Circle Theorems.
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Lesson Objectives
- Understand the basic parts of a circle relevant to circle theorems (chord, arc, segment, tangent)
- Apply the theorem relating the angle at the centre to the angle at the circumference
- Apply the theorem that angles subtended by the same arc in the same segment are equal
- Apply the theorem that the angle in a semicircle is a right angle
- Apply the properties of cyclic quadrilaterals to find unknown angles
- Apply tangent properties, including the alternate segment theorem, to find unknown angles and lengths
- Interpret circle diagrams correctly and identify which theorem applies to a given diagram
Introduction to Circle Theorems
Circle theorems are a set of rules describing the special angle and length relationships that occur between points, chords, and tangents on a circle. Unlike general triangle or polygon geometry, circle theorems rely on the unique symmetry of the circle: every point on the circumference is the same distance from the centre. These theorems are heavily examined in WAEC/NECO Mathematics because they combine geometric reasoning with algebraic angle-chasing, and they also appear in real-world contexts such as engineering designs involving circular arches, gear systems, and satellite dish alignment, where precise angle relationships around a curve must be calculated.
1. Angle at centre $= 2 \times$ angle at circumference (same arc)
2. Angles in the same segment (subtended by the same arc) are equal
3. Angle in a semicircle $= 90^\circ$
4. Opposite angles of a cyclic quadrilateral sum to $180^\circ$
5. Tangent $\perp$ radius at the point of contact
6. Two tangents drawn from the same external point are equal in length
7. Alternate segment theorem: tangent-chord angle $=$ angle in the alternate segment
• Chord: a straight line joining any two points on the circumference of a circle.
• Arc: a part of the circumference of a circle, lying between two points.
• Segment: the region enclosed between a chord and the arc it cuts off (major segment or minor segment).
• Tangent: a straight line that touches a circle at exactly one point, without crossing into the circle.
• Cyclic quadrilateral: a four-sided shape whose vertices all lie on the circumference of the same circle.
• Subtend: an arc or chord "subtends" an angle at a point when lines drawn from that point to the two ends of the arc/chord form that angle.
Quick Reference: Circle Theorems Summary
| Theorem | Statement | Applies To |
|---|---|---|
| Centre & Circumference | Angle at centre $= 2\times$ angle at circumference | Same arc, two points on major arc |
| Same Segment | Angles subtended by the same arc are equal | Two or more points in the same segment |
| Semicircle | Angle in a semicircle $= 90^\circ$ | Any point on the circle, diameter as one side |
| Cyclic Quadrilateral | Opposite angles sum to $180^\circ$ | Four points on the circle forming a quadrilateral |
| Tangent-Radius | Tangent is perpendicular to the radius at the point of contact | Any tangent line and its radius |
| Equal Tangents | Tangents from the same external point are equal in length | Two tangents drawn from one external point |
| Alternate Segment | Tangent-chord angle $=$ angle in the alternate segment | A tangent and a chord drawn from the same point of contact |
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Angle at the Centre and Angle at the Circumference
When an arc subtends an angle at the centre of a circle and also subtends an angle at the circumference (on the major arc), the angle at the centre is always exactly twice the angle at the circumference, provided both angles are subtended by the same arc. $$ \angle AOB = 2 \times \angle ACB $$
1. Identify the arc that is subtending both angles — it must be the same arc in both cases.
2. Identify the angle at the centre (formed by two radii to the ends of the arc).
3. Identify the angle at the circumference (formed at a point on the major arc).
4. Apply $\angle \text{centre} = 2 \times \angle \text{circumference}$, and solve for the unknown angle.
O is the centre; A and B are points on the circle; C is on the major arc. Angle AOB (120°) is twice angle ACB (60°), since both are subtended by the same arc AB.
Problem: In a circle with centre $O$, points $A$ and $B$ lie on the circumference such that $\angle AOB = 112^\circ$. Point $C$ lies on the major arc. Find $\angle ACB$.
