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 onlyNon-calculator
Circle theorems worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.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.
- 2.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.
- 3.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.
- 4.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.
- 5.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.
- 6.From an external point P, two tangents PA and PB touch a circle with centre O at A and B. Angle APB = 40°. Work out the size of angle AOB.
- 7.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.
- 8.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.
- 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.ABCD is a cyclic quadrilateral, with its vertices in that order around the circle. Angle DAB = 108°. Work out the size of angle BCD.
- 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 = 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.
- 12.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?
- 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 = 50°. Work out the size of angle OAB.
- 14.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.
- 15.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.
Answer key
- (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.
- (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.
- (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°.
- (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.
- (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°.
- (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.
- (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) 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.
- (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.
- (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.
- (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°.
- (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.
- (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) 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.
- (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.
Build your own mix at the worksheet builder.