Mathematics · Topic 1.1
Circle Theorems
Master the seven key circle theorems and their extended applications. Learn to identify, apply, and justify each theorem in both standard and extended-level problems, including multi-step reasoning chains required for eAssessment.
What You'll Learn
- State and apply all seven circle theorems with correct mathematical terminology
- Identify which theorem(s) to use from a given diagram
- Solve multi-step problems combining multiple circle theorems
- Apply the alternate segment theorem and intersecting chord/secant results (Extended)
- Construct logical chains of reasoning and justify every step
- Tackle eAssessment-style structured and extended response questions
eAssessment Focus
Criterion A: Apply circle theorems in unfamiliar, multi-step geometric configurations.
Criterion B: Discover relationships between angles and prove why theorems hold.
Criterion C: Communicate with correct theorem names — naming the theorem is mandatory.
Criterion D: Apply circle geometry to real-world contexts (e.g., satellite signals, Ferris wheels).
Key Vocabulary
| Term | Definition |
|---|---|
| Arc | Part of the circumference of a circle |
| Chord | A straight line connecting two points on the circumference |
| Tangent | A line that touches the circle at exactly one point; perpendicular to the radius at that point |
| Subtended angle | The angle formed at a point by two lines drawn to the endpoints of an arc |
| Inscribed angle | An angle formed by two chords with the vertex on the circumference |
| Central angle | An angle with its vertex at the centre of the circle |
| Cyclic quadrilateral | A quadrilateral with all four vertices on the circumference of a circle |
| Segment | A region of a circle bounded by a chord and an arc |
Mathematics · Topic 1.1
Core Circle Theorems
These four theorems form the foundation of circle geometry. You must know each one by name and be able to apply it with justification.
Theorem 1 — Angle at the Centre
Theorem
The angle at the centre is twice the angle at the circumference, when both are subtended by the same arc.
If an arc AB subtends angle AOB at the centre and angle ACB at the circumference, then ∠AOB = 2 × ∠ACB.
Example: If the central angle AOB = 124°, then the inscribed angle ACB = 124° ÷ 2 = 62°.
Reverse application: If the inscribed angle = 35°, then the central angle = 35° × 2 = 70°.
Theorem 2 — Angles in the Same Segment
Theorem
Angles subtended by the same arc in the same segment are equal.
If points C and D are on the same side of chord AB, then ∠ACB = ∠ADB. This is because both angles are half of the same central angle (by Theorem 1).
Example: If ∠ACB = 48° and D is also on the major arc, then ∠ADB = 48°.
Theorem 3 — Angle in a Semicircle
Theorem
The angle subtended by a diameter at the circumference is always 90°.
This is a special case of Theorem 1: the central angle for a diameter is 180° (a straight line), so the inscribed angle = 180° ÷ 2 = 90°.
Key recognition: Whenever you see a triangle inscribed in a circle where one side is a diameter, the angle opposite the diameter is 90°. This creates a right-angled triangle.
Theorem 4 — Cyclic Quadrilateral
Theorem
Opposite angles of a cyclic quadrilateral sum to 180°.
If ABCD is a cyclic quadrilateral (all four vertices on the circle), then ∠A + ∠C = 180° and ∠B + ∠D = 180°.
Example: In cyclic quadrilateral ABCD, ∠A = 72°. Find ∠C.
∠C = 180° − 72° = 108° (opposite angles in a cyclic quadrilateral sum to 180°).
∠C = 180° − 72° = 108° (opposite angles in a cyclic quadrilateral sum to 180°).
eAssessment Rule: You MUST name the theorem you are using. Writing "opposite angles = 180°" is not enough — you must write "opposite angles of a cyclic quadrilateral sum to 180°" or similar. Failure to name the theorem loses Criterion C marks.
Mathematics · Topic 1.1
Tangent Theorems
Tangent lines have special properties that are frequently tested. These two theorems complete the standard set of seven.