Formula: $\angle \text{centre} = 2 \times \angle \text{circumference}$
Solution:
Step 1: $\angle AOB$ is the angle at the centre, subtended by arc $AB$: $\angle AOB = 112^\circ$.
Step 2: $\angle ACB$ is the angle at the circumference, subtended by the same arc $AB$.
Step 3: $\angle AOB = 2 \times \angle ACB \Rightarrow 112 = 2 \times \angle ACB \Rightarrow \angle ACB = 56^\circ$.
Answer: $\angle ACB = 56^\circ$
Problem: $A$, $B$, and $C$ are points on the circumference of a circle with centre $O$. $\angle BAC = 35^\circ$ (angle at the circumference subtended by arc $BC$). Find $\angle BOC$.
Formula: $\angle \text{centre} = 2 \times \angle \text{circumference}$
Solution:
Step 1: $\angle BAC$ is at the circumference, subtended by arc $BC$: $\angle BAC = 35^\circ$.
Step 2: $\angle BOC$ is at the centre, subtended by the same arc $BC$.
Step 3: $\angle BOC = 2 \times 35^\circ = 70^\circ$.
Answer: $\angle BOC = 70^\circ$
This theorem only applies when both angles are subtended by the same arc, with the circumference angle taken from a point on the major arc (the larger arc). If the angle at the circumference is measured from the minor arc instead, the relationship does not directly apply in this simple form — always check which arc each angle "sees".
Practice for Angle at the Centre and Angle at the Circumference
- $\angle AOB = 96^\circ$ at the centre. Find the angle at the circumference subtended by the same arc.
- The angle at the circumference is $42^\circ$. Find the angle at the centre subtended by the same arc.
- $\angle POQ = 150^\circ$ at the centre. $R$ is on the major arc. Find $\angle PRQ$.
- The angle at the centre is three times a certain angle at the circumference. Explain why this statement is only true under specific conditions, using the theorem.
- $\angle XOY = 80^\circ$. Find $\angle XZY$, where $Z$ lies on the major arc.
Angles in the Same Segment
When two or more points lie in the same segment of a circle and are joined to the two ends of the same chord, the angles they form are always equal. This is a direct consequence of the centre-circumference theorem, since every such angle at the circumference is half of the same fixed angle at the centre. $$ \angle ACB = \angle ADB \quad \text{(C and D in the same segment)} $$
1. Identify the chord that is being subtended (the shared base of both angles).
2. Confirm that both angles are formed from points in the same segment (the same side of the chord).
3. State that the two angles are equal, since they are subtended by the same arc.
4. Substitute any known angle value to find the unknown one.
A and B are fixed points on the circle (chord AB). C and D lie in the same segment. Angle ACB and angle ADB are both subtended by arc AB, so both equal 60°.
Problem: $A$, $B$, $C$, and $D$ are points on a circle, with $C$ and $D$ on the same side of chord $AB$. If $\angle ACB = 48^\circ$, find $\angle ADB$.
Formula: $\angle ACB = \angle ADB$ (same segment)
Solution:
Step 1: $C$ and $D$ are both in the same segment relative to chord $AB$.
Step 2: Both angles are therefore subtended by the same arc $AB$.
Step 3: By the same segment theorem, $\angle ADB = \angle ACB = 48^\circ$.
Answer: $\angle ADB = 48^\circ$
The two points must be in the same segment (same side of the chord) for their angles to be equal. If one point is in the major segment and the other in the minor segment, their angles are instead supplementary (they add up to $180^\circ$), not equal — this related idea is used again in the cyclic quadrilateral theorem.
Practice for Angles in the Same Segment
- $\angle ACB = 55^\circ$, with $D$ in the same segment as $C$. Find $\angle ADB$.
- $\angle PRQ = 73^\circ$, with $S$ in the same segment as $R$. Find $\angle PSQ$.
- Three points $C$, $D$, and $E$ all lie in the same segment relative to chord $AB$. If $\angle ACB = 40^\circ$, state $\angle ADB$ and $\angle AEB$, giving reasons.
