Circle Geometry Theorems

Circle geometry is a large part of the Euclidean geometry section of Paper 2. You need to know the theorems, give the accepted reasons, and reproduce the examinable proofs exactly. This page covers the chord and angle theorems that every rider builds on.

Perpendicular from the centre to a chord

Theorem. The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord. (Reason: line from centre to midpt of chord)

Proof (examinable, IEB). Let $O$ be the centre and $M$ the midpoint of chord $A B$ . Join $O A$ , $O B$ and $O M$ .

In $△ O M A$ and $△ O M B$ :

$O A = O B$ (radii)
$A M = M B$ (given)
$O M$ is common

So $△ O M A \equiv △ O M B$ (SSS), and $\hat{M}_{1} = \hat{M}_{2}$ . But $\hat{M}_{1} + \hat{M}_{2} = 180°$ ( $∠$ s on a str line), so $\hat{M}_{1} = \hat{M}_{2} = 90°$ .

Result to use. The line drawn from the centre perpendicular to a chord bisects the chord. (Reason: line from centre $⊥$ chord). You use this in calculations; its proof (RHS congruence) is not one of the IEB examinable proofs.

Example. A chord of length 24 cm is in a circle of radius 13 cm. The distance from the centre to the chord is $1 3^{2} - 1 2^{2} = 5$ cm.

The angle at the centre

Theorem. The angle subtended by an arc at the centre is double the angle it subtends at any point on the circumference. (Reason: $∠$ at centre $= 2 \times ∠$ at circumference)

Proof (examinable). Join $C O$ and extend it to $D$ . In $△ A O C$ , $O A = O C$ (radii), so $\hat{A} = \hat{C}_{1}$ . The exterior angle $\hat{O}_{1} = \hat{A} + \hat{C}_{1} = 2 \hat{C}_{1}$ . Similarly $\hat{O}_{2} = 2 \hat{C}_{2}$ in $△ B O C$ . Adding: $\hat{A O B} = 2 (\hat{C}_{1} + \hat{C}_{2}) = 2 \hat{A C B}$ .

Example. If $\hat{A O B} = 124°$ , then any angle on the major arc subtended by $A B$ is $62°$ . An angle on the minor arc is half the reflex angle: $\frac{1}{2} (360° - 124°) = 118°$ .

Angles in the same segment

Theorem. Angles subtended by the same arc (or chord) in the same segment are equal. (Reason: $∠$ s in the same seg)

This follows from the angle at the centre: both angles equal half the same central angle.

Example. $A$ , $B$ , $C$ and $D$ lie on a circle with $C$ and $D$ on the same side of $A B$ . If $\hat{A C B} = 41°$ , then $\hat{A D B} = 41°$ .

Related results, used with the same reasoning:

Equal chords subtend equal angles at the circumference. (equal chords; equal $∠$ s)
Equal arcs subtend equal angles.

The angle in a semicircle

Theorem. The angle subtended by a diameter at the circumference is a right angle. (Reason: $∠$ in semi-circle)

This is the angle at the centre theorem with a central angle of 180°.

Converse. If a chord subtends a right angle at the circumference, the chord is a diameter. (Reason: chord subtends $90°$ )

Example. $A B$ is a diameter and $C$ is on the circle. If $\hat{C A B} = 35°$ , then $\hat{A C B} = 90°$ and $\hat{C B A} = 55°$ .

Right angles from this theorem combine with Pythagoras, and later with similar triangles, in many riders.

Key terms

Chord: A line segment joining two points on a circle.
Subtend: An arc or chord subtends an angle at a point when the lines from its endpoints meet there.
Segment: The region between a chord and an arc.
Circumference: The circle itself; an angle 'at the circumference' has its vertex on the circle.
Rider: A geometry problem that applies theorems to a new diagram.

Common mistakes

Writing reasons in your own words instead of the accepted form, such as ' $∠$ at centre $= 2 \times ∠$ at circumference'.
Using the angle at the centre theorem when the angles are not subtended by the same arc.
Forgetting to use the reflex angle when the circumference angle is on the minor arc.
Assuming a line is a diameter or a tangent because it looks like one. It must be given or proved.
Leaving out the construction in a theorem proof.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

$O$ is the centre of a circle with radius 13 cm. $A B$ is a chord of length 24 cm and $M$ is the midpoint of $A B$ . Calculate the length of $O M$ .

