Chapter 5: Circle Theorems
Circle theorems describe the angle relationships that arise whenever points, chords, and tangents interact within or around a circle. Once you have learned the seven key theorems, a wide family of exam questions becomes manageable — the skill lies in recognising which theorem (or combination of theorems) applies, and in providing the correct geometric reason.
Exam Board Coverage
All seven theorems are assessed at both Foundation and Extended tier by Cambridge (0580) and Edexcel (4MA1). The Alternate Segment Theorem and multi-step problems combining several theorems are more common at Extended level. Always state the theorem name as your reason — numerical reasoning alone does not earn method marks.
5.1 Circle Vocabulary
Key Terms
- Centre — the fixed point equidistant from all points on the circle, usually labelled $O$.
- Radius (pl. radii) — a straight line from the centre to any point on the circle.
- Diameter — a chord through the centre; length $= 2r$.
- Chord — a straight line segment joining two points on the circle (a diameter is a special chord).
- Arc — part of the circumference. A minor arc is the shorter part; a major arc is the longer part.
- Sector — a region bounded by two radii and an arc (like a slice of pizza).
- Segment — a region bounded by a chord and an arc. The minor segment is the smaller region; the major segment is the larger.
- Tangent — a straight line that touches the circle at exactly one point (the point of tangency).
- Cyclic polygon — a polygon whose vertices all lie on a circle (it is inscribed in the circle).
Circle with centre $O$. Points $P$, $Q$ on the circle; angle at centre $2\theta$, angle at circumference $\theta$.
Angle at centre $= 2\times$ angle at circumference
5.2 Theorem 1: Angle at the Centre
Theorem 1 — Angle at the Centre is Twice the Angle at the Circumference
The angle subtended at the centre of a circle by an arc is twice the angle subtended at the circumference by the same arc.
$$\angle AOB = 2 \times \angle APB$$
where $O$ is the centre and $P$ is any point on the major arc.
Example 5.1 — Angle at the centre
$O$ is the centre of a circle. Angle $AOB = 130°$. Point $P$ is on the major arc. Find angle $APB$.
Step 1 Angle at circumference $= \tfrac{1}{2} \times$ angle at centre.
Step 2 $\angle APB = \tfrac{1}{2} \times 130° = \mathbf{65°}$
Reason: angle at centre is twice angle at circumference.
5.3 Theorem 2: Angle in a Semicircle
Theorem 2 — Angle in a Semicircle
The angle subtended at the circumference by a diameter is always $\mathbf{90°}$.
Equivalently: if $AB$ is a diameter and $P$ is any other point on the circle, then $\angle APB = 90°$.
Proof: The angle at the centre on a diameter is $180°$ (a straight line). By Theorem 1, the angle at the circumference is half of $180° = 90°$.
AB is a diameter. P is any point on the circle. $\angle APB = 90°$.
Example 5.2 — Angle in a semicircle
$AB$ is a diameter of a circle. Point $C$ is on the circle. Angle $CAB = 38°$. Find angle $ABC$.
Step 1 By Theorem 2, $\angle ACB = 90°$ (angle in a semicircle).
Step 2 Angles in triangle $ABC$ sum to $180°$:
$$\angle ABC = 180° - 90° - 38° = \mathbf{52°}$$
5.4 Theorem 3: Angles in the Same Segment
Theorem 3 — Angles in the Same Segment are Equal
Angles subtended by the same chord on the same side of the chord (i.e., from the same arc) are equal.
If $P$ and $Q$ are both on the same arc of chord $AB$: $\angle APB = \angle AQB$.
P and Q are on the same (major) arc. $\angle APB = \angle AQB = \alpha$.
Example 5.3 — Same segment
$ABCD$ are four points on a circle. $\angle ADB = 47°$. Find $\angle ACB$.
Reason $C$ and $D$ are on the same arc of chord $AB$.
$\angle ACB = \angle ADB = \mathbf{47°}$ (angles in the same segment).
5.5 Theorem 4: Cyclic Quadrilateral
Theorem 4 — Opposite Angles in a Cyclic Quadrilateral Sum to 180°
If a quadrilateral $ABCD$ is inscribed in a circle (all four vertices on the circle), then:
$$\angle A + \angle C = 180° \qquad \text{and} \qquad \angle B + \angle D = 180°$$
The opposite angles are supplementary.