- $\angle XZY = 3k$ and $\angle XWY = 2k + 15$, with $Z$ and $W$ in the same segment relative to chord $XY$. Find $k$.
- Explain, using the centre-circumference theorem, why angles in the same segment must be equal.
Angle in a Semicircle
A special case of the centre-circumference theorem occurs when the chord subtending the angle is actually a diameter. Since a diameter subtends an angle of $180^\circ$ at the centre (a straight line), the angle it subtends at any point on the circumference must be exactly half of that: $90^\circ$. This means any triangle formed by a diameter and a third point on the circle is always right-angled. $$ \angle ACB = 90^\circ \quad \text{when } AB \text{ is a diameter} $$
1. Confirm that the chord in question is a diameter (it passes through the centre).
2. Identify the third point on the circumference forming the triangle with the diameter.
3. State that the angle at this third point is $90^\circ$, by the angle-in-a-semicircle theorem.
4. Use this right angle together with angle-sum rules to find any other missing angles in the triangle.
AB is a diameter through centre O. C is any point on the circle. Angle ACB is always 90°, regardless of where C is placed on the circle.
Problem: $AB$ is a diameter of a circle, and $C$ is a point on the circumference such that $\angle BAC = 34^\circ$. Find $\angle ABC$.
Formula: $\angle ACB = 90^\circ$ (angle in a semicircle); angle sum of a triangle $= 180^\circ$
Solution:
Step 1: Since $AB$ is a diameter, $\angle ACB = 90^\circ$.
Step 2: The three angles of triangle $ABC$ sum to $180^\circ$: $\angle BAC + \angle ABC + \angle ACB = 180^\circ$.
Step 3: Substitute known values: $34^\circ + \angle ABC + 90^\circ = 180^\circ$.
Step 4: Solve: $\angle ABC = 180^\circ - 124^\circ = 56^\circ$.
Answer: $\angle ABC = 56^\circ$
This theorem applies only when the chord is a full diameter (passing exactly through the centre). A chord that merely looks long, but does not pass through the centre, does not produce a $90^\circ$ angle at the circumference — always check that the chord is confirmed as a diameter, either directly stated or shown passing through the marked centre.
Practice for Angle in a Semicircle
- $AB$ is a diameter, and $\angle ABC = 27^\circ$. Find $\angle BAC$.
- $PQ$ is a diameter, and $\angle QPR = 61^\circ$. Find $\angle PRQ$ and $\angle PQR$.
- Explain why a triangle inscribed in a semicircle, with one side as the diameter, is always right-angled.
- $AB$ is a diameter of length $10$ cm. If $\angle ACB = 90^\circ$ and $AC = 6$ cm, find $BC$ using Pythagoras' theorem.
- $XY$ is a diameter. $\angle XZY$ is claimed to be $85^\circ$. State, with a reason, whether this is possible.
Cyclic Quadrilaterals
A cyclic quadrilateral is a four-sided shape whose four vertices all lie on the circumference of the same circle. In any cyclic quadrilateral, each pair of opposite angles sums to $180^\circ$ (they are supplementary). A closely related result is that the exterior angle at any vertex of a cyclic quadrilateral equals the interior opposite angle. $$ \angle A + \angle C = 180^\circ \qquad \angle B + \angle D = 180^\circ $$
1. Confirm that all four vertices of the quadrilateral lie on the circle.
2. Identify each pair of opposite angles (angles that do not share a side).
3. Set up the equation: (one angle) $+$ (its opposite angle) $= 180^\circ$.
4. Solve for the unknown angle, repeating for the other pair if needed.
A, B, C, and D lie on the circle, forming cyclic quadrilateral ABCD. Angle A + angle C = 180°, and angle B + angle D = 180°.
Problem: $ABCD$ is a cyclic quadrilateral with $\angle A = 85^\circ$. Find $\angle C$.
Formula: $\angle A + \angle C = 180^\circ$
Solution:
Step 1: $A$ and $C$ are opposite angles in cyclic quadrilateral $ABCD$.