Theorem 5 — Tangent-Radius Perpendicularity
Theorem
A tangent to a circle is perpendicular to the radius at the point of contact (90°).
This means that if you draw a radius to the point where a tangent touches the circle, the angle between them is always exactly 90°. This is useful for finding angles in triangles formed by radii and tangents.
Example: A tangent at point P and a radius OP create a right angle. If a line from O to an external point T makes angle OTP = 25°, then ∠OPT = 90°, so ∠TOP = 180° − 90° − 25° = 65° (angle sum of triangle).
Theorem 6 — Two Tangents from an External Point
Theorem
Two tangents drawn from the same external point to a circle are equal in length.
If tangents TA and TB are drawn from external point T to a circle with centre O, then TA = TB. The triangles OTA and OTB are congruent (RHS), so ∠OTA = ∠OTB.
Example: Two tangents from point T touch the circle at A and B. If TA = 12 cm and the radius is 5 cm, find the distance OT.
Using Pythagoras in triangle OAT (right angle at A): OT² = OA² + AT² = 5² + 12² = 25 + 144 = 169. So OT = 13 cm.
Using Pythagoras in triangle OAT (right angle at A): OT² = OA² + AT² = 5² + 12² = 25 + 144 = 169. So OT = 13 cm.
Theorem 7 — Alternate Segment Theorem
Theorem
The angle between a tangent and a chord at the point of contact equals the inscribed angle in the alternate segment.
If a tangent at point P and a chord PQ form angle α on one side, then the inscribed angle PRQ in the segment on the opposite side of PQ also equals α.
Example: A tangent at P makes a 55° angle with chord PQ. By the alternate segment theorem, the angle in the alternate segment (angle PRQ where R is on the opposite arc) = 55°.
Common Mistake: Students often confuse which angle is the "alternate segment" angle. The key is: the tangent-chord angle and the inscribed angle must be on opposite sides of the chord. Draw the chord, identify which segment the tangent angle opens into, then look at the other segment.
Mathematics · Topic 1.1 · Extended
Extended: Intersecting Chords & Secants
Extended-level content covers additional results about intersecting lines within and outside circles. These appear in the calculator section of the eAssessment.
Intersecting Chords Theorem
Formula
If two chords AB and CD intersect at point P inside the circle:AP × PB = CP × PD
Example: Chords AB and CD cross at P inside a circle. AP = 3, PB = 8, CP = 4. Find PD.
AP × PB = CP × PD → 3 × 8 = 4 × PD → 24 = 4PD → PD = 6
AP × PB = CP × PD → 3 × 8 = 4 × PD → 24 = 4PD → PD = 6
Intersecting Secants from External Point
Formula
If two secants from an external point P intersect the circle at A, B and C, D respectively:PA × PB = PC × PD
This also applies when one line is a tangent (in which case PA = PB, so the formula becomes PA² = PC × PD).
Tangent-Secant case: A tangent from P touches at T, and a secant from P passes through A and B. Then:
PT² = PA × PB
PT² = PA × PB
Example: A tangent PT = 6 cm and a secant from P passes through A (nearer) and B. If PA = 3 cm, find PB.
PT² = PA × PB → 36 = 3 × PB → PB = 12 cm. So AB = PB − PA = 12 − 3 = 9 cm.
PT² = PA × PB → 36 = 3 × PB → PB = 12 cm. So AB = PB − PA = 12 − 3 = 9 cm.
Extended Content: Intersecting chord and secant results are Extended level only. Standard-level students should focus on the seven core theorems. Extended-level students must know all results and be prepared to combine them with trigonometry.
Mathematics · Topic 1.1
Multi-Step Problem Strategy
eAssessment questions often require chaining multiple theorems together. Here is a systematic approach.
The 4-Step Strategy
- Identify all given information — label every known angle, length, and relationship on the diagram.
- Spot the features — look for diameters, tangents, cyclic quadrilaterals, same-segment angles, and chords.
