Mathematics Β· AQA 8300 Β§G8
Circle Theorems
- State and apply all 8 circle theorems correctly, citing the theorem by name
- Give full geometric reasons for every theorem applied in a multi-step solution
- Combine three or more circle theorems in a single problem to find unknown angles
- Write algebraic expressions for angles using circle theorems and solve equations
- Prove circle theorems from first principles using isosceles triangles and angle facts
π Core Concepts
Theorem 1 β Angle at the Centre
The angle subtended at the centre of a circle is twice the angle subtended at the circumference, when both angles are on the same arc.
DEFINITION β Theorem 1
If $O$ is the centre of a circle and $A$, $B$, $P$ are points on the circumference, then $\angle AOB = 2\,\angle APB$, provided $A$, $B$ and $P$ are on the same arc.
Angle at Centre Formula
$$\angle AOB = 2 \times \angle APB$$
$O$ = centre of circle
$A, B$ = endpoints of arc
$P$ = point on major/minor arc
Proof sketch: Join $OP$ and extend to $Q$. Triangle $OAP$ is isosceles ($OA = OP$ = radii), so $\angle OAP = \angle OPA$. Let these equal $\alpha$. Then $\angle AOQ = 2\alpha$ (exterior angle of triangle). Similarly let $\angle OBP = \angle OPB = \beta$, giving $\angle BOQ = 2\beta$. Hence $\angle AOB = 2(\alpha+\beta) = 2\angle APB$. $\square$
EXAM TIP
This theorem works regardless of whether $P$ is on the major or minor arc β but if $P$ is on the minor arc, the angle at the centre is the reflex angle. Always identify which arc $P$ lies on before applying the factor of 2.
COMMON MISTAKE
Do not confuse "angle at the centre" with "angle at the circumference" β many students write $\angle APB = 2\,\angle AOB$, getting the relationship backwards. The centre angle is always the larger one (for the same arc).
Theorem 2 β Angle in a Semicircle
This is a special case of Theorem 1. When the arc $AB$ is a diameter, the angle at the centre is $180Β°$, so the angle at the circumference must be $90Β°$.
DEFINITION β Theorem 2
The angle inscribed in a semicircle is always $90Β°$. Equivalently, if $AB$ is a diameter and $P$ is any other point on the circle, then $\angle APB = 90Β°$.
Angle in a Semicircle
$$\angle APB = 90Β° \quad \text{(AB is a diameter)}$$
$A, B$ = endpoints of diameter
$P$ = any point on circumference (not $A$ or $B$)
EXAM TIP
Whenever you see a diameter in a circle with a third point on the circumference, immediately mark that angle as $90Β°$. This often opens up Pythagoras or trigonometry β a common Grade 9 synoptic link.
Theorem 3 β Angles in the Same Segment
All angles inscribed in the same segment of a circle (i.e. subtended by the same chord, from the same side) are equal.
DEFINITION β Theorem 3
If $A$, $B$ are two points on a circle, and $P$, $Q$ are any other points on the same arc, then $\angle APB = \angle AQB$.
Angles in the Same Segment
$$\angle APB = \angle AQB \quad (P, Q \text{ same arc})$$
$A, B$ = chord endpoints
$P, Q$ = points on the same arc
Proof: By Theorem 1, $\angle AOB = 2\angle APB$ and also $\angle AOB = 2\angle AQB$. Therefore $\angle APB = \angle AQB$. $\square$
MEMORY TRICK
"Same arc, same angle." If two angles are standing on the same chord and looking at the same arc, they are twins.
COMMON MISTAKE
If $P$ and $Q$ are on opposite arcs (opposite sides of chord $AB$), the angles are supplementary, not equal. This leads directly into Theorem 4.
Theorem 4 β Cyclic Quadrilateral
A cyclic quadrilateral is a quadrilateral with all four vertices on a circle. Its opposite angles are supplementary.
DEFINITION β Theorem 4
In a cyclic quadrilateral $ABCD$: $\angle A + \angle C = 180Β°$ and $\angle B + \angle D = 180Β°$.
Cyclic Quadrilateral
$$\angle A + \angle C = 180Β° \qquad \angle B + \angle D = 180Β°$$
$A, B, C, D$ = vertices in order on the circle
Proof: $\angle A$ and $\angle C$ are both inscribed angles standing on arcs $BCD$ and $BAD$ respectively, and these two arcs together make the full circle ($360Β°$). By Theorem 1, the central angles for those arcs sum to $360Β°$. Since each inscribed angle is half its central angle, $\angle A + \angle C = \tfrac{1}{2}(360Β°) = 180Β°$. $\square$
EXAM TIP
An exterior angle of a cyclic quadrilateral equals the interior opposite angle. This is tested directly: "Find the exterior angle at $D$."