Step 2: $\angle A + \angle C = 180^\circ \Rightarrow 85^\circ + \angle C = 180^\circ$.
Step 3: $\angle C = 180^\circ - 85^\circ = 95^\circ$.
Answer: $\angle C = 95^\circ$
Problem: In cyclic quadrilateral $PQRS$, side $QR$ is extended to a point $T$. If $\angle SRT = 108^\circ$ (the exterior angle at $R$), find the interior opposite angle $\angle SPQ$.
Formula: Exterior angle of cyclic quadrilateral $=$ interior opposite angle
Solution:
Step 1: The exterior angle at $R$ is $\angle SRT = 108^\circ$.
Step 2: By the exterior angle property, this equals the interior angle opposite to $R$, which is $\angle SPQ$.
Step 3: $\angle SPQ = 108^\circ$.
Answer: $\angle SPQ = 108^\circ$
Opposite angles in a cyclic quadrilateral are supplementary ($180^\circ$), not equal. A common error is applying the same-segment "equal angles" rule to opposite vertices of a quadrilateral — remember that the equal-angle rule applies to angles subtended from the same segment, while the $180^\circ$ rule applies specifically to opposite vertices of a cyclic quadrilateral.
Practice for Cyclic Quadrilaterals
- $ABCD$ is a cyclic quadrilateral with $\angle B = 76^\circ$. Find $\angle D$.
- $WXYZ$ is a cyclic quadrilateral with $\angle W = 4k$ and $\angle Y = 2k$. Find $k$ and hence $\angle W$ and $\angle Y$.
- In cyclic quadrilateral $PQRS$, side $PQ$ is extended to $T$, forming exterior angle $\angle TQR = 95^\circ$. Find the interior opposite angle.
- A cyclic quadrilateral has angles $\angle A = 3x + 10$, $\angle B = 2x$, $\angle C = x + 50$, $\angle D = 4x - 20$. Use the opposite-angle property to form and solve an equation for $x$.
- Explain why a rectangle is always a cyclic quadrilateral, using the opposite-angle theorem.
Tangents to a Circle
A tangent is a straight line that touches a circle at exactly one point, called the point of contact. Tangents have three key properties used throughout SSCE geometry: the tangent is always perpendicular to the radius drawn to the point of contact; two tangents drawn from the same external point to a circle are always equal in length; and the angle between a tangent and a chord drawn from the point of contact equals the angle in the alternate segment (the alternate segment theorem). $$ OT \perp \text{tangent at } T \qquad PA = PB \qquad \angle(\text{tangent, chord}) = \angle(\text{alternate segment}) $$
1. Identify the point(s) of tangency and draw/imagine the radius to each.
2. If two tangents share an external point, mark them as equal in length, forming an isosceles or kite-shaped figure.
3. For angle problems, check whether a right angle (tangent-radius) or the alternate segment theorem is needed.
4. Combine the tangent property with angle-sum or isosceles-triangle rules to solve for unknowns.
P is an external point; PA and PB are tangents to the circle, touching at A and B. Radii OA and OB are perpendicular to the tangents (right angles marked), and PA = PB.
Problem: $PA$ and $PB$ are tangents from external point $P$ to a circle with centre $O$, touching at $A$ and $B$. If $\angle APB = 50^\circ$, find $\angle OAP$ and $\angle AOB$.
Formula: Tangent $\perp$ radius; angle sum of quadrilateral $= 360^\circ$
Solution:
Step 1: Since $PA$ is a tangent and $OA$ is the radius to the point of contact, $\angle OAP = 90^\circ$. By the same reasoning, $\angle OBP = 90^\circ$.
Step 2: Quadrilateral $OAPB$ has angle sum $360^\circ$: $\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ$.
Step 3: Substitute known angles: $90^\circ + 50^\circ + 90^\circ + \angle AOB = 360^\circ$.
Step 4: Solve: $\angle AOB = 360^\circ - 230^\circ = 130^\circ$.
