IGCSE Maths Circle Theorems: 3 Worked Examples

IGCSE Maths tests eight circle theorems, and they sit in the Extended content only. The examinable list is: angle at the centre, angle in a semicircle, angles in the same segment, cyclic quadrilateral, tangent perpendicular to radius, equal tangents from an external point, the alternate segment theorem, and the perpendicular from the centre bisecting a chord.

In Cambridge IGCSE Mathematics 0580 these live in the Geometry section of the Extended syllabus, so a Core candidate is never asked them. In the current 0580 assessment, Extended candidates sit Paper 2 without a calculator and Paper 4 with one, each two hours and 100 marks, and circle theorem questions appear on either. In Edexcel International GCSE Mathematics A (4MA1) they are Higher tier content, examined across Paper 1H and Paper 2H, each two hours and 100 marks, with a calculator allowed on both. If you are still deciding your tier, the Extended versus Core decision matters more than any single topic.

None of the eight is hard on its own.

What makes them one of the hardest IGCSE Maths topics is that examiners combine them, and that they expect the name of the theorem written out as a reason.

You find an angle at the centre from two radii by using the isosceles triangles first, then doubling. In the IGCSE question below the answers are $\angle ABC = 53^\circ$ and $\angle AOC = 106^\circ$. The radii step comes before the circle theorem, and it is the step most students leave out.

The question.

$A$, $B$ and $C$ lie on a circle with centre $O$. $\angle BAO = 32^\circ$ and $\angle BCO = 21^\circ$. Work out the obtuse angle $\angle AOC$, giving a reason for each step. (4 marks)

Step 1: mark what is equal.

$OA$, $OB$ and $OC$ are all radii of the same circle, so $OA = OB = OC$. Write that on the diagram before you write anything else.

Step 2: use triangle $OAB$.

It has $OA = OB$, so it is isosceles and the base angles are equal: $\angle OBA = \angle OAB = 32^\circ$.

Step 3: use triangle $OBC$.

Same argument with $OB = OC$: $\angle OBC = \angle OCB = 21^\circ$.

Step 4: add them.

$\angle ABC = 32^\circ + 21^\circ = 53^\circ$, because $OB$ splits $\angle ABC$ into those two parts.

Step 5: apply the theorem.

$\angle AOC = 2 \times 53^\circ = 106^\circ$, angle at centre is twice angle at circumference.

Where the marks sit.

A mark scheme for this typically gives an M1 for either base angle, an A1 for $\angle ABC = 53^\circ$, an M1 for doubling and an A1 for $106^\circ$. Because the doubling mark is a method mark, a student who gets $\angle ABC$ wrong but doubles it correctly still scores it on follow through.

You chain circle theorems by treating each angle you find as a new given. In this IGCSE question the chain runs alternate segment, then cyclic quadrilateral, then a subtraction, and the final answer is $\angle BDC = 29^\circ$. Multi-step questions like this are how examiners build a five mark item out of theorems that are worth two marks each.

The question.

$A$, $B$, $C$ and $D$ lie on a circle in that order. The straight line $SAT$ is the tangent to the circle at $A$, with $T$ on the opposite side of chord $AB$ from $C$ and $D$. $\angle TAB = 63^\circ$ and $\angle ABC = 88^\circ$. Work out $\angle BDC$. (5 marks)

Step 1: alternate segment.

The angle between the tangent $AT$ and the chord $AB$ equals the angle subtended by $AB$ in the alternate segment. $D$ lies in that alternate segment, so $$\angle ADB = \angle TAB = 63^\circ$$

Step 2: cyclic quadrilateral.

$ABCD$ is a cyclic quadrilateral, so $\angle ABC$ and $\angle ADC$ are opposite angles and sum to $180^\circ$: $$\angle ADC = 180^\circ - 88^\circ = 92^\circ$$

Step 3: split the angle at $D$.

$DB$ lies inside $\angle ADC$, so $\angle ADC = \angle ADB + \angle BDC$, which gives $$\angle BDC = 92^\circ - 63^\circ = 29^\circ$$

Step 4: check it.

$\angle BDC$ and $\angle BAC$ stand on the same arc $BC$, so they must be equal. If you can compute $\angle BAC$ another way and it does not come to $29^\circ$, you have gone wrong somewhere. That thirty second check is worth doing on the calculator paper, where you have the time.

Why students lose marks here.

They spot the alternate segment angle, write $63^\circ$, and then stop, because the question does not look like it has more in it. The giveaway is the mark allocation: five marks means at least three separate steps, never one theorem.

