All 8 GCSE Circle Theorems with Step-by-Step Angle Proofs
On the GCSE Maths Higher Tier, Circle Theorems are tested in multi-step geometric problems where half of the marks are awarded for calculating the unknown angles and the other half for writing the exact, word-perfect geometrical reason for each step. There are eight core circle theorems governing angles subtended by arcs, cyclic quadrilaterals, chords, and tangents (culminating in the Alternate Segment Theorem). Examiners strictly penalise vague language such as 'bowtie rule', 'arrow theorem', or 'opposite angles in a cyclic quadrilateral are equal' (they add up to 180°, they are not equal).
- 1. Angle at the Centre & 2. Angle in a Semicircle
- Theorem 1: The angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the remaining circumference (Angle at centre = 2 x Angle at circumference). Theorem 2 (special case when the central angle is a 180° diameter): The angle in a semicircle is a right angle (90°).
- 3. Angles in the Same Segment & 4. Cyclic Quadrilateral
- Theorem 3: Angles subtended by the same arc in the same segment of a circle are equal. Theorem 4: Opposite angles of a cyclic quadrilateral (a four-sided shape where all four vertices lie on the circumference) add up to 180° (they are supplementary).
- 5. Tangent & Radius (90°) & 6. Two Tangents from a Point
- Theorem 5: The angle between a tangent to a circle and the radius drawn to the point of contact is 90°. Theorem 6: Two tangents drawn to a circle from the same external point P are equal in length (PA = PB), forming an isosceles triangle PAB (and congruent right-angled triangles OAP and OBP when joined to the centre O).
Parallel Line Angles (Z, F, C Shapes) & Polygon Interior Angles
Switch between Alternate (Z), Corresponding (F), Co-interior (C), and Polygon Interior Angles. In the exam, always write the formal geometric name, not the letter!
Alternate Angles (Z-Shape): Equal
Both angles sit on opposite sides of the transversal line between the two parallel lines, forming a "Z" shape. Both equal 62°
- Write “Alternate angles are equal” (never write “Z-angles”).
- Write “Corresponding angles are equal” (never write “F-angles”).
- Write “Co-interior angles add up to 180°” (never write “C-angles”).
Key Concepts & Exam Definitions (6)
- 1. Angle at the Centre & 2. Angle in a Semicircle
- Theorem 1: The angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the remaining circumference (Angle at centre = 2 x Angle at circumference). Theorem 2 (special case when the central angle is a 180° diameter): The angle in a semicircle is a right angle (90°).
- 3. Angles in the Same Segment & 4. Cyclic Quadrilateral
- Theorem 3: Angles subtended by the same arc in the same segment of a circle are equal. Theorem 4: Opposite angles of a cyclic quadrilateral (a four-sided shape where all four vertices lie on the circumference) add up to 180° (they are supplementary).
- 5. Tangent & Radius (90°) & 6. Two Tangents from a Point
- Theorem 5: The angle between a tangent to a circle and the radius drawn to the point of contact is 90°. Theorem 6: Two tangents drawn to a circle from the same external point P are equal in length (PA = PB), forming an isosceles triangle PAB (and congruent right-angled triangles OAP and OBP when joined to the centre O).
- 7. Radius Perpendicular to a Chord
- A radius (or line from the centre O) that is perpendicular (at 90°) to a chord bisects the chord (cuts the chord into two equal halves). This allows you to form a right-angled triangle with the radius as the hypotenuse and half the chord as the base to use Pythagoras' Theorem or SOHCAHTOA.
- 8. The Alternate Segment Theorem
- The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle subtended by that chord in the alternate (opposite) segment.
- Algebraic Proof of Circle Theorems (Grade 9 Higher Tier)
- To prove any circle theorem algebraically (such as proving the angle at the centre is 2x or opposite angles in a cyclic quadrilateral sum to 180°): draw radii from the centre O to all vertices on the circumference to split the shape into isosceles triangles (since all radii are equal). Label the equal base angles x and y, and use 'angles in a triangle sum to 180°' and 'angles around a point sum to 360°'.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Write the exact exam reason for why angle ABC = 90° when AC is a diameter of the circle and B is on the circumference.
Write the exact exam reason for why the opposite angles A and C of a four-sided shape ABCD with all corners on a circle satisfy A + C = 180°.
Points A, B, and C lie on a circle with centre O. If reflex angle AOC = 250°, find obtuse angle ABC (where B is on the minor arc AC).
Tangents PA and PB are drawn from an external point P to a circle with centre O. If angle APB = 50°, find: (a) angle PAB, and (b) angle AOB.
State the Alternate Segment Theorem in full.