COMMON MISTAKE
Students often write "opposite angles in a cyclic quad are equal." They are supplementary (sum to $180Β°$), not equal β unless the quadrilateral is a rectangle.
Theorem 5 β Tangent Perpendicular to Radius
A tangent to a circle is a straight line that touches the circle at exactly one point (the point of tangency). The radius to that point is always perpendicular to the tangent.
DEFINITION β Theorem 5
If $T$ is the point of tangency and $O$ is the centre, then $OT \perp$ tangent, i.e. $\angle OTP = 90Β°$ for any point $P$ on the tangent.
TangentβRadius Angle
$$\text{radius} \perp \text{tangent} \implies \angle OTP = 90Β°$$
$O$ = centre
$T$ = point of tangency
$P$ = any point on tangent line
EXAM TIP
Tangentβradius right angles unlock right-angled triangles. Combine with Pythagoras or trigonometry: "find the length of the tangent" or "find the distance from external point to centre."
Theorem 6 β Tangents from an External Point
Two tangents drawn from the same external point to a circle are always equal in length. The line from the external point to the centre bisects the angle between the tangents.
DEFINITION β Theorem 6
If tangents from external point $P$ touch the circle at $A$ and $B$, then $PA = PB$.
Equal Tangents
$$PA = PB \quad \text{(tangents from external point } P\text{)}$$
$P$ = external point
$A, B$ = points of tangency
Proof: In triangles $OAP$ and $OBP$: $OA = OB$ (radii), $OP = OP$ (common), $\angle OAP = \angle OBP = 90Β°$ (Theorem 5). By RHS congruence, $\triangle OAP \cong \triangle OBP$, hence $PA = PB$. $\square$
MEMORY TRICK
"Hat on a cone" β imagine two tangent lines as the sides of an ice-cream cone. Both sides are the same length. The centre sits directly above the tip.
COMMON MISTAKE
This theorem only holds when both lines are genuine tangents (touching, not crossing). Do not apply it if one line is a secant.
Theorem 7 β Alternate Segment Theorem
The angle between a tangent to a circle and a chord drawn from the point of tangency equals the inscribed angle subtending the same chord from the opposite (alternate) segment.
DEFINITION β Theorem 7
If $TP$ is a tangent at $T$ and $TC$ is a chord, and $B$ is any point on the arc on the other side of $TC$ from the tangent, then $\angle PTC = \angle TBC$ (angle in the alternate segment).
Alternate Segment Theorem
$$\angle \text{(tangent--chord)} = \angle \text{(inscribed in alternate segment)}$$
Tangentβchord angle = angle looking at chord from the far arc
Proof sketch: Let $\angle PTC = \alpha$. Draw diameter $TD$. Then $\angle TCD = 90Β°$ (angle in semicircle). Also $\angle DTQ = 90Β°$ (tangent $\perp$ radius), so $\angle DTP = 90Β°$. Thus $\angle DTC = 90Β° - \alpha$. In triangle $TCD$: $\angle TDC = 90Β° - (90Β°-\alpha) = \alpha$. By the same-arc theorem, any inscribed angle in the alternate segment equals $\angle TDC = \alpha$. $\square$
EXAM TIP
Identify the tangentβchord angle first. The "alternate segment" is the region on the opposite side of the chord from the tangent. Any inscribed angle there equals the tangentβchord angle.
COMMON MISTAKE
Students sometimes use the angle in the same segment, not the alternate. Remember: it must be the segment on the other side of the chord from where the tangent is.
Theorem 8 β Perpendicular from Centre Bisects Chord
The perpendicular from the centre of a circle to any chord bisects that chord. Conversely, the perpendicular bisector of any chord passes through the centre.
DEFINITION β Theorem 8
If $O$ is the centre and $OM \perp AB$ (chord), then $AM = MB$, i.e. $M$ is the midpoint of $AB$.
