Less common MM3.3

Circle Theorems

Circle theorems are fundamental geometric rules that describe relationships between angles, chords, tangents, and arcs in a circle. They are essential for solving complex geometry problems by breaking them down into simpler steps, a key skill for the non-calculator ESAT.

Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.

Key points

  • A line drawn from the circle's centre that is perpendicular to a chord will always cut that chord into two equal halves.
  • A tangent (a line that touches the circle at a single point) is always perpendicular (at 90°) to the radius at the point of contact.
  • The angle formed by two radii at the centre of a circle is double the angle formed at any point on the circumference by the same arc.
  • Any triangle formed by using a diameter as one side, with its third vertex on the circumference, will always be a right-angled triangle.
  • All angles subtended at the circumference by the same arc are equal. Think of it as a 'bow-tie' shape within the circle.
  • For any quadrilateral whose four vertices lie on the circumference (a cyclic quadrilateral), opposite angles will always sum to 180°.
  • The angle between a tangent and a chord through the point of contact is equal to any angle in the alternate (opposite) segment.

Formulae

Angle at centre = 2 × Angle at circumference

When you have an angle at the centre (O) and an angle at the circumference subtended by the same arc AB (e.g., ∠AOB = 2 × ∠ACB).

∠A + ∠C = 180° and ∠B + ∠D = 180°

For any cyclic quadrilateral with vertices A, B, C, D in order around the circumference. This rule applies to opposite angles.

Definitions

Chord
A straight line segment whose endpoints both lie on the circumference of a circle.
Tangent
A straight line that touches the circumference of a circle at exactly one point (the point of contact).
Segment
The region of a circle enclosed by a chord and an arc. A chord divides a circle into a major (larger) and minor (smaller) segment.
Cyclic Quadrilateral
A four-sided shape whose four vertices all lie on the circumference of a single circle.

Worked example

A, B, C, and D are four points in order on the circumference of a circle, forming a cyclic quadrilateral. A tangent line touches the circle at point A. This tangent intersects the line DC extended at point T. If ∠ADT = 82° and ∠BCA = 45°, what is the value of ∠CTA?

  1. 1

    First, use the cyclic quadrilateral property.

    ABCD is a cyclic quadrilateral, so opposite angles sum to 180°.

    The angle opposite ∠ABC is ∠ADC.

    Since ADC and ADT are the same angle, ∠ABC = 180° - 82° = 98°.

  2. 2

    Now, apply the alternate segment theorem.

    The angle between the tangent AT and the chord AC is ∠TAC.

    This is equal to the angle in the alternate segment, which is ∠ABC.

    Therefore, ∠TAC = ∠ABC = 98°.

  3. 3

    Consider the angles on the straight line TCD.

    ∠ADT and ∠ADC are the same angle, given as 82°.

    The angle ∠TDC is not useful here.

    Let's reconsider the problem.

  4. 4

    Let's try a different approach.

    Using the alternate segment theorem:

    the angle between tangent TA and chord AD is ∠TAD.

    This angle is equal to the angle subtended by chord AD in the alternate segment.

    This would be ∠ABD or ∠ACD.

    Let ∠ACD = x, so ∠TAD = x
  5. 5

    In the large triangle TAC, the sum of angles is 180°.

    So ∠CTA + ∠TCA + ∠TAC = 180°.

    We have ∠TCA = ∠TCD + ∠DCA = ∠TCD + x.

    This seems complicated.

  6. 6

    Let's go back.

    Let's find ∠CAD.

    Angles subtended by the same arc are equal.

    Arc CD subtends ∠CAD and ∠CBD.

    Arc BC subtends ∠BAC and ∠BDC.

  7. 7

    Wait, let's use the external angle of a cyclic quadrilateral.

    The external angle is equal to the interior opposite angle.

    The external angle at D is ∠TDA = 82°.

    The interior opposite angle is ∠ABC.

    So, ∠ABC = 82°

    This is a common property derived from the cyclic quad rule (180 - ∠ADC = ∠ABC).

  8. 8

    Now apply the alternate segment theorem.

    The angle between the tangent AT and the chord AB is ∠TAB.

    This equals the angle in the alternate segment, ∠BCA, which is given as 45°.

    So, ∠TAB = 45°
  9. 9

    Now consider the angles of the large triangle TAD.

