Mastering Circle Theorems for Mathematics Examinations

What the examiner is testing

The examiner is testing your ability to identify and apply the correct circle theorem(s) to find unknown angles or lengths within a given diagram. Marks are awarded for correctly stating the theorem used and for accurate calculation of the required values.

The method

1. Identify the knowns: Mark all given angles and lengths directly onto the diagram.
2. Scan for key features: Look for chords, tangents, radii, diameters, and cyclic quadrilaterals. These features trigger specific circle theorems.
3. Match features to theorems: For each key feature found, recall the associated circle theorem(s). For example, a tangent meeting a radius at the point of contact implies a \(90^\circ\) angle.
4. Formulate a plan: Determine which unknown you need to find and which theorem(s) will lead you there. You might need to find an intermediate angle first.
5. Apply the theorem(s) and calculate: Use the identified theorem(s) to set up equations and solve for the unknown angles or lengths. State the theorem used clearly as part of your working.
6. Check for consistency: Ensure all angles in triangles sum to \(180^\circ\), angles on a straight line sum to \(180^\circ\), and angles around a point sum to \(360^\circ\).

Worked example

Question: In the diagram below, O is the centre of the circle. A, B and C are points on the circumference. Angle AOC = \(110^\circ\). Find angle ABC.

\[ \begin{tikzpicture} \coordinate (O) at (0,0); \draw (O) circle (2cm); \coordinate (A) at (1.732,1); \coordinate (B) at (-1.732,-1); \coordinate (C) at (-1.732,1); \draw (O) -- (A); \draw (O) -- (C); \draw (A) -- (B) -- (C); \node at (0.2,0.2) {O}; \node at (1.9,1.2) {A}; \node at (-1.9,-1.2) {B}; \node at (-1.9,1.2) {C}; \draw pic [draw, angle radius=0.5cm, "110$^\circ$"] {angle = C--O--A}; \end{tikzpicture} \]

Solution:
Angle AOC is the angle at the centre.
Angle ABC is the angle at the circumference subtended by the same arc AC.
The angle at the centre is twice the angle at the circumference.
$$ \text{Angle ABC} = \frac{\text{Angle AOC}}{2} $$
$$ \text{Angle ABC} = \frac{110^\circ}{2} $$
$$ \text{Angle ABC} = 55^\circ $$
Sanity check: The angle at the circumference should be smaller than the angle at the centre, and \(55^\circ\) is indeed smaller than \(110^\circ\).

Worked example: a harder one

Question: A, B, C and D are points on the circumference of a circle with centre O. TA is a tangent to the circle at A. Angle ATD = \(30^\circ\). Angle BCD = \(100^\circ\). Find angle BAD.

\[ \begin{tikzpicture} \coordinate (O) at (0,0); \draw (O) circle (2cm); \coordinate (A) at (0,2); \coordinate (B) at (-1.732, -1); \coordinate (C) at (1.732, -1); \coordinate (D) at (2,0); \draw (A) -- (B) -- (C) -- (D) -- (A); \coordinate (T) at (-2,2); \draw (T) -- (A); \node at (0.2,0.2) {O}; \node at (0.2,2.2) {A}; \node at (-1.9,-1.2) {B}; \node at (1.9,-1.2) {C}; \node at (2.2,0.2) {D}; \node at (-2.2,2.2) {T}; \draw pic [draw, angle radius=0.5cm, "30$^\circ$"] {angle = D--A--T}; \draw pic [draw, angle radius=0.5cm, "100$^\circ$"] {angle = D--C--B}; \end{tikzpicture} \]

Initial thought: Angle ATD is given, perhaps use the tangent-radius theorem.
Why it fails: We don't have a radius drawn to A, and even if we did, it wouldn't directly help find angle BAD without more information about point T's position relative to the circle. This theorem is for angles at the point of tangency with a radius.

Correct approach:
1. Identify cyclic quadrilateral: ABCD is a cyclic quadrilateral because all its vertices lie on the circumference.
2. Apply cyclic quadrilateral theorem: Opposite angles in a cyclic quadrilateral sum to \(180^\circ\).
Angle BAD + Angle BCD = \(180^\circ\)
Angle BAD + \(100^\circ\) = \(180^\circ\)
Angle BAD = \(180^\circ - 100^\circ\)
Angle BAD = \(80^\circ\)
3. Alternative check (using alternate segment theorem):
Angle ATD is the angle between the tangent TA and the chord AD.
By the alternate segment theorem, the angle between a tangent and a chord is equal to the angle in the alternate segment.
Therefore, Angle ATD = Angle ABD.
So, Angle ABD = \(30^\circ\).
This gives us part of angle BAD, but not the whole angle. This path is more complex and requires finding angle DBC. Sticking with the cyclic quadrilateral theorem is more direct here.