Perpendicular Bisector of Chord
$$OM \perp AB \implies AM = MB$$
$O$ = centre
$M$ = foot of perpendicular
$A, B$ = chord endpoints
Proof: Triangles $OAM$ and $OBM$: $OA = OB$ (radii), $OM = OM$ (common), $\angle OMA = \angle OMB = 90Β°$ (given). By RHS congruence, $\triangle OAM \cong \triangle OBM$, so $AM = MB$. $\square$
EXAM TIP
This theorem is frequently used in calculation problems: if you know the chord length and the perpendicular distance from the centre to the chord, you can find the radius using Pythagoras: $r^2 = d^2 + \left(\tfrac{l}{2}\right)^2$.
MEMORY TRICK
"The centre is fair β it always meets a chord at its midpoint when it drops a perpendicular."
πΊοΈ Visual Notes
Circle Theorems
Centre Angles
- Angle at centre = 2 Γ circumference
- Semicircle gives 90Β°
- Proof uses isosceles triangles
- Apply: reflex angle if P on minor arc
Inscribed Angles
- Same segment βΉ equal angles
- Opposite arcs βΉ supplementary
- Cyclic quad: opp. β s = 180Β°
- Exterior β = interior opposite
Tangents
- Tangent β₯ radius at contact
- Two tangents from P are equal
- Alternate segment theorem
- Right-angled triangle formed
Chords
- Perp from centre bisects chord
- Converse: perp bisector β centre
- Equal chords equidistant from centre
- Use Pythagoras with radius
Key Proofs
- All proofs use radii = isosceles β³
- Congruence (RHS, SAS, ASA)
- Exterior angle of triangle
- Angles on straight line / full turn
Grade 9 Links
- Combine 3+ theorems per problem
- Algebra: form and solve equations
- Trig in circle diagrams
- Formal proof required in exams
Quick-Reference: All 8 Theorems
| # | Theorem Name | Statement | Key Condition |
|---|---|---|---|
| 1 | Angle at Centre | Centre angle = 2 Γ circumference angle | Same arc |
| 2 | Angle in Semicircle | Angle in semicircle = 90Β° | AB is diameter |
| 3 | Same Segment | Inscribed angles in same segment are equal | Same arc, same side |
| 4 | Cyclic Quadrilateral | Opposite angles sum to 180Β° | All 4 vertices on circle |
| 5 | TangentβRadius | Tangent β₯ radius at point of contact | At point of tangency |
| 6 | Equal Tangents | Two tangents from same point are equal | External point |
| 7 | Alternate Segment | Tangentβchord angle = inscribed angle opposite | Alternate segment |
| 8 | Perpendicular Bisector | Perp from centre bisects chord | Perpendicular condition |
Problem-Solving Strategy
Identify all given angles, radii, tangents and chords
β
Look for: diameter (T2), same arc (T1/T3), cyclic quad (T4), tangent (T5/T6/T7), chord + perp (T8)
β
Apply theorems in logical order, marking each angle found
β
Form and solve algebraic equation if unknowns remain
β
State full geometric reason for every step
Theorem 1 vs Theorem 3 β Key Differences
| Feature | Theorem 1 (Centre) | Theorem 3 (Same Segment) |
|---|---|---|
| Vertex location | One vertex at centre $O$ | Both vertices on circumference |
| Relationship | Centre angle = 2 Γ circumference | Both inscribed angles are equal |
| Proof method | Isosceles triangle + exterior angle | Follows from Theorem 1 (both = half centre angle) |
| Typical use | Finding angle at centre / circumference | Identifying equal angles from same arc |
βοΈ Worked Examples
Grade 4β5 Β· Single Theorem
$O$ is the centre of a circle. Points $A$, $B$ and $C$ lie on the circumference. $\angle ACB = 38Β°$. Find $\angle AOB$.
1
Identify the theorem
$A$, $C$, $B$ are all on the circumference and $O$ is the centre. $\angle ACB$ is the angle at the circumference; $\angle AOB$ is the angle at the centre, both subtended by arc $AB$.
2
Apply Theorem 1
Angle at centre = $2 \times$ angle at circumference (same arc):
$$\angle AOB = 2 \times \angle ACB = 2 \times 38Β° = 76Β°$$
3
State the reason
"The angle at the centre is twice the angle at the circumference, subtended by the same arc."
$\angle AOB = 76Β°$
Grade 6β7 Β· Two Theorems
$ABCD$ is a cyclic quadrilateral. $\angle DAB = 3x + 10$ and $\angle BCD = 5x - 30$. Find the value of $x$ and calculate all four angles if $\angle ABC = 112Β°$.