Answer: $\angle OAP = 90^\circ$, $\angle AOB = 130^\circ$
TQ is a chord from the point of tangency T; the tangent-chord angle marked x on the left equals angle TRQ (marked x) in the alternate segment, where R lies on the arc on the far side of chord TQ.
Problem: A tangent touches a circle at $T$. $TQ$ is a chord, and the tangent-chord angle on one side of $TQ$ is $42^\circ$. $R$ is a point on the arc in the alternate segment. Find $\angle TRQ$.
Formula: Tangent-chord angle $=$ angle in the alternate segment
Solution:
Step 1: The tangent-chord angle is given as $42^\circ$.
Step 2: By the alternate segment theorem, this equals the angle subtended by chord $TQ$ at any point in the alternate segment, including $R$.
Step 3: $\angle TRQ = 42^\circ$.
Answer: $\angle TRQ = 42^\circ$
In the alternate segment theorem, the angle you compare with the tangent-chord angle must be measured in the alternate (opposite) segment, not the same one. If you pick a point on the same side as the marked tangent-chord angle, you get the supplementary angle instead, not an equal one.
Practice for Tangents to a Circle
- $PA$ and $PB$ are tangents from external point $P$. If $\angle APB = 64^\circ$, find $\angle AOB$ (where $O$ is the centre).
- A tangent touches a circle at $T$, and radius $OT = 5$ cm. If $P$ is an external point with $PT = 12$ cm, find $OP$ using Pythagoras' theorem (recall $OT \perp PT$).
- The tangent-chord angle at a point of contact is $58^\circ$. Find the angle in the alternate segment.
- Two tangents from an external point $P$ touch a circle at $A$ and $B$. If $PA = 9$ cm, find $PB$, giving a reason.
- A tangent-chord angle is marked as $x + 10$, and the angle in the alternate segment is marked as $2x - 5$. Find $x$.
Cumulative Practice Exercises
- $\angle AOB = 88^\circ$ at the centre of a circle. Find the angle at the circumference subtended by the same arc.
- $\angle PCQ = 39^\circ$, with $D$ in the same segment as $C$ relative to chord $PQ$. Find $\angle PDQ$.
- $AB$ is a diameter, and $\angle BAC = 22^\circ$. Find $\angle ABC$.
- $ABCD$ is a cyclic quadrilateral with $\angle A = 102^\circ$. Find $\angle C$.
- $PA$ and $PB$ are tangents from external point $P$, with $\angle APB = 46^\circ$. Find $\angle AOB$.
- A tangent-chord angle is $63^\circ$. Find the angle in the alternate segment.
- $\angle XOY = 140^\circ$ at the centre. $Z$ is on the major arc. Find $\angle XZY$.
- $WXYZ$ is a cyclic quadrilateral. Side $WX$ is extended to $T$, forming exterior angle $\angle TXY = 99^\circ$. Find the interior opposite angle.
- $MN$ is a diameter of a circle, and $\angle NMK = 48^\circ$. Find $\angle MKN$ and $\angle MNK$.
- Two tangents from an external point $P$ touch a circle at $A$ and $B$, with $\angle OAP = 90^\circ$ and $\angle APB = 54^\circ$. Find $\angle AOB$.
- $\angle ACB = 3k+5$ and $\angle ADB = k+35$, with $C$ and $D$ in the same segment relative to chord $AB$. Find $k$.
- A cyclic quadrilateral has $\angle A = 3x + 15$ and the opposite angle $\angle C = 2x + 25$. Find $x$, then state $\angle A$ and $\angle C$.
Solutions to Cumulative Exercises
-
Step 1: $\angle AOB$ is at the centre, subtended by arc $AB$: $88^\circ$.
Step 2: By the centre-circumference theorem, the circumference angle is half: $88^\circ \div 2 = 44^\circ$.
Answer: $44^\circ$ -
Step 1: $C$ and $D$ are in the same segment relative to chord $PQ$, so they subtend equal angles.
Step 2: $\angle PDQ = \angle PCQ = 39^\circ$.
