Circle Geometry Riders — Grade 11 Maths (CAPS)

The biggest block of marks in Paper 2. How the circle theorems are proved rather than just quoted, every converse and when you need it, the reason wording a marker expects, how to read a circle diagram, and four full riders worked from the given information to the conclusion.

Also known as: circle geometry riders · circle theorem proofs · converse circle theorems · grade 11 euclidean geometry · cyclic quadrilateral proof

In Grade 10 a rider chained together facts about parallel lines, triangles and quadrilaterals. Grade 11 adds the circle, and the questions get considerably harder: a single rider will move from a tangent to a cyclic quadrilateral to a pair of similar triangles.

Two things change from Grade 10. You must be able to prove the theorems, not only apply them — a proof is examinable and worth up to six marks on its own. And you must know each theorem's converse, because half of the marks in a rider come from proving that a shape is cyclic, or that a line is a tangent.

1The starting point: what you may assume

You accept everything proved in earlier grades, plus one new axiom:

A tangent to a circle is perpendicular to the radius drawn to the point of contact.

Everything else on this page is built from that and from the earlier grades' work.

2Proving a theorem — the chord bisection

O is the centre and OM ⊥ AB. The theorem claims AM = MB — and the proof is a congruence argument, not an observation.

Worked ExampleProve: the line drawn from the centre of a circle perpendicular to a chord bisects the chord.

1
Construction: join OA and OB.
Almost every circle proof begins by joining radii. Say so — the construction earns a mark.
2
In △OMA and △OMB: OA = OB
radii
3
OM = OM
common
4
OM̂A = OM̂B = 90°
given, OM ⊥ AB
5
∴ △OMA ≡ △OMB
RHS
6
∴ AM = MB
≡△s

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Notice the shape of the proof: construction → congruence → conclusion. The perpendicular-from-the-centre theorem, its converse and the equal-tangents theorem are all proved this way; the angle theorems use a construction and isosceles triangles instead. Learn the shapes and the individual proofs become much shorter to remember.

3Proving the angle-at-the-centre theorem

AÔB at the centre is twice AP̂B at the circumference, standing on the same arc AB.

Worked ExampleProve: the angle subtended by an arc at the centre is double the angle subtended by the same arc at the circle.

1
Construction: join PO and produce it to a point Q on the circle.
The proof splits the angle into two isosceles triangles.
2
Let AP̂O = x₁ and BP̂O = x₂.
Naming the parts is what makes the algebra work.
3
OA = OP
radii — so △OAP is isosceles
4
∴ OÂP = x₁
∠s opp equal sides
5
AÔQ = x₁ + x₁ = 2x₁
ext ∠ of a △ equals the sum of the two opposite interior ∠s
6
Similarly BÔQ = 2x₂.
The same argument in △OBP
7
∴ AÔB = 2x₁ + 2x₂ = 2(x₁ + x₂) = 2 × AP̂B

Two of the other theorems fall straight out of this one. Angles in the same segment are equal because each is half the same central angle. And the angle in a semicircle is 90° because the central angle is a straight angle of 180°.

4Every theorem and its converse

A converse swaps the given and the conclusion. You use the theorem to find angles; you use the converse to prove a shape is cyclic or prove a line is a tangent.

Line from centre ⊥ chord bisects it. Converse: the line from the centre to the midpoint of a chord is perpendicular to it — and the perpendicular bisector of a chord passes through the centre.
∠ at centre = 2 × ∠ at circumference. Used constantly; no converse examined.
∠s in the same segment are equal. Converse: if a line segment subtends equal angles at two points on the same side of it, those four points are concyclic.
Opp ∠s of a cyclic quad are supplementary. Converse: if the opposite angles of a quadrilateral add to 180°, the quadrilateral is cyclic.
Ext ∠ of a cyclic quad = the opposite interior ∠. Converse: if the exterior angle equals the opposite interior angle, the quadrilateral is cyclic.
Two tangents from the same point are equal.
Tan–chord ∠ = ∠ in the alternate segment. Converse: if the angle between a line and a chord equals the angle in the alternate segment, that line is a tangent.

⚠️

Using a theorem when the question needs its converse. If you are asked to prove ABCD is cyclic, you may not begin “opp ∠s of a cyclic quad” — that assumes what you are proving. You must show the opposite angles add to 180° and then conclude the quadrilateral is cyclic, quoting the converse.

5How to read a circle diagram

Find the centre. If O is marked, every line from it to the circle is a radius, and all radii are equal — that is a free equality in any proof.
Find the tangents. A tangent gives you a 90° at the point of contact, and a tan–chord angle with every chord drawn from it.
Find the cyclic quadrilaterals. Any four points on the circle joined in order form one. Learners routinely miss the second cyclic quad in a diagram.
Mark every angle you find onto the diagram as you go. A rider is a chain, and the angle you found in 8.1 is almost always the way into 8.2.
Look for the isosceles triangles. Two radii and a chord always make one.

6Rider 1 — a straight chain

Worked ExampleO is the centre. AP̂B = 35° with A, B and P on the circle. Determine AÔB and the size of OÂB.

1
AÔB = 2 × 35° = 70°
∠ at centre = 2 × ∠ at circumference
2
OA = OB
radii
3
∴ OÂB = OB̂A
∠s opp equal sides
4
OÂB = (180° − 70°) ÷ 2 = 55°
∠s of a △

7Rider 2 — proving a quadrilateral is cyclic

Worked ExampleIn quadrilateral PQRS, P̂ = 105° and R̂ = 75°. Prove that PQRS is cyclic.