1
Apply Theorem 4 (cyclic quadrilateral)
Opposite angles in a cyclic quadrilateral sum to $180Β°$:
$$\angle DAB + \angle BCD = 180Β°$$
$$(3x + 10) + (5x - 30) = 180Β°$$
2
Solve for $x$
$$8x - 20 = 180$$
$$8x = 200$$
$$x = 25$$
3
Find all angles
$\angle DAB = 3(25) + 10 = 85Β°$$\angle BCD = 5(25) - 30 = 95Β°$
$\angle ABC = 112Β°$ (given)
$\angle CDA = 180Β° - 112Β° = 68Β°$ (opposite angles, cyclic quad)
4
Verify
Check: $85Β° + 95Β° = 180Β°$ β and $112Β° + 68Β° = 180Β°$ β
$x = 25$; angles: $85Β°$, $112Β°$, $95Β°$, $68Β°$
Grade 9 Β· Multi-theorem + Algebra
$PT$ and $PS$ are tangents from external point $P$ to a circle with centre $O$, touching at $T$ and $S$. The chord $TS$ divides the circle into two segments. Point $Q$ lies on the major arc $TS$. $\angle TPS = 48Β°$. Find: (a) $\angle TOS$; (b) $\angle TQS$; (c) $\angle PTS$. Give a geometric reason for each answer.
1
Find $\angle TOS$ using quadrilateral $PTOS$
In quadrilateral $PTOS$: $\angle OTP = \angle OSP = 90Β°$ (tangent β₯ radius, Theorem 5).Angles in a quadrilateral sum to $360Β°$: $$\angle TOS = 360Β° - 90Β° - 90Β° - 48Β° = 132Β°$$
2
Find $\angle TQS$ using Theorem 1
$\angle TOS = 132Β°$ is the angle at the centre. $Q$ is on the major arc, so the angle at the centre for the minor arc is $360Β° - 132Β° = 228Β°$ (reflex). The angle at circumference on the major arc is:
$$\angle TQS = \frac{360Β° - 132Β°}{2}$$
Wait β we use the non-reflex angle correctly. Since $Q$ is on the major arc and the chord $TS$ subtends $\angle TOS = 132Β°$ at the centre (for the minor arc), the angle at the circumference from the major arc is:
$$\angle TQS = \frac{132Β°}{2} = 66Β°$$
(Angle at centre = 2 Γ angle at circumference, same minor arc, $Q$ on major arc.)
3
Find $\angle PTS$ using Theorem 6 and symmetry
Since $PT = PS$ (equal tangents, Theorem 6), triangle $PTS$ is isosceles. Also $\angle TPS = 48Β°$, so:
$$\angle PTS = \angle PST = \frac{180Β° - 48Β°}{2} = 66Β°$$
4
Confirm with Alternate Segment Theorem
Notice $\angle PTS = 66Β°$ and $\angle TQS = 66Β°$. This confirms the alternate segment theorem: the angle between the tangent $PT$ and chord $TS$ equals the inscribed angle in the alternate segment ($\angle TQS$). β
(a) $\angle TOS = 132Β°$ (angles in quad $PTOS = 360Β°$, tangentβ₯radius)
(b) $\angle TQS = 66Β°$ (angle at centre = 2 Γ angle at circumference)
(c) $\angle PTS = 66Β°$ (isosceles triangle, equal tangents from $P$) β confirmed by alternate segment theorem
(b) $\angle TQS = 66Β°$ (angle at centre = 2 Γ angle at circumference)
(c) $\angle PTS = 66Β°$ (isosceles triangle, equal tangents from $P$) β confirmed by alternate segment theorem
β Exam Questions
Q11 mark
$O$ is the centre of a circle. $A$ and $B$ are points on the circumference. $\angle AOB = 110Β°$. Find the size of $\angle ACB$ where $C$ is any point on the major arc.
Answer: $\angle ACB = 55Β°$
Mark scheme: $\angle ACB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 110Β° = 55Β°$ [1 mark β angle at centre = twice angle at circumference]
Mark scheme: $\angle ACB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 110Β° = 55Β°$ [1 mark β angle at centre = twice angle at circumference]
Q22 marks
$AB$ is a diameter of a circle. $C$ is a point on the circumference. $AC = 8\text{ cm}$ and $BC = 15\text{ cm}$. Find the diameter $AB$. Give your answer to 3 significant figures.