Answer: $39^\circ$ -
Step 1: Since $AB$ is a diameter, $\angle ACB = 90^\circ$.
Step 2: Angle sum of triangle $ABC$: $22^\circ + \angle ABC + 90^\circ = 180^\circ \Rightarrow \angle ABC = 68^\circ$.
Answer: $68^\circ$ -
Step 1: $A$ and $C$ are opposite angles in cyclic quadrilateral $ABCD$.
Step 2: $\angle A + \angle C = 180^\circ \Rightarrow 102^\circ + \angle C = 180^\circ \Rightarrow \angle C = 78^\circ$.
Answer: $78^\circ$ -
Step 1: $\angle OAP = \angle OBP = 90^\circ$ (tangent $\perp$ radius).
Step 2: Quadrilateral $OAPB$: $90^\circ+46^\circ+90^\circ+\angle AOB=360^\circ \Rightarrow \angle AOB = 134^\circ$.
Answer: $134^\circ$ -
Step 1: By the alternate segment theorem, tangent-chord angle equals the angle in the alternate segment.
Step 2: The angle in the alternate segment $= 63^\circ$.
Answer: $63^\circ$ -
Step 1: $\angle XOY$ is at the centre, subtended by arc $XY$: $140^\circ$.
Step 2: $\angle XZY = 140^\circ \div 2 = 70^\circ$.
Answer: $70^\circ$ -
Step 1: The exterior angle at $X$ ($\angle TXY = 99^\circ$) equals the interior opposite angle at $Z$.
Step 2: $\angle WZY = 99^\circ$.
Answer: $99^\circ$ -
Step 1: Since $MN$ is a diameter, $\angle MKN = 90^\circ$.
Step 2: Angle sum of triangle $MNK$: $48^\circ + \angle MNK + 90^\circ = 180^\circ \Rightarrow \angle MNK = 42^\circ$.
Answer: $\angle MKN = 90^\circ$, $\angle MNK = 42^\circ$ -
Step 1: Quadrilateral $OAPB$: $90^\circ + 54^\circ + 90^\circ + \angle AOB = 360^\circ$.
Step 2: $\angle AOB = 360^\circ - 234^\circ = 126^\circ$.
Answer: $126^\circ$ -
Step 1: Since $C$ and $D$ are in the same segment, $\angle ACB = \angle ADB$.
Step 2: $3k+5 = k+35 \Rightarrow 2k = 30 \Rightarrow k = 15$.
Answer: $k = 15$ -
Step 1: $A$ and $C$ are opposite angles, so $\angle A + \angle C = 180^\circ$.
Step 2: $(3x+15)+(2x+25)=180 \Rightarrow 5x+40=180 \Rightarrow 5x=140 \Rightarrow x=28$.
Answer: $x=28$, $\angle A = 3(28)+15 = 99^\circ$, $\angle C = 2(28)+25 = 81^\circ$
Conclusion & Summary
This lesson covered the core circle theorems examined at SSCE level: the centre-circumference relationship, equal angles in the same segment, the right angle in a semicircle, the supplementary opposite angles of a cyclic quadrilateral, and the key tangent properties including the alternate segment theorem. Circle theorem questions are rarely solved using just one theorem — most require you to identify which theorem applies to each part of a diagram and chain several results together, so practising diagram recognition is just as important as memorising the theorem statements themselves.
Key Takeaways:
1. Centre & Circumference: $\angle \text{centre} = 2 \times \angle \text{circumference}$
2. Same Segment: angles subtended by the same arc, in the same segment, are equal
3. Semicircle: $\angle \text{circumference} = 90^\circ$ when the chord is a diameter
4. Cyclic Quadrilateral: opposite angles sum to $180^\circ$
5. Tangents: tangent $\perp$ radius; equal tangents from one external point; tangent-chord angle $=$ angle in alternate segment
Always start a circle theorem question by labelling every angle you can find directly from the diagram — the correct theorem to apply usually becomes clear once the diagram is fully labelled.
Video Resource
Watch this video for more examples of circle theorem problems.
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