x and y, and leads to the required proof?
- 14.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?
- 15.ABCD is a cyclic quadrilateral, with its vertices in that order around the circle. Angle DAB = (2x + 10)° and angle DCB = (3x − 5)°. Work out the size of angle DAB.
Answer key
- (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.
- (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) 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.
- (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) 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) (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.
- (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.
- (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.
- (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) 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) 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.
- (b) 2x + 2y = 180° — Method: use the fact that OA, OB and OC are all radii, so triangles OAC and OBC are each isosceles, and set up the angle sum of triangle ABC as a whole. Working: since OA = OC, triangle OAC is isosceles, so angle OAC = angle OCA = x. Since OB = OC, triangle OBC is isosceles, so angle OBC = angle OCB = y. The angle at A in triangle ABC is x, the angle at B is y, and the angle at C is angle OCA plus angle OCB, which is x + y. The three angles of triangle ABC sum to 180°, giving x + y + (x + y) = 180°, which is 2x + 2y = 180°. Answer: 2x + 2y = 180°. Dividing this by 2 gives x + y = 90°, and since angle ACB = x + y, this proves angle ACB = 90°: the point C was never restricted to any particular position on the circle, so the proof holds for every choice of C. Include ALL THREE angles of the triangle in the sum, do not use a quadrilateral's 360° total, and remember angle ACB is the SUM x + y, not x alone.
- (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) 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°.
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