Answer: $AB = 17\text{ cm}$
Working: $\angle ACB = 90Β°$ (angle in a semicircle) [1 mark]. By Pythagoras: $AB^2 = AC^2 + BC^2 = 64 + 225 = 289$, so $AB = 17\text{ cm}$ [1 mark].
Working: $\angle ACB = 90Β°$ (angle in a semicircle) [1 mark]. By Pythagoras: $AB^2 = AC^2 + BC^2 = 64 + 225 = 289$, so $AB = 17\text{ cm}$ [1 mark].
Q33 marks
$ABCD$ is a cyclic quadrilateral. $\angle DAB = 2x + 15Β°$, $\angle ABC = 3x - 5Β°$, and $\angle BCD = x + 45Β°$. Find $\angle CDA$.
Answer: $\angle CDA = 115Β°$
Working:
Opposite angles sum to 180Β°: $\angle DAB + \angle BCD = 180Β°$ [1 mark]
$(2x+15) + (x+45) = 180 \Rightarrow 3x + 60 = 180 \Rightarrow x = 40$ [1 mark]
$\angle ABC = 3(40) - 5 = 115Β°$; $\angle CDA = 180Β° - 115Β° = 65Β°$
Wait β let's recalculate: $\angle CDA = 180Β° - \angle ABC = 180Β° - 115Β° = 65Β°$ [1 mark]
(Check: $\angle DAB = 95Β°$, $\angle BCD = 85Β°$, sum $= 180Β°$ β)
Working:
Opposite angles sum to 180Β°: $\angle DAB + \angle BCD = 180Β°$ [1 mark]
$(2x+15) + (x+45) = 180 \Rightarrow 3x + 60 = 180 \Rightarrow x = 40$ [1 mark]
$\angle ABC = 3(40) - 5 = 115Β°$; $\angle CDA = 180Β° - 115Β° = 65Β°$
Wait β let's recalculate: $\angle CDA = 180Β° - \angle ABC = 180Β° - 115Β° = 65Β°$ [1 mark]
(Check: $\angle DAB = 95Β°$, $\angle BCD = 85Β°$, sum $= 180Β°$ β)
Q44 marks
$PT$ is a tangent to a circle at $T$. $O$ is the centre. $\angle TPO = 32Β°$. Find: (a) $\angle OTP$; (b) $\angle TOP$; (c) the length $PT$ if $OP = 12\text{ cm}$, giving your answer to 3 significant figures.
Answers:
(a) $\angle OTP = 90Β°$ β tangent is perpendicular to radius at point of tangency [1 mark]
(b) $\angle TOP = 180Β° - 90Β° - 32Β° = 58Β°$ β angles in triangle sum to 180Β° [1 mark]
(c) Using trigonometry in right-angled triangle $OTP$:
$\tan 32Β° = \dfrac{OT}{PT}$, so $PT = \dfrac{OT}{\tan 32Β°}$
Also $\sin 32Β° = \dfrac{OT}{OP}$, so $OT = 12\sin 32Β° = 6.359\ldots$
$PT = 12\cos 32Β° = 10.2\text{ cm}$ (3 s.f.) [2 marks]
(a) $\angle OTP = 90Β°$ β tangent is perpendicular to radius at point of tangency [1 mark]
(b) $\angle TOP = 180Β° - 90Β° - 32Β° = 58Β°$ β angles in triangle sum to 180Β° [1 mark]
(c) Using trigonometry in right-angled triangle $OTP$:
$\tan 32Β° = \dfrac{OT}{PT}$, so $PT = \dfrac{OT}{\tan 32Β°}$
Also $\sin 32Β° = \dfrac{OT}{OP}$, so $OT = 12\sin 32Β° = 6.359\ldots$
$PT = 12\cos 32Β° = 10.2\text{ cm}$ (3 s.f.) [2 marks]
Q55 marks
$P$, $Q$, $R$ and $S$ are points on a circle. $PR$ is a diameter. $\angle QPR = 27Β°$ and $\angle RPS = 41Β°$. A tangent at $P$ makes an angle with chord $PQ$. Find: (a) $\angle PQR$; (b) $\angle PSR$; (c) $\angle QPS$; (d) the angle between the tangent at $P$ and chord $PS$.
