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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Answer key: Circle theorems worksheet — GCSE Higher
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- (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) 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.
- (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) 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) 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) 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) 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.
- (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) 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.
- (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) 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) 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) 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) 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.
- (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.
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