ABCD cyclic quadrilateral: $\angle A + \angle C = 180°$, $\angle B + \angle D = 180°$.
Example 5.4 — Cyclic quadrilateral
$PQRS$ is a cyclic quadrilateral. Angle $P = 112°$ and angle $Q = 78°$. Find angles $R$ and $S$.
Step 1 Opposite angles: $\angle P + \angle R = 180°$
$\angle R = 180° - 112° = \mathbf{68°}$ (opposite angles in cyclic quadrilateral)
Step 2 $\angle Q + \angle S = 180°$
$\angle S = 180° - 78° = \mathbf{102°}$
5.6 Theorem 5: Tangent Perpendicular to Radius
Theorem 5 — Tangent is Perpendicular to the Radius at the Point of Contact
If a tangent touches a circle at point $T$, and $OT$ is the radius to $T$, then $OT \perp$ tangent: the angle between the radius and the tangent is $\mathbf{90°}$.
Radius OP meets the tangent at $P$ at exactly 90°.
Example 5.5 — Tangent and radius
$O$ is the centre of a circle. $PT$ is a tangent at $T$. $\angle TOP = 55°$. Find $\angle TPO$.
Step 1 $\angle OTP = 90°$ (tangent perpendicular to radius).
Step 2 Angle sum in triangle $OTP$: $\angle TPO = 180° - 90° - 55° = \mathbf{35°}$.
5.7 Theorem 6: Two Tangents from an External Point
Theorem 6 — Two Tangents from an External Point are Equal in Length
If two tangent lines from an external point $P$ touch the circle at $T_1$ and $T_2$, then $PT_1 = PT_2$.
Furthermore, $OP$ bisects angle $T_1 O T_2$ and also angle $T_1 P T_2$ (the quadrilateral $OT_1PT_2$ has two pairs of equal angles).
From external point E, $ET_1 = ET_2$. Right angles at $T_1$ and $T_2$.
Example 5.6 — Two tangents
From an external point $P$, two tangents $PA$ and $PB$ touch a circle with centre $O$. $\angle APB = 64°$. Find $\angle AOB$.
Step 1 $\angle OAP = \angle OBP = 90°$ (tangent ⊥ radius at both touch points).
Step 2 Angle sum in quadrilateral $OAPB$: $360° = 90° + 90° + 64° + \angle AOB$
Step 3 $\angle AOB = 360° - 244° = \mathbf{116°}$
5.8 Theorem 7: Alternate Segment Theorem Extended
Theorem 7 — Alternate Segment Theorem (Tangent-Chord Angle)
The angle between a tangent to a circle and a chord drawn from the point of tangency equals the angle in the alternate segment (the segment on the opposite side of the chord).
If $TA$ is a tangent at $T$ and $TB$ is a chord, then: $\angle ATB = \angle TXB$ for any point $X$ in the alternate segment.
Tangent at P, chord PQ. Angle between tangent and chord (α) equals inscribed angle at R (α) in the alternate segment.
Example 5.7 — Alternate segment theorem
$PQ$ is a tangent to a circle at $T$. $TA$ is a chord. $\angle ATQ = 58°$. Point $B$ is in the alternate segment. Find $\angle TBA$.
Step 1 By the alternate segment theorem: $\angle TBA = \angle ATQ = \mathbf{58°}$
Reason: angle between tangent $PQ$ and chord $TA$ equals angle in alternate segment.
5.9 Perpendicular from Centre to Chord
Perpendicular from the Centre Bisects the Chord
If a line is drawn from the centre $O$ of a circle perpendicular to a chord $AB$, it bisects the chord: the perpendicular meets $AB$ at its midpoint $M$, so $AM = MB$.
Conversely, the perpendicular bisector of any chord passes through the centre.
OM ⊥ chord AB. M is the midpoint of AB: $AM = MB$.
Example 5.8 — Perpendicular from centre
A circle has centre $O$ and radius $10\,\text{cm}$. A chord $AB$ is $16\,\text{cm}$ long. Find the perpendicular distance from $O$ to the chord.