Answers:
(a) $\angle PQR = 90Β°$ β $PR$ is a diameter, angle in semicircle [1 mark]
Therefore in triangle $PQR$: $\angle PQR + \angle QPR + \angle QRP = 180Β°$
$\angle QRP = 180Β° - 90Β° - 27Β° = 63Β°$
(b) $\angle PSR$ is subtended by the same arc $PR$ as $\angle PQR$? No β $\angle PSR$ and $\angle PQR$ subtend arc $PR$ from the same side. Since $PR$ is a diameter, $\angle PSR = 90Β°$ also (angle in semicircle). [1 mark]
(c) $\angle QPS = \angle QPR + \angle RPS = 27Β° + 41Β° = 68Β°$ [1 mark]
(d) By alternate segment theorem, angle between tangent at $P$ and chord $PS$ = inscribed angle in alternate segment = $\angle PRS$.
In triangle $PRS$: $\angle PSR = 90Β°$, $\angle RPS = 41Β°$, so $\angle PRS = 180Β° - 90Β° - 41Β° = 49Β°$.
Angle between tangent and $PS$ = $49Β°$ [2 marks]
(a) $\angle PQR = 90Β°$ β $PR$ is a diameter, angle in semicircle [1 mark]
Therefore in triangle $PQR$: $\angle PQR + \angle QPR + \angle QRP = 180Β°$
$\angle QRP = 180Β° - 90Β° - 27Β° = 63Β°$
(b) $\angle PSR$ is subtended by the same arc $PR$ as $\angle PQR$? No β $\angle PSR$ and $\angle PQR$ subtend arc $PR$ from the same side. Since $PR$ is a diameter, $\angle PSR = 90Β°$ also (angle in semicircle). [1 mark]
(c) $\angle QPS = \angle QPR + \angle RPS = 27Β° + 41Β° = 68Β°$ [1 mark]
(d) By alternate segment theorem, angle between tangent at $P$ and chord $PS$ = inscribed angle in alternate segment = $\angle PRS$.
In triangle $PRS$: $\angle PSR = 90Β°$, $\angle RPS = 41Β°$, so $\angle PRS = 180Β° - 90Β° - 41Β° = 49Β°$.
Angle between tangent and $PS$ = $49Β°$ [2 marks]
Q66 marks
$O$ is the centre of a circle. $A$, $B$, $C$, $D$ are on the circumference. $\angle BAD = 5yΒ°$, $\angle BCD = (3y + 20)Β°$, and $\angle BOD = (14y - 10)Β°$. (a) Show that $y = 20$. (b) Find $\angle BAD$. (c) Prove that $BD$ cannot be a diameter.
(a) $ABCD$ is a cyclic quadrilateral, so $\angle BAD + \angle BCD = 180Β°$ [1 mark]
$5y + 3y + 20 = 180 \Rightarrow 8y = 160 \Rightarrow y = 20$ [1 mark] β
(b) $\angle BAD = 5(20) = 100Β°$ [1 mark]
(c) $\angle BOD = 14(20) - 10 = 280 - 10 = 270Β°$. By Theorem 1 (angle at centre = 2 Γ circumference), if $BD$ were a diameter, $\angle BAD$ would have to be $90Β°$ (angle in semicircle). But $\angle BAD = 100Β° \neq 90Β°$, so $BD$ is not a diameter [2 marks]. Alternatively: angle at centre would be $180Β°$ for a diameter, but $\angle BOD = 270Β°$ (reflex), whose non-reflex counterpart is $90Β°$, giving circumference angle $45Β°$, not $100Β°$. Contradiction. [full credit] [1 mark for conclusion]
$5y + 3y + 20 = 180 \Rightarrow 8y = 160 \Rightarrow y = 20$ [1 mark] β
(b) $\angle BAD = 5(20) = 100Β°$ [1 mark]
(c) $\angle BOD = 14(20) - 10 = 280 - 10 = 270Β°$. By Theorem 1 (angle at centre = 2 Γ circumference), if $BD$ were a diameter, $\angle BAD$ would have to be $90Β°$ (angle in semicircle). But $\angle BAD = 100Β° \neq 90Β°$, so $BD$ is not a diameter [2 marks]. Alternatively: angle at centre would be $180Β°$ for a diameter, but $\angle BOD = 270Β°$ (reflex), whose non-reflex counterpart is $90Β°$, giving circumference angle $45Β°$, not $100Β°$. Contradiction. [full credit] [1 mark for conclusion]
β Grade 9 Model Answers
IMPORTANT β Why This Answer Earns Full Marks
Q6 above is a classic Grade 9 question. It tests: algebraic manipulation, cyclic quadrilateral theorem, the angle-at-centre theorem, and logical deduction (proof by contradiction). Each step below is annotated with why it earns the mark.