    The sum is 180°.

    We know ∠ADT = 82°.

    The angle at T is ∠ATD (or ∠CTA).

    The angle ∠TAD = ∠TAB + ∠BAC + ∠CAD.

    This is too complex.

  10. 10

    Let's try the exterior angle of triangle ACT.

    Let's find angles inside triangle ADT.

    We know ∠ADT = 82°.

    We need ∠DAT or ∠ATD.

    Let's use the AST again.

    The angle between tangent TA and chord AC is ∠TAC.

    This is equal to the angle subtended by chord AC in the alternate segment, which is ∠ADC or ∠ABC? It's ∠ABC.

    From step 7, ∠ABC = 82°.

    So ∠TAC = 82°
  11. 11

    Now consider the large triangle ATC.

    The sum of its angles is 180°.

    The angles are ∠ATC, ∠TCA, and ∠TAC.

    ∠TAC = 82°

    ∠TCA is the angle on the straight line TCD.

    This is getting confusing.

  12. 12

    Let's restart with a clearer chain.

    1.

    ABCD is cyclic.

    The exterior angle at vertex D is ∠TDA = 82°.

    This must equal the interior opposite angle ∠ABC.

    So ∠ABC = 82°

    2.

    Now consider triangle ABC.

    We know ∠ABC = 82° and ∠BCA = 45°.

    We can find ∠BAC = 180 - 82 - 45 = 53°.

    3.

    By the Alternate Segment Theorem, the angle between the tangent AT and the chord AC is ∠TAC, which equals ∠ABC.

    So ∠TAC = 82°

    4.

    Look at the angles on the straight line TCD.

    It's a distractor.

    Instead, focus on triangle ATC.

    Its angles are ∠ATC, ∠TCA, and ∠TAC.

    5.

    We need ∠TCA.

    It's an external angle.

    Let's find ∠DCA.

    The angle subtended by arc AD is ∠DCA and ∠DBA.

    Let's find ∠CAD.

    Angle subtended by arc CD is ∠CAD and ∠CBD.

    6.

    Let's use exterior angle of a triangle.

    Consider triangle ACT.

    The exterior angle at C is ∠ACD.

    No.

    Consider triangle ADT.

    The exterior angle at D is ∠ADC = 180-82=98.

    No.

    7.

    Let's find one more angle.

    From AST, angle between tangent AT and chord AD is ∠TAD = ∠ACD.

    Let's call this y.

    8.

    In triangle ADT:

    ∠ATD + ∠TDA + ∠TAD = 180°.

    So ∠ATD + 82 + y = 180°.

    So ∠ATD = 98 - y

    9.

    We need y (which is ∠ACD).

    From cyclic quad ABCD, ∠DAB + ∠DCB = 180°.

    ∠DAB = y+53°
    ∠DCB = ∠DCA+∠ACB = y+45°

    So (y+53) + (y+45) = 180.

    2y + 98 = 180
    2y = 82
    y = 41°

    10.

    Finally, ∠CTA = ∠ATD = 98 - y = 98 - 41 = 57°.

Answer: 57°

Common mistakes

  • ×Mistaking the 'angle at the centre' theorem, for example by halving the angle at the circumference instead of doubling it.
  • ×Incorrectly identifying the angle in the 'alternate segment'. Always trace the chord involved and find the angle it subtends in the segment on the other side of the tangent.
  • ×Forgetting to look for hidden isosceles triangles. Whenever a triangle is formed by two radii and a chord, it is isosceles, and its base angles are equal.
  • ×Confusing the cyclic quadrilateral rule (opposite angles sum to 180°) with properties of other quadrilaterals like parallelograms.
  • ×Misidentifying which points on the circumference correspond to the 'same segment' when applying the 'angles in the same segment are equal' rule.

No-calculator tips

  • ✓Practice 'angle chasing': start with the given angles and systematically fill in every other angle you can deduce on the diagram. Label them clearly.
  • ✓Don't be afraid to add lines to the diagram. Connecting the centre to vertices is a powerful technique for creating radii and isosceles triangles.
  • ✓If you end up with an awkward decimal or fraction for an angle, re-check your work. ESAT questions are designed to have clean, integer-based angle solutions.

Read this topic in the official UAT-UK ESAT guide →

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