- Chain theorems — apply one theorem to find a new angle, then use that angle with another theorem.
- Name every theorem — write the theorem name in brackets after each calculation.
Summary Table: All Seven Theorems
| # | Theorem | Key Statement |
|---|---|---|
| 1 | Angle at centre | Central angle = 2 × inscribed angle (same arc) |
| 2 | Same segment | Inscribed angles from same arc are equal |
| 3 | Semicircle | Angle in a semicircle = 90° |
| 4 | Cyclic quad | Opposite angles sum to 180° |
| 5 | Tangent-radius | Tangent ⊥ radius at point of contact |
| 6 | Two tangents | Equal length from same external point |
| 7 | Alternate segment | Tangent-chord angle = inscribed angle in opposite segment |
Common Theorem Chains
Chain A
Angle at centre → same segment → find all inscribed angles
Chain B
Tangent-radius (90°) → triangle angle sum → alternate segment
Chain C
Angle in semicircle (90°) → Pythagoras or trigonometry
Chain D
Cyclic quad opposite angles → same segment → isosceles triangle
Exam Tip: In eAssessment extended response questions, marks are awarded for the logical chain of reasoning, not just the final answer. Even if your final answer is wrong, correct intermediate steps with named theorems will earn method marks.
Mathematics · Topic 1.1
Worked Examples
These examples demonstrate the multi-step reasoning and theorem-naming expected at Grade 10 eAssessment level.
EXAMPLE 1A central angle AOB = 110°. Point C is on the major arc. Find angle ACB.
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Full Solution
By the angle at centre theorem, the inscribed angle is half the central angle when subtended by the same arc.
C is on the major arc, so ∠ACB is subtended by the minor arc AB.
∠ACB = ∠AOB ÷ 2 = 110° ÷ 2 = 55°
Note: If C were on the minor arc, the reflex angle at O would be 360° − 110° = 250°, giving ∠ACB = 250° ÷ 2 = 125°.
C is on the major arc, so ∠ACB is subtended by the minor arc AB.
∠ACB = ∠AOB ÷ 2 = 110° ÷ 2 = 55°
Note: If C were on the minor arc, the reflex angle at O would be 360° − 110° = 250°, giving ∠ACB = 250° ÷ 2 = 125°.
EXAMPLE 2In cyclic quadrilateral ABCD, ∠A = 3x + 10° and ∠C = 2x + 20°. Find x and both angles.
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Full Solution
By the cyclic quadrilateral theorem, opposite angles sum to 180°.
∠A + ∠C = 180°
(3x + 10) + (2x + 20) = 180
5x + 30 = 180
5x = 150
x = 30
∠A = 3(30) + 10 = 100°
∠C = 2(30) + 20 = 80°
Check: 100 + 80 = 180° ✓
∠A + ∠C = 180°
(3x + 10) + (2x + 20) = 180
5x + 30 = 180
5x = 150
x = 30
∠A = 3(30) + 10 = 100°
∠C = 2(30) + 20 = 80°
Check: 100 + 80 = 180° ✓
EXAMPLE 3A tangent at P makes a 40° angle with chord PQ. Find the inscribed angle PRQ in the alternate segment.
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Full Solution
By the alternate segment theorem, the angle between a tangent and a chord equals the inscribed angle in the alternate segment.
The tangent-chord angle = 40°.
Therefore ∠PRQ = 40° (alternate segment theorem).
Justification: R is in the segment on the opposite side of chord PQ from the tangent-chord angle, which is why it is called the "alternate" segment.
The tangent-chord angle = 40°.
Therefore ∠PRQ = 40° (alternate segment theorem).
Justification: R is in the segment on the opposite side of chord PQ from the tangent-chord angle, which is why it is called the "alternate" segment.
EXAMPLE 4AB is a diameter. Tangent at B meets line AC extended at point T. ∠CAB = 32°. Find ∠ABT.
