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.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.
- 2.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.
- 3.A, B and C are points on a circle with centre O. B is on the major arc AC. Angle AOC = 116°. Work out the size of angle ABC.
- 4.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.
- 5.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.
- 6.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?
- 7.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.
- 8.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.
- 9.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.
- 10.A ferris wheel's circular frame has centre O. A support strut runs from O to a point A on the rim, and a second strut runs from O to point B on the rim, with angle AOB = 76°. A cabin is mounted at point C on the major arc AB, and D is a separate point on the minor arc AB. Work out the difference between angle ACB and angle ADB.
- 11.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.
- 12.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.
- 13.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.
- 14.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.
- 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
- (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) 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.
- (b) 58° — Method: the angle at the centre is TWICE the angle at the circumference when both stand on the same arc. Working: angle AOC is the angle at the centre standing on arc AC, and angle ABC is the angle at the circumference standing on the same arc AC, with B on the major arc, so angle AOC = 2 × angle ABC. Rearranging, angle ABC = 116 ÷ 2 = 58 degrees. Answer: 58°. Halve the centre angle, do not double it, and use the angle AOC exactly as given, 116°: you do not need its reflex angle here, because B sits on the major arc, which is precisely the arrangement the theorem is stated for.
- (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) 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.
- (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.
- (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) 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) 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.
- (a) 104° — Method: find each inscribed angle separately using the angle-at-the-centre theorem: for a point on the arc NOT cut off by the given central angle, halve that central angle; for a point on the OTHER arc, halve the REFLEX central angle instead; then subtract the smaller from the larger. Working: for C on the major arc, angle ACB = 76 ÷ 2 = 38 degrees. For D on the minor arc, D sees the reflex angle at the centre, 360 − 76 = 284 degrees, so angle ADB = 284 ÷ 2 = 142 degrees. The difference is 142 − 38 = 104 degrees. Answer: 104°. Both inscribed angles need the theorem applied separately, using the correct arc's central angle each time (the reflex angle for D), and the question asks for the DIFFERENCE between the two, not either angle on its own and not their sum, 180°, which is simply the opposite-angle total for the cyclic quadrilateral ACBD.
- (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) 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) 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.
- (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.
- (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.
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