中文（繁體）
中文（簡體）
English
S5 Maths: Circle Geometry prep guide | Central angles, tangents and cyclic quadrilaterals
Students add random construction lines and write the reason as circle property. This lesson demands a named theorem every time and builds all angles from three right triangles.
Objectives — Students can: (1) use the angle at the centre and angles in the same segment to find angles; (2) use the tangent-radius right angle and equal tangents to find lengths and angles; (3) use opposite angles of a cyclic quadrilateral and the exterior angle to finish a proof.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Pythagoras | Half chord when the radius is 13 and the distance is 5 | 12 |
| Isosceles triangles | Base angles in a triangle with two radii | Equal |
| Angle sum of a triangle | The three angles added | 180 degrees |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-10 min | Centre and circumference | An inscribed angle of 35 degrees | Gives 70 degrees |
| 10-22 min | Same segment and semicircle | Equal angles in one segment, 90 degrees in a semicircle | 90 degrees |
| 22-34 min | Chord and distance | Radius 13, distance 5 | Chord 24 |
| 34-46 min | Tangents | External point 13 from the centre, radius 5 | Tangent 12 |
| 46-56 min | Cyclic quadrilateral | Angles of 110 and 85 | Then 70 and 95 |
| 56-60 min | Exit check | Three questions collected and marked | Independent work |
03 Teaching script (read it out)
Scene 1: The angle at the centre is twice the angle at the circumference
Two angles standing on the same chord satisfy one being double the other. An inscribed angle of 35 degrees gives a central angle of 70 degrees.
Three corollaries belong together: angles in the same segment are equal, the angle in a semicircle is a right angle, and opposite angles of a cyclic quadrilateral are supplementary.
Every step needs a named reason such as angle at centre twice angle at circumference, or angles in the same segment. Writing circle property alone scores nothing.
When the central angle of 70 degrees is given, the answer must run the other way, back to 35 degrees.
Three corollaries belong together: angles in the same segment are equal, the angle in a semicircle is a right angle, and opposite angles of a cyclic quadrilateral are supplementary.
Every step needs a named reason such as angle at centre twice angle at circumference, or angles in the same segment. Writing circle property alone scores nothing.
When the central angle of 70 degrees is given, the answer must run the other way, back to 35 degrees.
Task: draw one circle, mark two angles in the same segment and state why they are equal.
Scene 2: Chords and tangents: three right triangles
A perpendicular from the centre to a chord bisects it. With radius 13 and distance 5, the half chord is the root of 169 minus 25, which is 12, so the chord is 24.
A tangent is perpendicular to the radius at the point of contact, so the external point, the contact point and the centre make a right triangle: with 13 and 5 the tangent is 12. Tangents from one external point are equal in length and the line to the centre bisects the angle between them.
The alternate segment theorem says the angle between a tangent and a chord equals the angle in the alternate segment.
A tangent is perpendicular to the radius at the point of contact, so the external point, the contact point and the centre make a right triangle: with 13 and 5 the tangent is 12. Tangents from one external point are equal in length and the line to the centre bisects the angle between them.
The alternate segment theorem says the angle between a tangent and a chord equals the angle in the alternate segment.
Warn: the moment a tangent appears, draw the radius. Without it no angle can be justified.
Scene 3: Cyclic quadrilaterals and the circumradius
Opposite angles of a cyclic quadrilateral are supplementary, so 110 degrees forces 70, and 85 forces 95. An exterior angle equals the interior opposite angle, which is the same fact written differently.
The circumradius comes from the area formula: sides 13, 14 and 15 give a semiperimeter of 21, an area of 84, so the radius is the product of the sides over four times the area, which is 8.125.
Three moves for any proof: mark what is given, choose the theorem, then write the conclusion with its reason.
The circumradius comes from the area formula: sides 13, 14 and 15 give a semiperimeter of 21, an area of 84, so the radius is the product of the sides over four times the area, which is 8.125.
Three moves for any proof: mark what is given, choose the theorem, then write the conclusion with its reason.
Task: explain why the exterior angle result and the supplementary result are one theorem.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Find an angle | Centre and circumference | Double the inscribed angle | 70 degrees |
| Find a length | Chord and distance | Half chord is the root of r squared minus d squared | 24 |
| Find a length | Tangent from an external point | Root of OP squared minus r squared | 12 |
| Proof | Cyclic quadrilateral | Supplementary opposite angles | 70 and 95 |
Sanity check (teach this): An inscribed angle must be smaller than its central angle, and 35 against 70 is fine. The half chord 12 must be shorter than the radius 13, otherwise the square root goes negative. The four angles of the cyclic quadrilateral must total 360, and 110 plus 85 plus 70 plus 95 does. The circumradius 8.125 must exceed half the longest side, which is 7.5.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Chord when the radius is 13 and the distance is 5 | 24 | Twice the root of 169 minus 25 |
| Central angle for an inscribed angle of 35 degrees | 70 degrees | Double it |
| Tangent from a point 13 from the centre with radius 5 | 12 | Root of 169 minus 25 |
| Angle opposite 110 degrees in a cyclic quadrilateral | 70 degrees | 180 minus 110 |
| Circumradius of a 13-14-15 triangle | 8.125 | Product over four times 84 |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Reason written as circle property | A proof with conclusions but no theorem names | “Every conclusion needs a named theorem.” |
| Assuming the tangent right angle | Using Pythagoras without stating the tangent radius fact | “State that the tangent is perpendicular to the radius first.” |
| Doubling the wrong angle | Multiplying 70 by two to get 140 | “The centre angle is the larger one.” |
| Writing opposite angles as equal | Answering 110 for the opposite angle | “Opposite angles add to 180, they are not equal.” |
07 Homework (paper, section and question numbers)
- Secondary mathematics chapter index — All chapters
- DSE mathematics practice paper — Circle questions 17 to 21
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Chord when the radius is 13 and the distance is 5 | 24 | All correct: move to probability |
| Relation between a tangent and the radius | They meet at a right angle | Wrong: redo the tangent steps |
| Sum of opposite angles in a cyclic quadrilateral | 180 degrees | Two wrong: restart act one |
09 Tutor FAQ
Why is the angle in a semicircle a right angle
It subtends a straight angle at the centre, so the inscribed angle is half of 180.
Why does the perpendicular from the centre bisect a chord
Two radii form congruent right triangles with the two halves of the chord.
When do I use the alternate segment theorem
When a tangent and a chord form an angle that has to move to the other segment.
Find a tutor — freeTutor Registration →
A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
