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Year 10 Maths Advanced Circle Geometry

Angle Properties of Circles

20 practice questions 2 video lessons Theory + worked examples

The angle properties of circles link angles at the centre and at the circumference. The angle at the centre is twice the angle at the circumference standing on the same arc, angles in the same segment are equal, the angle in a semicircle is \(90^\circ\), and opposite angles of a cyclic quadrilateral add to \(180^\circ\). This is NSW Year 10 Advanced maths (NESA outcome MA5‑CIR‑P‑01).

You will find unknown angles in circles with a reason for each step and combine them with isosceles triangles formed by radii. You will use the exterior angle of a cyclic quadrilateral and solve equations for angles. Harder questions find all four angles of a cyclic quadrilateral from expressions, and prove that the angle at the centre is twice the angle at the circumference. It is part of Year 10 Advanced Circle Geometry.

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Theory

An angle at the centre of a circle is twice the angle at the circumference standing on the same arc. From this come the angle in a semicircle, equal angles on the same arc, and the supplementary opposite angles of a cyclic quadrilateral. This Year 10 Advanced maths page shows how to use and prove these results, with reasons at every step.

An angle at the centre has its vertex at the centre \(O\). An angle at the circumference has its vertex on the circle. Both stand on (are subtended by) the arc between the ends of their arms.

To find the arc an angle stands on, follow its two arms to the circle: the arc between those points on the far side from the vertex is the one. An angle at the circumference that stands on the major arc is obtuse, and its matching angle at the centre is a reflex angle.

A cyclic quadrilateral has all four vertices on a circle. Its exterior angle is made by producing (extending) one side.

Every answer needs a reason, written as a short theorem name such as "angles on the same arc" or "opposite angles of a cyclic quadrilateral".

It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.

Angle at the centre and angles at the circumference on the same arcA circle with centre O. The red arc AB is at the bottom. The angle AOB at the centre is 2 theta. Points C and D on the major arc each make an angle theta with A and B, standing on the same arc. 2θ θ θ O A B C D
Angles standing on the same arc \(AB\): the angle at the centre is twice the angle at the circumference, and angles at the circumference are equal.
Cyclic quadrilateral with an exterior angleA cyclic quadrilateral ABCD. The angle at A is alpha and the opposite angle at C is beta, with alpha plus beta equal to 180 degrees. Side BC is produced to E, and the exterior angle DCE equals alpha. α β α A B C D E
Cyclic quadrilateral: \(\alpha + \beta = 180^\circ\), and the exterior angle \(\angle DCE\) equals the interior opposite angle \(\alpha\).

The theorems, all for angles standing on the same arc:

\[\angle AOB = 2\angle ACB \qquad \angle ACB = \angle ADB\]
∠AOB=2∠ACB

For a cyclic quadrilateral \(ABCD\):

\[\angle A + \angle C = 180^\circ \qquad \angle B + \angle D = 180^\circ\]
∠A+∠C=180
TheoremReason to write
Centre angle \(= 2 \times\) circumference angleangle at the centre is twice the angle at the circumference
Angle in a semicircle \(= 90^\circ\)angle in a semicircle
Angles on the same arc are equalangles standing on the same arc
Opposite angles add to \(180^\circ\)opposite angles of a cyclic quadrilateral
Exterior angle \(=\) interior opposite angleexterior angle of a cyclic quadrilateral
Older facts still apply. Radii are equal (isosceles triangles), angles in a triangle add to \(180^\circ\), in a quadrilateral to \(360^\circ\), at a point to \(360^\circ\), and the exterior angle of a triangle equals the sum of the two interior opposite angles.

How to find an angle in a circle

  1. Mark the given angles and any equal radii on the diagram.
  2. Trace the arms of each angle to find the arc it stands on. Angles on the same arc are linked: equal at the circumference, double at the centre.
  3. Look for a diameter (right angle), a cyclic quadrilateral (supplementary pairs) or an isosceles triangle with two radii.
  4. Calculate one angle at a time and write the reason beside each step.

Problem types

  1. Expressions in a pronumeral: use a theorem to write an equation, e.g. centre angle \(= 2\times\) circumference angle or opposite angles add to \(180\). Solve, then substitute back to find the angle.
  2. Two unknowns: each pair of opposite angles of a cyclic quadrilateral gives one equation. Solve the two equations simultaneously, then check that each pair adds to \(180^\circ\).
  3. Reflex angles: an obtuse angle at the circumference pairs with the reflex angle at the centre. Subtract from \(360^\circ\) to get the non-reflex angle.
  4. Quadrilateral with the centre as a vertex (e.g. \(AOBC\)): let the circumference angle be \(x\), write the reflex angle at \(O\) as \(360^\circ - 2x\), and use the angle sum of \(360^\circ\).
  5. Proofs: join the centre to the points on the circle, mark equal radii, name the base angles \(\alpha\), \(\beta\), and use the exterior angle of a triangle or an angle sum. To prove the centre angle is twice the circumference angle, produce \(CO\) to \(P\) and use the exterior angles at \(O\). To prove the cyclic quadrilateral result, the two angles at \(O\) (\(2x\) and \(2y\)) make a revolution.
Example 1 — The reflex angle at the centre
\(A\), \(B\), \(C\) and \(D\) lie on a circle with centre \(O\), with \(B\) on the minor arc \(AC\) and \(D\) on the major arc. \(\angle ABC = 118^\circ\). Find
(i) the non-reflex \(\angle AOC\)
(ii) \(\angle ADC\).
Solution

\(\angle ABC\) stands on the major arc \(AC\), so the angle at the centre on that arc is the reflex angle \(AOC\).