Diagram: circle centre $O$ , chord $A B$ with midpoint $M$ , $O M$ and $O A$ drawn.

Worked solution

Hidden - try the question first

Solution

$O M ⊥ A B$ (line from centre to midpt of chord)
$A M = 12$ cm and $O A = 13$ cm (radius)
$O M = 1 3^{2} - 1 2^{2} = 5$ cm (Pythagoras)

Question 2

5 marks total

knowledge

Prove the theorem that states that the line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Worked solution

Hidden - try the question first

Solution

Given $M$ the midpoint of $A B$ ; join $O A$ , $O B$ and $O M$
$O A = O B$ (radii); $A M = M B$ (given); $O M$ common
$△ O M A \equiv △ O M B$ (SSS)
$\hat{M}_{1} = \hat{M}_{2}$ ( $\equiv$ triangles)
$\hat{M}_{1} + \hat{M}_{2} = 180°$ ( $∠$ s on a str line), so $\hat{M}_{1} = \hat{M}_{2} = 90°$

Question 3

6 marks total

complex

$A$ , $B$ , $C$ and $D$ lie on a circle with centre $O$ . $C$ lies on the major arc $A B$ and $D$ lies on the minor arc $A B$ . $\hat{A O B} = 124°$ .

Calculate, with reasons:

(a) $\hat{A C B}$ (2)

(b) $\hat{A D B}$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $M$ ( $A M = M B$ at centre $= 2 \times ∠$ at circumference)
$= 62°$
(b) Reflex $\hat{A O B} = 360° - 124° = 236°$ ( $∠$ s round a pt)
$\hat{A D B} = 118°$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
(c) $O A = O B$ (radii), so $\hat{O A B} = \hat{O B A}$
$\hat{O A B} = \frac{1}{2} (180° - 124°) = 28°$ (sum of $∠$ s in $△$ )

Question 4

9 marks total

problem-solving

$A B$ is a diameter of a circle with centre $O$ and $A B = 26$ cm. Chord $C E$ is perpendicular to $A B$ and cuts it at $D$ , with $D$ between $O$ and $B$ and $O D = 5$ cm. $C$ lies on the circle.

(a) Calculate the length of $C E$ . (3)

(b) Write down the size of $\hat{A C B}$ , with a reason. (1)

(d) Calculate $B C$ and use it to confirm your answer to (b). (2)

Worked solution

Hidden - try the question first

Solution

(a) $M$ (radius), so $C D = 1 3^{2} - 5^{2} = 12$
$C D = D E$ (line from centre $⊥$ chord)
$C E = 24$ cm
(b) $\hat{A C B} = 90°$ ( $∠$ in semi-circle)
(c) $A D = 13 + 5 = 18$
$A C^{2} = 1 8^{2} + 1 2^{2} = 468$
$A C = 468 = 613$ cm
(d) $D B = 8$ , so $B C^{2} = 8^{2} + 1 2^{2} = 208$
$A C^{2} + B C^{2} = 468 + 208 = 676 = 2 6^{2} = A B^{2}$ , so $\hat{A C B} = 90°$ (converse Pythagoras)

FAQs

Which circle geometry proofs are examinable in IEB Grade 12?

From Grade 11 (acute-angle case only): the line from the centre to the midpoint of a chord is perpendicular to the chord, the angle at the centre theorem, opposite angles of a cyclic quadrilateral, and the tan-chord theorem. From Grade 12: the proportionality theorem, equiangular triangles are similar, and the perpendicular from the right angle to the hypotenuse. Converses are not examinable as proofs, and there is no concurrency or proof by contradiction.

Do I have to use the exact wording for reasons?

Use the accepted abbreviated reasons. Markers accept standard abbreviations, but vague reasons such as 'circle theorem' are not accepted, so they lose the reason mark.

How do I get better at geometry riders?

Mark every equal angle and length on the diagram before writing anything, and look for the theorem each pair of equal angles comes from. Practise with the memo reasons in front of you at first.

Perpendicular from the centre to a chord

The angle at the centre

Angles in the same segment

The angle in a semicircle

Key terms

Common mistakes

Practice questions

FAQs

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