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Full Solution
Step 1: Since AB is a diameter, ∠ACB = 90° (angle in a semicircle).
Step 2: In triangle ACB: ∠ABC = 180° − 90° − 32° = 58° (angle sum of triangle).
Step 3: The tangent at B is perpendicular to radius OB. Since AB is a diameter, OB is a radius, so ∠ABT + ∠ABC = 90° − ... wait, let me reconsider. The tangent at B meets the radius OB at 90° (tangent-radius theorem). Since AB is a diameter, the tangent at B is perpendicular to AB.
Step 4: ∠ABT = 90° − ∠ABC = 90° − 58° = 32°
Note: This equals ∠CAB, which confirms the alternate segment theorem! The tangent-chord angle ∠ABT = inscribed angle ∠ACB's complement pair.
Step 2: In triangle ACB: ∠ABC = 180° − 90° − 32° = 58° (angle sum of triangle).
Step 3: The tangent at B is perpendicular to radius OB. Since AB is a diameter, OB is a radius, so ∠ABT + ∠ABC = 90° − ... wait, let me reconsider. The tangent at B meets the radius OB at 90° (tangent-radius theorem). Since AB is a diameter, the tangent at B is perpendicular to AB.
Step 4: ∠ABT = 90° − ∠ABC = 90° − 58° = 32°
Note: This equals ∠CAB, which confirms the alternate segment theorem! The tangent-chord angle ∠ABT = inscribed angle ∠ACB's complement pair.
EXAMPLE 5Two tangents from point T touch a circle at A and B. ∠ATB = 50°. Find the reflex angle AOB.
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Full Solution
Step 1: By the tangent-radius theorem, ∠OAT = 90° and ∠OBT = 90°.
Step 2: Quadrilateral OATB has angle sum = 360°.
∠AOB + ∠OAT + ∠ATB + ∠OBT = 360°
∠AOB + 90° + 50° + 90° = 360°
∠AOB = 360° − 230° = 130°
Step 3: The reflex angle AOB = 360° − 130° = 230°
Step 2: Quadrilateral OATB has angle sum = 360°.
∠AOB + ∠OAT + ∠ATB + ∠OBT = 360°
∠AOB + 90° + 50° + 90° = 360°
∠AOB = 360° − 230° = 130°
Step 3: The reflex angle AOB = 360° − 130° = 230°
EXAMPLE 6Chords AB and CD intersect inside a circle at point P. AP = 5, PB = 6, CP = 3. Find PD and the length CD.
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Full Solution
By the intersecting chords theorem: AP × PB = CP × PD.
5 × 6 = 3 × PD
30 = 3 × PD
PD = 10
CD = CP + PD = 3 + 10 = 13
5 × 6 = 3 × PD
30 = 3 × PD
PD = 10
CD = CP + PD = 3 + 10 = 13
EXAMPLE 7Points A, B, C, D lie on a circle. ∠BAC = 35° and ∠BDC = 35°. Explain why this is consistent with circle theorems.
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Full Solution
By the angles in the same segment theorem, angles subtended by the same arc at the circumference are equal.
Both ∠BAC and ∠BDC are inscribed angles subtended by arc BC. Since A and D are on the same side of chord BC (i.e., in the same segment), the theorem guarantees ∠BAC = ∠BDC.
The fact that both equal 35° is consistent with (and indeed predicted by) the same segment theorem. ✓
Both ∠BAC and ∠BDC are inscribed angles subtended by arc BC. Since A and D are on the same side of chord BC (i.e., in the same segment), the theorem guarantees ∠BAC = ∠BDC.
The fact that both equal 35° is consistent with (and indeed predicted by) the same segment theorem. ✓
Mathematics · Topic 1.1
Practice Q&A
Attempt each question before revealing the model answer. Name every theorem you use.
CALCULATEThe central angle AOB = 148°. Find the inscribed angle ACB where C is on the major arc.
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Model Answer
By the angle at centre theorem: the inscribed angle is half the central angle.