1
P̂ + R̂ = 105° + 75° = 180°
The two angles are opposite each other in the quadrilateral.
2
∴ PQRS is a cyclic quadrilateral
converse: opp ∠s supplementary

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Two marks, two lines. Short riders like this are free marks if you know the converse wording — and lost entirely if you write “because they add to 180°” without naming the converse.

8Rider 3 — using a tangent

Worked ExampleO is the centre, TA is a tangent at A, and AB is a chord. The angle between TA and AB is 64°. Determine the angle in the alternate segment, and explain why OÂT = 90°.

1
The angle in the alternate segment = 64°
tan–chord ∠
2
OÂT = 90°
tangent ⊥ radius — this is the axiom, not something to prove
3
So if OÂB is asked for: OÂB + BÂT = OÂT = 90°, so OÂB = 90° − 64° = 26°.
OA ⊥ AT, and OÂB and BÂT together make up that right angle

9Rider 4 — a two-step chain

Worked ExampleO is the centre, and AB is a chord with M its midpoint. OM is drawn. If AB = 24 cm and the radius is 13 cm, determine OM.

1
OM ⊥ AB
converse: line from centre to midpt of chord ⊥ chord
2
AM = 24 ÷ 2 = 12 cm
M is the midpoint
3
In △OMA: OA² = OM² + AM²
Pythagoras — OM̂A = 90°
4
13² = OM² + 12² → OM² = 169 − 144 = 25
5
OM = 5 cm
Length, so take the positive root.

10The reason wording markers expect

∠ at centre = 2 × ∠ at circumference · ∠s in same seg · ∠ in semi-circle
opp ∠s of cyclic quad · ext ∠ of cyclic quad · converse opp ∠s cyclic quad
tan ⊥ radius · tan–chord ∠ · tans from same pt · converse tan–chord ∠
line from centre ⊥ chord · line from centre to midpt of chord · radii

⚠️

Writing a reason that is merely true instead of the one that justifies the step. “Both angles are 40°” is an observation. “∠s in the same seg” is a reason. Only the second earns the mark.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
State the axiom about a tangent and a radius that all of circle geometry is built on.
Show answer ▾Hide answer ▴
A tangent to a circle is perpendicular to the radius drawn to the point of contact ✓✓
2
O is the centre and AP̂B = 28°, with A, B and P on the circle. Determine AÔB, with a reason.
Show answer ▾Hide answer ▴
AÔB = 2 × 28° = 56° ✓ (∠ at centre = 2 × ∠ at circumference) ✓
3
In cyclic quadrilateral ABCD, Â = 118°. Determine Ĉ, and state the theorem you used.
Show answer ▾Hide answer ▴
Ĉ = 180° − 118° ✓ = 62° ✓ (opp ∠s of a cyclic quad are supplementary) ✓
4
In quadrilateral KLMN, K̂ = 96° and M̂ = 84°. Prove that KLMN is cyclic.
Show answer ▾Hide answer ▴
K̂ + M̂ = 96° + 84° ✓ = 180° ✓; ∴ KLMN is cyclic (converse: opp ∠s supplementary) ✓
5
O is the centre of a circle of radius 17 cm. A chord PQ is 30 cm long and M is its midpoint. Determine OM.
Show answer ▾Hide answer ▴
OM ⊥ PQ ✓ (line from centre to midpt of chord); PM = 15 cm ✓; OM² = 17² − 15² = 289 − 225 = 64 ✓; OM = 8 cm ✓
6
TA is a tangent at A and AB is a chord. The tangent–chord angle TÂB = 52°. Write down the size of the angle in the alternate segment, with a reason, and then determine OÂB.
Show answer ▾Hide answer ▴
The angle in the alternate segment = 52° ✓ (tan–chord ∠) ✓; OÂB = 90° − 52° = 38° ✓ (tan ⊥ radius)
7
Explain why you may NOT quote “opp ∠s of a cyclic quad” as a reason inside a proof that a quadrilateral IS cyclic.
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That reason assumes the quadrilateral is already cyclic ✓, which is exactly what is being proved ✓. You must show the opposite angles add to 180° and then conclude using the CONVERSE ✓.
8
In △OMA, O is the centre, OA is a radius and OM ⊥ AB where AB is a chord. Prove that AM = MB.
Show answer ▾Hide answer ▴
Join OB ✓ (construction); OA = OB (radii) ✓; OM = OM (common); OM̂A = OM̂B = 90° (given) ✓; ∴ △OMA ≡ △OMB (RHS); ∴ AM = MB (≡△s) ✓
9
State the converse of the tangent–chord theorem and say what it is used to prove.
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If the angle between a line and a chord equals the angle in the alternate segment ✓, then that line is a tangent to the circle ✓. It is used to PROVE that a given line is a tangent ✓.
10
O is the centre and AÔB = 96°. OA and OB are radii. Determine OÂB, with reasons.
Show answer ▾Hide answer ▴
OA = OB (radii) ✓, so △OAB is isosceles and OÂB = OB̂A (∠s opp equal sides) ✓; OÂB = (180° − 96°) ÷ 2 = 42° ✓