Full Model Answer for Q6
Grade 9 Β· Full Annotated Answer
$ABCD$ is cyclic. $\angle BAD = 5yΒ°$, $\angle BCD = (3y+20)Β°$, $\angle BOD = (14y-10)Β°$ where $O$ is the centre. (a) Show $y=20$. (b) Find $\angle BAD$. (c) Prove $BD$ is not a diameter.
1
Part (a) β State the theorem [M1]
"Opposite angles in a cyclic quadrilateral sum to 180Β°" β this is the reason that must be stated explicitly. Without it, the M1 method mark is lost even if the algebra is correct.
$$\angle BAD + \angle BCD = 180Β°$$
2
Part (a) β Algebra [A1]
$$5y + (3y + 20) = 180$$
$$8y + 20 = 180$$
$$8y = 160$$
$$y = 20 \quad \checkmark$$
The question says "show that", so both the algebraic steps and the final value must be shown β do not just write $y = 20$.
3
Part (b) β Substitute [B1]
$$\angle BAD = 5 \times 20 = 100Β°$$
Simple substitution, but must follow from a correct $y$.
4
Part (c) β Proof structure [M1 + A1 + reason]
Substitute $y = 20$ into $\angle BOD$: $14(20) - 10 = 270Β°$. This is a reflex angle.If $BD$ were a diameter, the angle at the centre over chord $BD$ would be $180Β°$ (a straight line). But the angle is $270Β°$ (or equivalently, the non-reflex angle is $360Β° - 270Β° = 90Β°$).
By Theorem 1: inscribed angle from major arc $= \frac{1}{2} \times 90Β° = 45Β°$.
But $\angle BAD = 100Β° \neq 45Β°$.
5
Conclusion β Contradiction [A1]
"If $BD$ were a diameter, then $\angle BAD$ would be $45Β°$ (angle at circumference = half non-reflex central angle). However $\angle BAD = 100Β°$. This is a contradiction, therefore $BD$ cannot be a diameter." [QED]The key examiner requirement: explicit contradiction with a stated theorem. Full marks require naming the circle theorem used.
(a) Using cyclic quad theorem: $8y + 20 = 180 \Rightarrow y = 20$ β
(b) $\angle BAD = 100Β°$
(c) If $BD$ were a diameter, $\angle BAD = 45Β°$ (half the non-reflex central angle of $90Β°$). But $\angle BAD = 100Β°$. Contradiction. $BD$ is not a diameter. $\square$
(b) $\angle BAD = 100Β°$
(c) If $BD$ were a diameter, $\angle BAD = 45Β°$ (half the non-reflex central angle of $90Β°$). But $\angle BAD = 100Β°$. Contradiction. $BD$ is not a diameter. $\square$
EXAM TIP β Proof Questions
Every proof mark requires: (1) the theorem name stated, (2) algebraic/numerical working, (3) a clear logical conclusion. Examiners use mark schemes requiring "fully correct proof with reasons" β vague statements like "by circle theorems" score zero.
π Revision Sheet
Key Definitions
| Term | Definition |
|---|---|
| Chord | A line segment joining two points on a circle |
| Arc | Part of the circumference of a circle |
| Tangent | A line touching the circle at exactly one point |
| Cyclic quad | A quadrilateral with all vertices on a circle |
| Inscribed angle | Angle with vertex on the circumference |
| Subtend | An arc/chord "subtends" an angle when it forms that angle at a point |
Essential Formulae
$$\angle AOB = 2\,\angle APB \quad \text{(same arc)}$$
$$\angle \text{in semicircle} = 90Β°$$
$$\angle A + \angle C = 180Β° \quad \text{(cyclic quad)}$$
$$OT \perp \text{tangent} \implies \angle OTP = 90Β°$$
$$PA = PB \quad \text{(equal tangents)}$$
$$\text{tangentβchord} = \text{alt. segment angle}$$
$$OM \perp AB \implies AM = MB$$
Memory Hooks
- T1: "Centre is boss β double the circumference angle"
- T2: "Diameter makes a right angle β always"
- T3: "Same arc, same angle β twins on the circle"
- T4: "Opposite in a cyclic quad β they add to a straight line"
- T5: "Tangent meets radius at a corner β always 90Β°"
- T6: "Two tangents from a hat β always equal sides"
- T7: "Slip into the other segment β same angle"
- T8: "Centre drops fair β hits chord at midpoint"
Exam Tips
- Always name the theorem β never just "circle theorem"
- Mark all known angles on the diagram as you go
- Look for isosceles triangles formed by two radii
- A tangentβradius pair always creates a right angle
- For proofs: state the theorem, show the algebra, state the conclusion
- Check if the question gives a diameter β Theorem 2 applies immediately
- Algebra questions: form an equation, solve, then verify the answer
- Multi-step: work systematically around the diagram, one angle at a time
π Flashcards
Click a card to reveal the answer. Use these to test your recall of all 8 theorems and key applications.
