Circle Theorems

SPEC G10 GEOMETRY & MEASURES

The eight named theorems — with the exact mark-scheme phrasings examiners require

What you will learn

By the end of this lesson you will:

State each of the eight circle theorems using the exact mark-scheme phrasing

Identify which theorem applies from a given diagram

Use one or more theorems to find unknown angles in multi-step problems

Provide a reason (the named theorem) for every angle you calculate

Apply the alternate segment theorem to tangent-chord configurations

Why this matters

Higher-tier circle theorem questions are worth 3–5 marks each. Typical mark allocation: M1 for each angle calculated, B1 for each theorem named. Without the theorem name, you can't get full marks — even with the correct angle.

Recap from Lesson 4.09

Three properties from the previous lesson reappear in this lesson as Theorems 5, 6 and 7. Make sure these are solid before moving on.

TheoremProperty

5Tangent is perpendicular to the radius at the point of contact

6Two tangents drawn from an external point are equal in length

7The perpendicular from the centre to a chord bisects the chord

The other 5 theorems

Theorems 1–4 deal with angles at the centre, in semicircles, in segments, and in cyclic quadrilaterals. Theorem 8 (the alternate segment theorem) connects a tangent and a chord to an angle in the alternate segment. We'll take each in turn.

Theorem 1: Angle at Centre

Theorem 1

The angle subtended at the centre by an arc is twice the angle subtended at any point on the circumference by the same arc. Both angles "look at" the same red arc — the centre angle is double the circumference angle.

Angle at centre is twice angle at circumference

Theorem 2: Angle in a Semicircle

Theorem 2

The angle subtended by a diameter at any point on the circumference is 90°. Whenever the question gives you a diameter (or you can spot one), the angle at any other point on the circle made with the diameter's endpoints is a right angle.

Angle in a semicircle is 90°

A special case of Theorem 1

A diameter subtends a 180° angle at the centre. By Theorem 1, the angle at the circumference is 180 ÷ 2 = 90°. Theorem 2 is just Theorem 1 applied to the special case where the chord is a diameter.

Theorem 3: Angles in the Same Segment

Theorem 3

Angles subtended by the same chord on the same side of it are equal. Both P and Q "look at" the same chord AB from the same segment, so APB = AQB.

Angles in the same segment are equal

⚠️ Same side of the chord matters. If P is in the major segment and Q is in the minor segment, the angles are not equal — in fact they sum to 180° (related to Theorem 4).

Theorem 4: Cyclic Quadrilateral

Theorem 4

A cyclic quadrilateral is a four-sided shape whose vertices all lie on a circle. The opposite angles of a cyclic quadrilateral sum to 180°:

A + C = 180° and B + D = 180°

Opposite angles of cyclic quadrilateral sum to 180°

Spotting it

Four points on a circle connected in order = cyclic quadrilateral. The four angles at the vertices sum to 360° (true for any quadrilateral); the extra property is that opposite pairs sum to 180°.

Theorems 5, 6, 7 (from Lesson 4.09)

Theorem 5

Tangent radius at point of contact. Whenever a tangent appears, draw the radius to the point of contact — the 90° angle is your route to a solution.

Tangent perpendicular to radius

Theorem 6

Two tangents from an external point are equal. If P is outside the circle and PT₁, PT₂ are tangents to the circle, then PT₁ = PT₂.

Tangents from external point are equal

Theorem 7

Perpendicular from centre bisects chord. If you drop a perpendicular from the centre to a chord, it hits the chord at its midpoint — creating a right-angled triangle with the radius as hypotenuse (Pythagoras territory).

Perpendicular from centre bisects chord

Theorem 8: Alternate Segment Theorem

Theorem 8 — Alternate Segment

The angle between a tangent and a chord drawn from the point of contact equals the angle in the alternate segment. That is, the angle Q in the segment on the other side of the chord from the tangent angle.

Alternate segment theorem

⚠️ "Alternate" means opposite side of the chord. Identify the angle between tangent and chord first, then look in the segment on the OPPOSITE side — the angle subtended by the chord at any point on that opposite arc equals the tangent-chord angle.

Proving Theorem 1

Spec G10 says apply and prove the standard circle theorems. The angle-at-centre proof is the canonical example — it appears as a 4–5 mark "Show that..." question and underpins Theorems 2 and 4 too.

Proof

Construction: Draw the line PO and extend it to meet the far side of the circle.

Step 1: OP and OA are both radii, so triangle OAP is isosceles with OA = OP. Therefore the base angles are equal: OAP = OPA. Call this angle a.

Step 2: Similarly, OB = OP (both radii), so triangle OBP is isosceles. OBP = OPB. Call this b.

Step 3: APB = a + b — this is the angle at the circumference.