A "show that" circle question is answered by writing every intermediate angle with its reason and finishing at the printed value, never by working backwards from it. The IGCSE example below asks you to show that $\angle CTA = 22^\circ$, and there are two valid routes to it. Both score full marks; one is three lines shorter.

The question.

$AB$ is a diameter of a circle with centre $O$. $C$ is a point on the circle with $\angle CAB = 34^\circ$. The tangent to the circle at $C$ meets $AB$ extended at $T$. Show that $\angle CTA = 22^\circ$. (4 marks)

Route 1, the semicircle route.

• $\angle ACB = 90^\circ$, angle in a semicircle.
• $\angle ABC = 180^\circ - 90^\circ - 34^\circ = 56^\circ$, angle sum of a triangle.
• $OB = OC$ are radii, so triangle $OBC$ is isosceles and $\angle OCB = 56^\circ$.
• $\angle COB = 180^\circ - 2 \times 56^\circ = 68^\circ$.
• $\angle OCT = 90^\circ$, tangent is perpendicular to the radius.
• In triangle $OCT$:$\angle OTC = 180^\circ - 90^\circ - 68^\circ = 22^\circ$, and $\angle CTA$ is the same angle because $A$, $O$, $B$ and $T$ are collinear.

Route 2, the alternate segment route.

• $\angle BCT = \angle CAB = 34^\circ$, alternate segment theorem.
• $\angle ACB = 90^\circ$ and $\angle ABC = 56^\circ$ as above, so $\angle CBT = 180^\circ - 56^\circ = 124^\circ$, angles on a straight line.
• In triangle $BCT$:$\angle BTC = 180^\circ - 124^\circ - 34^\circ = 22^\circ$.

What the mark scheme rewards.

Cambridge and Edexcel both accept any correct method, so route 2 is not penalised for being shorter. What is penalised is asserting $\angle CTA = 22^\circ$ and then checking the other angles are consistent with it. In a "show that" the printed answer is not evidence, so the final line has to be the conclusion of your working, not its starting point. If you want the shorthand the examiners use for all this, see how IGCSE Maths mark schemes work.

Marks in IGCSE circle theorem questions are won on the reasons and on the intermediate angles, not on the final number. A question worth four or five marks usually gives one mark per correctly reasoned step, so a student who writes only the final answer, even a correct one, can score two out of five while a student with a wrong answer and full working scores four.

The command words tell you what is being marked.

"Work out" and "Calculate" want the value plus enough working to show the method. "Give a reason for your answer" makes the reason itself a mark, usually a B1. "Show that" fixes the destination and marks the journey. "Give the exact value" or "Give your answer in terms of $\pi$" appears when a circle theorem question ends in an arc or sector calculation.

The four ways students actually lose marks:

• Writing a reason that is nearly right. "Angles in a triangle" instead of "angles in the same segment" scores nothing, because the mark scheme is checking for the theorem, not for a true statement.
• Assuming a line through the centre is a diameter when the diagram only shows it passing near $O$. The diagrams are printed NOT TO SCALE, so unless a point is labelled as the centre or a line is stated to be a diameter, you cannot use it.
• Assuming a tangent. A line touching the circle in the picture is only a tangent if the question says so.
• Stopping at the first angle found, when the mark allocation clearly signals more steps.

On the non-calculator paper, circle theorem work is pure arithmetic with $180$ and $360$, so there is no calculator disadvantage. It is the sector and arc follow-ups ($\text{arc} = \frac{\theta}{360} \times 2\pi r$) that are usually reserved for the calculator paper.

Practise IGCSE circle theorems by working through past paper questions with the reasons written out in full, then repeating the same questions two weeks later from a blank diagram. Recognition is the skill being tested, not calculation, and recognition only builds from seeing many different diagrams of the same eight theorems.

A four week routine that works.

Week one: eight index cards, one theorem each, statement on the front and the exam reason on the back. Week two: ten past paper questions, full reasons, self marked against the official mark scheme. Week three: the same ten questions again with the working covered. Week four: mixed questions where you do not know in advance which theorem applies, because that is the real exam condition.

Use the mark scheme as a vocabulary list.

Read three or four mark schemes for circle theorem questions and you will notice the reasons repeat almost word for word. Copy those phrasings onto your cards rather than inventing your own. The method for extracting this systematically is in how to use IGCSE Maths past papers.

One diagnostic.

Draw a circle with a centre, two chords and a tangent, label three angles at random, and ask yourself which theorems could apply. If you can name three candidate theorems in under thirty seconds, you are exam ready on this topic. If you can name none, the problem is recognition, not maths, and more calculation practice will not fix it.