∠ACB = 148° ÷ 2 = 74°
∠ACB = 148° ÷ 2 = 74°
CALCULATEIn cyclic quadrilateral PQRS, ∠P = 105° and ∠Q = 82°. Find ∠R and ∠S.
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Model Answer
By the cyclic quadrilateral theorem: opposite angles sum to 180°.
∠R = 180° − ∠P = 180° − 105° = 75°
∠S = 180° − ∠Q = 180° − 82° = 98°
Check: 105 + 82 + 75 + 98 = 360° ✓
∠R = 180° − ∠P = 180° − 105° = 75°
∠S = 180° − ∠Q = 180° − 82° = 98°
Check: 105 + 82 + 75 + 98 = 360° ✓
JUSTIFYAB is a diameter of a circle. C is a point on the circumference. Prove that triangle ACB is right-angled.
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Model Answer
By the angle in a semicircle theorem: the angle subtended by a diameter at any point on the circumference is 90°.
AB is a diameter, so the angle ∠ACB is subtended by the diameter at point C on the circumference.
Therefore ∠ACB = 90°, making triangle ACB right-angled at C. ✓
Justification of theorem: The central angle for a diameter is 180° (straight line through centre). By Theorem 1, the inscribed angle = 180°/2 = 90°.
AB is a diameter, so the angle ∠ACB is subtended by the diameter at point C on the circumference.
Therefore ∠ACB = 90°, making triangle ACB right-angled at C. ✓
Justification of theorem: The central angle for a diameter is 180° (straight line through centre). By Theorem 1, the inscribed angle = 180°/2 = 90°.
CALCULATEA tangent at P and a chord PQ make an angle of 65°. Find the angle subtended by PQ at the circumference in the alternate segment.
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Model Answer
By the alternate segment theorem: the angle between a tangent and a chord equals the inscribed angle in the alternate segment.
The inscribed angle in the alternate segment = 65°.
The inscribed angle in the alternate segment = 65°.
MULTI-STEPTwo tangents from T touch a circle (centre O, radius 8 cm) at A and B. ∠ATB = 60°. Find the length TA.
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Model Answer
Step 1: By the two tangents theorem, TA = TB and ∠OTA = ∠OTB = 60° ÷ 2 = 30° (by symmetry).
Step 2: By the tangent-radius theorem, ∠OAT = 90°.
Step 3: In right triangle OAT: tan(30°) = OA / TA
tan(30°) = 8 / TA
TA = 8 / tan(30°) = 8 / (1/√3) = 8√3 ≈ 13.9 cm
Step 2: By the tangent-radius theorem, ∠OAT = 90°.
Step 3: In right triangle OAT: tan(30°) = OA / TA
tan(30°) = 8 / TA
TA = 8 / tan(30°) = 8 / (1/√3) = 8√3 ≈ 13.9 cm
EXTENDEDA tangent from P touches a circle at T. A secant from P passes through A and B (A is nearer). PT = 8, PA = 4. Find AB.
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Model Answer
By the tangent-secant theorem: PT² = PA × PB.
8² = 4 × PB
64 = 4PB
PB = 16
AB = PB − PA = 16 − 4 = 12
8² = 4 × PB
64 = 4PB
PB = 16
AB = PB − PA = 16 − 4 = 12
MULTI-STEPAB is a diameter. A tangent at A meets CB extended at T. ∠ABC = 55°. Find ∠BAT and ∠ATB.
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Model Answer
Step 1: ∠ACB = 90° (angle in a semicircle — AB is a diameter).
Step 2: In triangle ABC: ∠BAC = 180° − 90° − 55° = 35° (angle sum).
Step 3: By the tangent-radius theorem, the tangent at A is perpendicular to the radius OA. Since AB is a diameter, the tangent at A ⊥ AB, so ∠BAT = 90°.