Step 4: The angle at the centre AOB can be split by the line PO. The exterior angle at O on each side equals the sum of the two opposite interior angles of its triangle. So the part of AOB beyond line PO on the A-side = 2a, and on the B-side = 2b.

Conclusion: AOB = 2a + 2b = 2(a + b) = 2 × APB. ∎

Why this proof matters

Theorem 2 follows directly: a diameter gives AOB = 180°, so APB = 90°. Theorem 4 follows by applying Theorem 1 twice to a cyclic quadrilateral. Knowing this proof unlocks "show that" questions on all three theorems.

The Eight Theorems — Reference

No.TheoremMark-scheme phrase

1Angle at centre = 2× angle at circumference (same arc)"Angle at centre is twice angle at circumference"

2Angle in a semicircle is 90°"Angle in a semicircle is 90°"

3Angles in same segment are equal"Angles in the same segment are equal"

4Opposite angles of cyclic quad sum to 180°"Opposite angles of cyclic quadrilateral sum to 180°"

5Tangent ⊥ radius at point of contact"Tangent perpendicular to radius"

6Two tangents from external point equal"Tangents from external point are equal"

7Perpendicular from centre bisects chord"Perpendicular from centre bisects chord"

8Alternate segment theorem"Alternate segment theorem"

Memorise the phrases verbatim

The mark scheme awards B1 each for the named theorem — but only if the phrase matches what's expected. Generic phrases like "because of the angles" or "circle theorem" do not score. Use the exact wording above.

Worked Example 1

01Cyclic Quadrilateral 4 marks

A, B, C and D lie on the circumference of a circle (in that order). BAD = 75° and ABC = 95°. Find BCD and ADC, giving a reason for each answer.

M

ABCD is a cyclic quadrilateral. Use Theorem 4: opposite angles sum to 180°. The pairs are (A, C) and (B, D).

C

BCD = 180° − BAD = 180° − 75° = 105° (opposite angles of cyclic quadrilateral sum to 180°).

ADC = 180° − ABC = 180° − 95° = 85° (opposite angles of cyclic quadrilateral sum to 180°).

E

BCD = 105° and ADC = 85°. Check: 75 + 95 + 105 + 85 = 360° ✓ (sum of interior angles of any quadrilateral). Both reasons must be written for full marks — the calculation alone is M1 each, with B1 each for the theorem name.

Worked Example 2

02Multi-step: Theorems 1 + 2 4 marks

In a circle with centre O, AB is a diameter. C is a point on the circumference. The angle AOC (between radii OA and OC, on the same side as B) is 130°. Find ACB and ABC, giving reasons.

M

Two strategies. (1) AB is a diameter → Theorem 2 gives ACB straight away. (2) For ABC: triangle OBC has OB = OC (both radii), so it's isosceles; and BOC = 180° − AOC since AOB is a straight line.

C

ACB = 90° (angle in a semicircle is 90°).

BOC = 180° − 130° = 50° (angles on a straight line).

Triangle OBC: OB = OC (radii), so OBC = OCB. Sum to 180°: OBC = (180° − 50°) ÷ 2 = 65°.

ABC = 65°.

E

ACB = 90° (angle in semicircle is 90°) and ABC = 65° (base angles of isosceles triangle are equal). Check: 90 + 65 + BAC = 180 in triangle ABC, so BAC = 25°. Cross-check via Theorem 1: AOC = 130° subtends arc AC (not the one through B); the angle at the circumference on the OTHER side, ABC, should be half of 130 — and 65 = 130 ÷ 2 ✓.

Always look for triangles with two radii

Whenever you see two radii sharing a vertex at O, the triangle they form is isosceles. Pair this with the angle-in-a-semicircle property and a vast number of diagrams unlock.

Worked Example 3

03Alternate Segment Theorem 3 marks

PT is a tangent to a circle at T. A chord TB is drawn from T. The angle between PT and TB (measured at T) is 58°. Q is a point on the major arc (the alternate segment from the tangent angle). Find TQB.

M

The configuration — tangent meeting a chord at the point of contact, plus an angle in the alternate segment — is the textbook setup for Theorem 8.

C

By the alternate segment theorem, TQB = (angle between tangent PT and chord TB) = 58°.

E

TQB = 58° (alternate segment theorem). Without naming the theorem, you'd lose half the marks even with the right answer. The phrase to write: "angle between tangent and chord = angle in alternate segment" or just "alternate segment theorem".

Exam Strategy: How to Attack Any Diagram

Circle theorem questions can look intimidating, but they all yield to the same systematic approach:

The 5-step routine

Identify the configuration. Mark the centre, any radii, any tangents, any chords. Spot diameters — they often unlock Theorem 2.