β Common Mistakes
MISTAKE 1 β Inverting Theorem 1
What students do wrong: Writing "angle at circumference = 2 Γ angle at centre", reversing the doubling factor.Why marks are lost: The angle relationship is stated backwards; any numerical answer derived from it is wrong.
How to avoid it: Memorise the hierarchy: centre is boss, bigger angle. If the inscribed angle is 30Β°, the central angle is 60Β° β never the other way.
MISTAKE 2 β Not stating the theorem name
What students do wrong: Writing "by circle theorems, $\angle x = 70Β°$" without specifying which theorem.Why marks are lost: The mark scheme requires a named reason. "Circle theorem" alone scores 0 for the reason mark.
How to avoid it: Always write the full theorem name: "angle at the centre is twice the angle at the circumference, subtended by the same arc."
MISTAKE 3 β Misidentifying the alternate segment
What students do wrong: Using the angle in the same segment (same side of the chord as the tangent) instead of the alternate (opposite) segment.Why marks are lost: The theorem says "alternate" β the other segment. Using the same-side angle gives a supplementary or unrelated angle.
How to avoid it: After drawing the chord from the point of tangency, shade the segment on the opposite side from the tangent. Any angle in that shaded region equals the tangentβchord angle.
MISTAKE 4 β Cyclic quad: equal vs supplementary
What students do wrong: Writing "opposite angles in a cyclic quadrilateral are equal" instead of supplementary.Why marks are lost: This error produces wrong equations ($\angle A = \angle C$ rather than $\angle A + \angle C = 180Β°$), giving a wrong value of $x$ and cascading errors throughout.
How to avoid it: Think "they balance to a straight line, $180Β°$". Only in a rectangle (special cyclic quad) are opposite angles equal (both $90Β°$).
MISTAKE 5 β Forgetting the reflex angle in Theorem 1
What students do wrong: When the point $P$ is on the minor arc, students use the non-reflex angle at the centre instead of the reflex angle.Why marks are lost: The angle at the centre relevant to $P$ on the minor arc is the reflex central angle. Using the non-reflex angle gives an answer that is the supplement of the correct answer.
How to avoid it: Always check: is $P$ on the major arc (use non-reflex $\angle AOB$) or the minor arc (use reflex $\angle AOB$)?
MISTAKE 6 β Applying tangent-from-point to a secant
What students do wrong: Using $PA = PB$ (Theorem 6) when one or both lines from $P$ are secants (cutting through the circle) rather than tangents.Why marks are lost: The equal tangent theorem only holds when both lines are tangent to the circle. A secant creates a different relationship and the lengths are not equal.
How to avoid it: Verify each line touches the circle at exactly one point. If a line crosses the circle at two points, it is a secant β do not apply Theorem 6.
β Final Checklist
Click each item when you are confident with it. Your progress is saved automatically.
- I can state Theorem 1: angle at centre = 2 Γ angle at circumference (same arc)
- I can state Theorem 2: angle in a semicircle = 90Β°, and recognise when AB is a diameter
- I can state Theorem 3: angles in the same segment are equal
- I can state Theorem 4: opposite angles in a cyclic quadrilateral sum to 180Β°
- I can state Theorem 5: tangent is perpendicular to the radius at the point of contact
- I can state Theorem 6: tangents from an external point are equal in length
- I can state Theorem 7: alternate segment theorem (tangentβchord = inscribed angle in alternate segment)
- I can state Theorem 8: the perpendicular from the centre bisects the chord
- I can prove Theorem 1 using isosceles triangles and the exterior angle of a triangle
- I can solve problems that require using three or more theorems in sequence
- I can form and solve algebraic equations using circle theorems
- I can combine circle theorems with trigonometry (e.g. right-angled triangles with tangentβradius)
- I always state the full theorem name as a reason in my working
- I can identify whether a point is on the major or minor arc before applying Theorem 1
- I can write a proof by contradiction to show a given line is or is not a diameter