Step 4: In triangle ABT: ∠ATB = 180° − ∠ABT − ∠BAT. Now ∠ABT = ∠ABC = 55° (same angle). So ∠ATB = 180° − 55° − 90° = 35°.
Step 2: In triangle ABC: ∠BAC = 180° − 90° − 55° = 35° (angle sum).
Step 3: By the tangent-radius theorem, the tangent at A is perpendicular to the radius OA. Since AB is a diameter, the tangent at A ⊥ AB, so ∠BAT = 90°.
Step 4: In triangle ABT: ∠ATB = 180° − ∠ABT − ∠BAT. Now ∠ABT = ∠ABC = 55° (same angle). So ∠ATB = 180° − 55° − 90° = 35°.
EXPLAINA student claims that a quadrilateral with angles 70°, 110°, 85°, 95° is cyclic. Is this correct? Justify.
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Model Answer
For a quadrilateral to be cyclic, opposite angles must sum to 180° (by the cyclic quadrilateral theorem).
Check: 70° + 85° = 155° ≠ 180°. Also: 110° + 95° = 205° ≠ 180°.
Neither pair of opposite angles sums to 180°, so the quadrilateral is NOT cyclic. The student is incorrect.
Note: All four angles sum to 360° (70 + 110 + 85 + 95 = 360 ✓), which is true of any quadrilateral. But this alone does not make it cyclic.
Check: 70° + 85° = 155° ≠ 180°. Also: 110° + 95° = 205° ≠ 180°.
Neither pair of opposite angles sums to 180°, so the quadrilateral is NOT cyclic. The student is incorrect.
Note: All four angles sum to 360° (70 + 110 + 85 + 95 = 360 ✓), which is true of any quadrilateral. But this alone does not make it cyclic.
Mathematics · Review
Flashcard Review
Tap each card to reveal the answer. Try to answer from memory first.
State the angle at centre theorem.
The angle at the centre is twice the angle at the circumference when both are subtended by the same arc.
Tap to reveal
What does the same segment theorem say?
Angles subtended by the same arc in the same segment are equal.
Tap to reveal
Angle in a semicircle?
The angle subtended by a diameter at the circumference is always 90°. (Special case of the angle at centre theorem.)
Tap to reveal
Cyclic quadrilateral property?
Opposite angles of a cyclic quadrilateral sum to 180°.
Tap to reveal
What angle does a tangent make with the radius?
90° — the tangent is perpendicular to the radius at the point of contact.
Tap to reveal
Two tangents from an external point — what is special?
They are equal in length. The line from the external point to the centre bisects the angle between the tangents.
Tap to reveal
State the alternate segment theorem.
The angle between a tangent and a chord at the point of contact equals the inscribed angle in the alternate segment.
Tap to reveal
Intersecting chords formula?
If chords AB and CD intersect at P: AP × PB = CP × PD.
Tap to reveal
Tangent-secant formula?
If tangent PT and secant PAB are from the same point P: PT² = PA × PB.
Tap to reveal
Central angle = 160°. Inscribed angle = ?
80° (half the central angle, by the angle at centre theorem).
Tap to reveal
How do you prove the angle in a semicircle is 90°?
A diameter creates a central angle of 180°. By the angle at centre theorem, the inscribed angle = 180° ÷ 2 = 90°.
Tap to reveal
Can a non-cyclic quadrilateral have opposite angles summing to 180°?
No. A quadrilateral is cyclic if and only if its opposite angles sum to 180°. This is both necessary and sufficient.
Tap to reveal
Why must you name theorems in eAssessment?
Criterion C (Communicating) requires correct mathematical terminology. Unnamed calculations lose communication marks.
Tap to reveal
What is a chord?
A straight line segment joining two points on the circumference of a circle. A diameter is the longest chord.
Tap to reveal
What is a tangent?
A line that touches the circle at exactly one point. It is perpendicular to the radius at that point of contact.
Tap to reveal
Mathematics · Practice Test
Practice Test — 20 Questions
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