Step 1 The perpendicular from $O$ bisects $AB$, so $AM = 8\,\text{cm}$.
Step 2 In right-angled triangle $OAM$: $OA = 10$ (radius), $AM = 8$.
Step 3 $OM^2 = OA^2 - AM^2 = 100 - 64 = 36 \Rightarrow OM = \mathbf{6\,\text{cm}}$
5.10 Combining Theorems
Harder exam questions require you to apply two or more circle theorems in sequence. The strategy is:
- Identify every circle theorem that might apply to the diagram.
- Find any angle you can find, even if it is not the final answer.
- Use each intermediate result to unlock the next.
- State the theorem name at each step as your reason.
Example 5.9 — Combining theorems
$O$ is the centre of a circle. $A$, $B$, $C$, $D$ are on the circle. $\angle AOC = 140°$. $ABCD$ is a cyclic quadrilateral. Find $\angle ABC$ and $\angle ADC$.
Step 1 Angle at centre $\angle AOC = 140°$. The angle at the circumference on the major arc is $\tfrac{1}{2} \times 140° = 70°$.
So any inscribed angle on the major arc $= 70°$. But here we need $\angle ADC$ (on the major arc).
$\angle ADC = \tfrac{1}{2} \times 140° = \mathbf{70°}$ (angle at centre = twice angle at circumference).
Step 2 $ABCD$ is cyclic, so opposite angles sum to $180°$:
$\angle ABC = 180° - 70° = \mathbf{110°}$ (opposite angles in cyclic quadrilateral).
Exercises
In each question, $O$ denotes the centre of the circle unless otherwise stated. State your geometric reason for each angle found.
Exercise 5.1
$O$ is the centre. $\angle BOC = 96°$. $A$ is on the major arc. Find $\angle BAC$.
Show Solution
$\angle BAC = \tfrac{1}{2} \times 96° = \mathbf{48°}$ (angle at centre is twice angle at circumference)
Exercise 5.2
$AB$ is a diameter. $\angle BAC = 53°$. Find $\angle ABC$.
Show Solution
$\angle ACB = 90°$ (angle in a semicircle).
$\angle ABC = 180° - 90° - 53° = \mathbf{37°}$ (angle sum of triangle)
Exercise 5.3
$P$, $Q$, $R$, $S$ are concyclic. $\angle PQR = 68°$ and $\angle PRS = 34°$. Find $\angle QPS$ and $\angle QRS$.
Show Solution
$PQRS$ cyclic quadrilateral: $\angle QPS + \angle QRS = 180°$.
$\angle PSR + \angle PQR = 180° \Rightarrow \angle PSR = 180° - 68° = 112°$. Also $\angle PRS = 34°$.
In triangle $PRS$: $\angle RPS = 180° - 112° - 34° = 34°$...
Simpler: opposite angles $\angle PQR + \angle PSR = 180°$, so $\angle PSR = 112°$.
Opposite angles $\angle QPS + \angle QRS = 180°$. Need more information to find each individually — the question gives $\angle PRS = 34°$, so $\angle QRS = \angle QRP + \angle PRS$. Without $\angle QRP$ we cannot find this further. The key facts are $\angle PSR = \mathbf{112°}$ and the sum $\angle QPS + \angle QRS = 180°$.
Exercise 5.4
$ABCD$ is a cyclic quadrilateral. $\angle DAB = 85°$ and $\angle ABC = 110°$. Find $\angle BCD$ and $\angle CDA$.
Show Solution
Opposite angles sum to $180°$:
$\angle BCD = 180° - 85° = \mathbf{95°}$
$\angle CDA = 180° - 110° = \mathbf{70°}$
Exercise 5.5
$OT$ is a radius and $PT$ is a tangent at $T$. $\angle OPT = 28°$. Find $\angle TOP$.
Show Solution
$\angle OTP = 90°$ (tangent ⊥ radius)
$\angle TOP = 180° - 90° - 28° = \mathbf{62°}$
Exercise 5.6
From external point $P$, two tangents touch the circle at $A$ and $B$. $\angle APB = 50°$. Find $\angle AOB$.