Look for "two radii" triangles. Two radii from O make an isosceles triangle.

Look for the chord/arc subtended. Centre and circumference both "looking at" the same arc → Theorem 1. Two circumference points on the same side → Theorem 3.

Look for cyclic quadrilaterals. Four labelled points connected in order → Theorem 4.

Look for tangent + chord configurations. → Theorem 8 (alternate segment).

If stuck

Add the radius to the point of contact on any tangent. Add the perpendicular from centre to chord on any chord that isn't a diameter. These auxiliary lines convert the diagram into right triangles or isosceles triangles you already know how to solve.

Common Errors

Wrong answerCorrect versionTheorem

Angle at centre = angle at circumferenceCentre = twice circumferenceT1

Angle in semicircle = 180°= 90°T2

Angles in different segments are equalSame side of the chord onlyT3

Opposite angles of cyclic quad = 90°= 180°T4

Adjacent angles of cyclic quad sum to 180°Opposite angles do (adjacent need not)T4

Tangent intersects radius at < 90°Always exactly 90°T5

Tangent-chord angle = angle in same segment= angle in alternate segmentT8

Correct angle, no theorem namedAlways name the theoremAll

⚠️ EXAMINER ALERT — reasons missing. Students compute correct angles but do not name the theorem. The theorem name is the mark, not a decoration. Every angle you state should be followed by the named theorem in brackets or after a "because"/"since".

Writing Your Reasoning

The single biggest leak of marks on circle theorem questions is failing to write reasons. Here's how to phrase them.

StepBadGood

State the angle"x = 50°"" ABC = 50°"

Give the reason"because of circle theorem""(angles in the same segment are equal)"

Multi-step"x = 50, y = 100, so z = 30"" 1 = 50° (T3); 2 = 100° (T1); so 3 = 30° (sum of angles in )"

The structure

Every line of working that produces an angle should follow the pattern: "Angle X = ... ° (reason)". Reasons can be a named theorem or a basic fact like "angles on a straight line", "angles in a triangle sum to 180°", "isosceles triangle".

Worked Example 4

04Multi-Theorem Reasoning 5 marks

In a circle, P, Q and R are three points on the circumference. The chord PR passes through the centre O (so PR is a diameter). Q is on the circle, with QPR = 35°. The tangent to the circle at R meets the line PQ produced at point T. Find PRT and PTR, giving full reasons.

M

PR is a diameter → Theorem 2 gives PQR = 90°. Then triangle PQR fully solved gives PRQ. The tangent at R is perpendicular to OR (Theorem 5), and OR lies along PR — so the tangent at R is perpendicular to PR. That gives PRT = 90°. Triangle PRT is then a right triangle and PTR follows from angle sum.

C

PQR = 90° (angle in a semicircle is 90°).

In PQR: PRQ = 180° − 35° − 90° = 55° (angles in a triangle).

PRT = 90° (tangent perpendicular to radius; PR is along the radius OR).

In PRT: PTR = 180° − 35° − 90° = 55° (angles in a triangle, using TPR = QPR = 35°).

E

PRT = 90° and PTR = 55°. Two named theorems used (T2, T5) plus standard angle-sum facts. Each named theorem earns B1, plus M1 marks for the chained calculations and A1 for the final answers.

Examiner Alerts

⚠️ EXAMINER ALERT — reasons missing.Students compute the angles correctly but fail to give a reason or theorem name. The theorem name is the mark. Every angle in your working should be paired with its theorem.

⚠️ EXAMINER ALERT — wrong theorem applied.Students see a tangent and immediately write "alternate segment theorem" without checking that the angle is actually between a tangent and a chord (drawn from the point of contact). Identify the configuration first: which two lines meet, where, and at what angle?

⚠️ EXAMINER ALERT — cyclic quadrilateral confusion.Students apply "opposite angles sum to 180°" to adjacent angles or apply it to a quadrilateral whose vertices aren't actually on the circle. Check first that all four vertices lie on the circumference; only then is it cyclic.

Summary

The eight theorems — in one place

1

Angle at centre is twice angle at circumference (same arc)

2

Angle in a semicircle is 90°

3

Angles in the same segment are equal (same side of chord)

4

Opposite angles of cyclic quadrilateral sum to 180°

5

Tangent perpendicular to radius (at point of contact)

6

Tangents from external point are equal

7

Perpendicular from centre bisects chord

8

Alternate segment theorem

The exam routine

Identify the configuration

Find unknowns step by step

Name the theorem at every step

Top exam tip

The theorem name is the mark. A correct angle without a reason is M1; an angle with the named reason is M1 + B1.

⚠️ Final reminder: memorise the exact mark-scheme phrasings. "Because of the circle theorem" or "by the rules" do not score — only the named theorem does.