How do you start the algebraic proof that 'the angle at the centre is twice the angle at the circumference'?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Points A, B, C, and D lie in order around the circumference of a circle with centre O. Angle ADC = 116° and angle OAB = 32°. Calculate the size of angle OBC, giving full geometrical reasons for every step of your working. (Higher Tier) [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Angle ABC = 180° - 116° = 64°, because opposite angles in a cyclic quadrilateral (ABCD) sum to 180° [1 mark for 64°, 1 mark for reason]. 2. In triangle OAB, OA = OB because both are radii of the circle, so triangle OAB is isosceles [1 mark]. 3. Therefore, angle OBA = angle OAB = 32°, because base angles in an isosceles triangle are equal [1 mark]. 4. Since angle ABC = angle OBA + angle OBC, we have angle OBC = 64° - 32° = 32° [1 mark].
- Q2
Line PCQ is a tangent to a circle at point C. Points A, B, and C lie on the circumference of the circle such that AB = AC. Given that angle ACP (between the tangent CP and chord AC) = 63°, calculate the size of angle CAB, giving full reasons for each step. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Angle ABC = 63° by the Alternate Segment Theorem [1 mark for 63°]. 2. Reason: The angle between a tangent and a chord equals the angle subtended by the chord in the alternate segment [1 mark]. 3. Since AB = AC, triangle ABC is an isosceles triangle, so the base angles ABC and ACB are equal: angle ACB = angle ABC = 63° (base angles of an isosceles triangle are equal) [1 mark]. 4. Using angles in a triangle sum to 180°, angle CAB = 180° - (63° + 63°) = 180° - 126° = 54° [1 mark].
- Q3
A circle with centre O has a radius of 13 cm. A straight chord AB has a length of 24 cm. M is the midpoint of the chord AB. Calculate the perpendicular distance OM from the centre O to the chord AB, giving reasons for your geometric steps. (Higher Tier) [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. Because M is the midpoint of chord AB, the line OM from the centre bisects the chord, so angle OMA = 90° (a line from the centre that bisects a chord is perpendicular to the chord), and AM = 24 ÷ 2 = 12 cm [1 mark]. 2. In the right-angled triangle OMA, the hypotenuse is the radius OA = 13 cm and the base is AM = 12 cm [1 mark]. 3. By Pythagoras' Theorem: OM² + 12² = 13² -> OM² = 169 - 144 = 25 -> OM = √25 = 5 cm [1 mark].
- Q4
Prove algebraically that opposite angles in a cyclic quadrilateral ABCD add up to 180°. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Let O be the centre of the circle and draw radii OB and OD to vertices B and D. Let angle BAD = x° and angle BCD = y° [1 mark]. 2. Using the theorem that 'the angle at the centre is twice the angle at the circumference', the obtuse/reflex angle BOD subtended by arc BCD is 2x° [1 mark]. 3. Similarly, the other angle BOD at the centre (subtended by arc BAD) is 2y° [1 mark]. 4. Since angles around a point (centre O) add up to 360°: 2x + 2y = 360° -> divide both sides by 2: x + y = 180°, which proves that opposite angles in a cyclic quadrilateral sum to 180° [1 mark].
Common Exam Questions (4)
What are the exact mark-scheme wordings for the 8 GCSE Circle Theorems?
Never use nicknames like 'bowtie' or 'arrowhead'. Memorise these exact 8 phrases: 1. The angle at the centre is twice the angle at the circumference. 2. The angle in a semicircle is 90°. 3. Angles in the same segment are equal. 4. Opposite angles in a cyclic quadrilateral sum to 180°. 5. The angle between a tangent and a radius is 90°. 6. Tangents from an external point to a circle are equal in length. 7. A radius perpendicular to a chord bisects the chord. 8. The angle between a tangent and a chord equals the angle in the alternate segment (Alternate Segment Theorem).
How do you spot the Alternate Segment Theorem in a busy circle diagram?
Look for a tangent line touching the circle at a point P, and a triangle drawn inside the circle with one vertex at that exact contact point P (all three vertices of the triangle must touch the circumference, not the centre). The angle between the tangent and one side (chord) of the triangle is equal to the interior angle at the opposite vertex of the triangle.
What is the most common hidden trick in Circle Theorem exam questions?
Isosceles triangles formed by two radii (or two tangents). Whenever you see two lines drawn from the centre O to the circumference (like OA and OB), mark them with dashes immediately because both are radii of the circle (OA = OB). Therefore, triangle OAB is isosceles, and its two base angles are equal.
What is the common trap with Cyclic Quadrilaterals?
A cyclic quadrilateral must have all four of its vertices (corners) touching the circumference of the circle. If one of the four corners is at the centre O instead of on the circumference, it is not a cyclic quadrilateral (it is an arrowhead for 'angle at the centre is twice the angle at the circumference', often combined with 'angles around a point sum to 360°').
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