Show Solution
$\angle OAP = \angle OBP = 90°$ (tangent ⊥ radius)
Angle sum of quadrilateral: $\angle AOB = 360° - 90° - 90° - 50° = \mathbf{130°}$
Exercise 5.7
A circle has centre $O$ and radius $13\,\text{cm}$. A chord is $24\,\text{cm}$ long. Find the perpendicular distance from $O$ to the chord.
Show Solution
Half chord $= 12\,\text{cm}$. Pythagoras: $d^2 = 13^2 - 12^2 = 169 - 144 = 25$, $d = \mathbf{5\,\text{cm}}$
Exercise 5.8 Extended
$TA$ is a tangent at $T$. Chord $TB$ makes an angle of $42°$ with the tangent. $X$ is a point on the major arc. Find $\angle TXB$. Also find the angle $\angle TYB$ where $Y$ is in the minor arc.
Show Solution
By alternate segment theorem: $\angle TXB = 42°$ (angle in alternate/major segment).
For $Y$ on the minor arc: The angle at $Y$ subtends the same chord but from the other segment. Since arc angles on the same chord are supplementary: $\angle TYB = 180° - 42° = \mathbf{138°}$.
Exercise 5.9 Extended
$O$ is the centre. $\angle AOB = 110°$ (reflex). Point $C$ is on the major arc. Find $\angle ACB$.
Show Solution
The reflex angle $\angle AOB = 110°$ is the angle at the centre for the minor arc. For points on the major arc, the angle at the centre on the major arc side is $360° - 110° = 250°$ (reflex). But the standard theorem states: angle at circumference $= \tfrac{1}{2}$ × angle at centre (same arc).
The non-reflex angle at centre for the major arc = $360° - 110° = 250°$ — no, this is reflex.
Simpler: non-reflex $\angle AOB = 110°$. $C$ is on the major arc. Then $\angle ACB = \tfrac{1}{2} \times (360°-110°)/1$... Actually: angle at circumference from major arc $= \tfrac{1}{2}$ reflex angle at centre $= \tfrac{1}{2}\times250° = 125°$. But that's >90°, meaning $\angle ACB = \mathbf{125°}$... Wait — the standard result is: if the angle at centre is the reflex $250°$, the inscribed angle is $\tfrac{250°}{2} = 125°$. Since $C$ is on the major arc, it sees the minor arc, giving $\angle ACB = \tfrac{1}{2}\times110° = \mathbf{55°}$.
Correct answer: $\angle ACB = \tfrac{1}{2} \times 110° = \mathbf{55°}$ (C on major arc, angle at centre for minor arc is 110°).
Exercise 5.10 Extended
$O$ is the centre. $\angle AOC = 136°$. $B$ and $D$ lie on the circle on opposite arcs of chord $AC$. Find $\angle ABC$ and $\angle ADC$.
Show Solution
$B$ on major arc: $\angle ABC = \tfrac{1}{2}\times136° = \mathbf{68°}$ (angle at centre = twice angle at circumference)
$D$ on minor arc: $\angle ADC = \tfrac{1}{2}\times(360°-136°) = \tfrac{1}{2}\times224° = \mathbf{112°}$
Check: $68° + 112° = 180°$ ✓ (opposite angles in cyclic quadrilateral $ABCD$)
Exam Tips
Tip 1 — Always state the theorem name as a reason
Writing just "$\angle APB = 65°$" earns no method mark. Write "$\angle APB = 65°$ (angle at centre is twice angle at circumference)". The reason is worth as many marks as the answer.
Tip 2 — Mark equal lengths on the diagram
All radii are equal. Mark them with a tick. Equal tangent lengths from an external point should also be marked. This often reveals isosceles triangles that you can exploit.
Tip 3 — Isosceles triangles from two radii
Any triangle formed by two radii and a chord is isosceles (the two radii are equal). The base angles are equal. This is a frequently overlooked source of angle relationships.
Tip 4 — Reflex angles and major/minor arcs
Theorem 1 uses the angle at the centre that is on the same side as the arc being considered. If the angle at the centre is reflex (greater than $180°$), multiply the circumference angle by 2 and you get a reflex angle at the centre — or equivalently, the obtuse/reflex inscribed angle comes from the reflex central angle.