Practice

1. O is the centre of the circle. A, B and C are points on the circumference. Angle OAB = \(35^\circ\). Find angle ACB.
   \[ \begin{tikzpicture} \coordinate (O) at (0,0); \draw (O) circle (2cm); \coordinate (A) at (1.732,1); \coordinate (B) at (0,2); \coordinate (C) at (-1.732, -1); \draw (O) -- (A); \draw (O) -- (B); \draw (A) -- (B) -- (C); \node at (0.2,0.2) {O}; \node at (1.9,1.2) {A}; \node at (0.2,2.2) {B}; \node at (-1.9,-1.2) {C}; \draw pic [draw, angle radius=0.5cm, "35$^\circ$"] {angle = B--A--O}; \end{tikzpicture} \]
2. A, B, C and D are points on the circumference of a circle. Angle ADC = \(70^\circ\). Find angle ABC.
   \[ \begin{tikzpicture} \coordinate (O) at (0,0); \draw (O) circle (2cm); \coordinate (A) at (0,2); \coordinate (B) at (-1.732, -1); \coordinate (C) at (0,-2); \coordinate (D) at (1.732, 1); \draw (A) -- (B) -- (C) -- (D) -- (A); \node at (0.2,2.2) {A}; \node at (-1.9,-1.2) {B}; \node at (0.2,-2.2) {C}; \node at (1.9,1.2) {D}; \draw pic [draw, angle radius=0.5cm, "70$^\circ$"] {angle = C--D--A}; \end{tikzpicture} \]
3. P, Q, R are points on the circumference of a circle with centre O. PR is a diameter. Angle QPR = \(20^\circ\). Find angle PRQ.
   \[ \begin{tikzpicture} \coordinate (O) at (0,0); \draw (O) circle (2cm); \coordinate (P) at (-2,0); \coordinate (R) at (2,0); \coordinate (Q) at (0,2); \draw (P) -- (Q) -- (R) -- (P); \node at (-2.2,0.2) {P}; \node at (0.2,2.2) {Q}; \node at (2.2,0.2) {R}; \node at (0.2,0.2) {O}; \draw pic [draw, angle radius=0.5cm, "20$^\circ$"] {angle = R--P--Q}; \end{tikzpicture} \]
4. A, B, C are points on the circumference of a circle. TA is a tangent to the circle at A. Angle TAC = \(65^\circ\). Find angle ABC.
   \[ \begin{tikzpicture} \coordinate (O) at (0,0); \draw (O) circle (2cm); \coordinate (A) at (0,2); \coordinate (B) at (-1.732, -1); \coordinate (C) at (1.732, -1); \draw (A) -- (B) -- (C) -- (A); \coordinate (T) at (-2,2); \draw (T) -- (A); \node at (0.2,2.2) {A}; \node at (-1.9,-1.2) {B}; \node at (1.9,-1.2) {C}; \node at (-2.2,2.2) {T}; \draw pic [draw, angle radius=0.5cm, "65$^\circ$"] {angle = C--A--T}; \end{tikzpicture} \]

Answers:
1. Angle ACB = \(55^\circ\)
* Triangle OAB is isosceles because OA and OB are radii.
* Angle OBA = Angle OAB = \(35^\circ\).
* Angle AOB = \(180^\circ - (35^\circ + 35^\circ) = 180^\circ - 70^\circ = 110^\circ\).
* Angle ACB is the angle at the circumference subtended by arc AB. Angle AOB is the angle at the centre subtended by arc AB.
* Angle ACB = Angle AOB / 2 = \(110^\circ / 2 = 55^\circ\).
2. Angle ABC = \(110^\circ\)
* ABCD is a cyclic quadrilateral.
* Opposite angles in a cyclic quadrilateral sum to \(180^\circ\).
* Angle ABC + Angle ADC = \(180^\circ\).
* Angle ABC + \(70^\circ = 180^\circ\).
* Angle ABC = \(110^\circ\).
3. Angle PRQ = \(70^\circ\)
* PR is a diameter, so angle PQR is the angle in a semicircle.
* Angle PQR = \(90^\circ\).
* In triangle PQR, the sum of angles is \(180^\circ\).
* Angle QPR + Angle PQR + Angle PRQ = \(180^\circ\).
* \(20^\circ + 90^\circ + \text{Angle PRQ} = 180^\circ\).
* \(110^\circ + \text{Angle PRQ} = 180^\circ\).
* Angle PRQ = \(70^\circ\).
4. Angle ABC = \(65^\circ\)
* By the alternate segment theorem, the angle between a tangent (TA) and a chord (AC) is equal to the angle in the alternate segment (angle ABC).
* Therefore, Angle ABC = Angle TAC.
* Angle ABC = \(65^\circ\).

The three mistakes that lose marks

1. Confusing angle at centre with angle at circumference: A common error is to assume the angle at the centre is half the angle at the circumference, or to mix up which angle is which. For example, if angle at centre is \(100^\circ\), writing angle at circumference as \(200^\circ\) instead of \(50^\circ\).
2. Incorrectly identifying alternate segment: When using the alternate segment theorem, students often pick the wrong angle in the segment. For instance, if TA is a tangent and AC is a chord, the angle between them (TAC) is equal to the angle in the opposite segment, which is angle ABC, not angle ADC or any other angle.
3. Forgetting properties of isosceles triangles formed by radii: When two radii form a triangle with a chord, that triangle is isosceles. Students often forget this, leading to incorrect calculations of base angles. For example, if OA and OB are radii, triangle OAB is isosceles, so Angle OAB = Angle OBA.

30-second recap

Circle theorems link angles and lengths within circles based on specific geometric features like tangents, chords, and diameters. The key is to identify these features and apply the correct theorem, stating it clearly. Look for cyclic quadrilaterals and isosceles triangles formed by radii.