Angle ABC of 118 degrees at a point B on the minor arcA circle with centre O. B is at the bottom on the minor arc AC and angle ABC is 118 degrees. The reflex angle AOC at the centre faces B. D is at the top on the major arc and angle ADC is unknown. 118° ? O A B C D

(i) Angle at the centre is twice the angle at the circumference:

\(\text{reflex }\angle AOC\)\(=\)\(2 \times 118^\circ = 236^\circ\)
\(\angle AOC\)\(=\)\(360^\circ - 236^\circ\)
\(\)\(=\)\(\mathbf{124^\circ}\)
∠AOC=360−236=124

(ii) \(ABCD\) is a cyclic quadrilateral, so opposite angles are supplementary:

\(\angle ADC\)\(=\)\(180^\circ - 118^\circ = \mathbf{62^\circ}\)
∠ADC=180−118=62

Check: \(\angle ADC\) is half of \(124^\circ\), as it stands on the minor arc.

Example 2 — Angles on the same arc
\(A\), \(B\), \(C\) and \(D\) lie on a circle and the chords \(AC\) and \(BD\) meet at \(X\). \(\angle ABD = 30^\circ\) and \(\angle BDC = 44^\circ\). Find \(\angle AXB\), giving reasons.
Solution
Chords AC and BD crossing at XFour points A, B, C, D on a circle. Chords AC and BD cross at X. Angle ABD is 30 degrees and angle BDC is 44 degrees. Angle AXB is unknown. 30° 44° ? X A B C D

\(\angle BDC\) and \(\angle BAC\) both stand on arc \(BC\), so they are equal (angles on the same arc):

\(\angle BAX\)\(=\)\(\angle BDC = 44^\circ\)

In \(\triangle ABX\) (angle sum of a triangle):

\(\angle AXB\)\(=\)\(180^\circ - 44^\circ - 30^\circ\)
\(\)\(=\)\(\mathbf{106^\circ}\)
∠AXB=180−44−30=106

Name each angle with three letters: the middle letter is the vertex, and the outer letters show the arc it stands on.

Example 3 — Cyclic quadrilateral with algebra
\(ABCD\) is a cyclic quadrilateral with \(\angle DAB = (5p - 20)^\circ\) and \(\angle BCD = (2p + 18)^\circ\). \(BC\) is produced to \(E\). Find \(p\) and \(\angle DCE\).
Solution
Cyclic quadrilateral ABCD with BC produced to ECyclic quadrilateral ABCD. The angle at A is 5p minus 20 degrees and the angle at C is 2p plus 18 degrees. BC is produced to E and the exterior angle DCE is unknown. (5p − 20)° (2p + 18)° ? A B C D E

Opposite angles of a cyclic quadrilateral are supplementary:

\((5p - 20) + (2p + 18)\)\(=\)\(180\)
\(7p - 2\)\(=\)\(180\)
\(p\)\(=\)\(\mathbf{26}\)
7p−2=180,p=26

So \(\angle DAB = 5(26) - 20 = 110^\circ\). The exterior angle of a cyclic quadrilateral equals the interior opposite angle:

\(\angle DCE\)\(=\)\(\angle DAB = \mathbf{110^\circ}\)

Check: \(\angle BCD = 2(26) + 18 = 70^\circ\), and \(70^\circ + 110^\circ = 180^\circ\) on the straight line \(BCE\).

Example 4 — Prove the angle in a semicircle
\(AB\) is a diameter of a circle with centre \(O\) and \(C\) is any other point on the circle. Prove that \(\angle ACB = 90^\circ\).
Solution

Join \(OC\). Then \(OA = OB = OC\) (radii), so \(\triangle OAC\) and \(\triangle OBC\) are isosceles.

Angle in a semicircleA semicircle on diameter AB with centre O. C is on the arc. OC is joined, making isosceles triangles OAC and OBC with base angles alpha and beta. α α β β O A B C

Let \(\angle OAC = \alpha\) and \(\angle OBC = \beta\). Base angles of isosceles triangles: \(\angle OCA = \alpha\) and \(\angle OCB = \beta\).

Angle sum of \(\triangle ABC\):

\(\alpha + \beta + (\alpha + \beta)\)\(=\)\(180^\circ\)
\(\alpha + \beta\)\(=\)\(90^\circ\)
\(\therefore\ \angle ACB\)\(=\)\(\alpha + \beta = 90^\circ\)
2(α+β)=180

The angle in a semicircle is a right angle.

Common pitfalls

Check the angles stand on the same arc. \(\angle AOB = 2\angle ACB\) only when both angles stand on arc \(AB\), with \(C\) on the opposite side from the arc. If \(C\) is on the minor arc, use the reflex angle at \(O\).
A cyclic quadrilateral needs all four vertices on the circle. If one vertex is the centre \(O\), the shape is not cyclic and the opposite angles do not add to \(180^\circ\).
Opposite, not adjacent. In \(ABCD\) the supplementary pairs are \(A\) with \(C\) and \(B\) with \(D\). The exterior angle at \(C\) equals the interior angle at \(A\), the vertex across the quadrilateral.
Answer the question asked. After solving for \(x\), substitute back. A question asking for \(\angle DAB\) wants degrees, not the value of the pronumeral.

Frequently asked questions

What is the angle at the centre theorem?

The angle at the centre of a circle is twice the angle at the circumference when both stand on the same arc.

Why is the angle in a semicircle 90 degrees?

The diameter makes an angle of 180 degrees at the centre, and the angle at the circumference on the same arc is half of that, 90 degrees. It can also be proved by joining the centre to the third point and using two isosceles triangles.

What are the properties of a cyclic quadrilateral?

All four vertices lie on a circle. Opposite angles add to 180 degrees, and an exterior angle equals the interior opposite angle.

How do you know which arc an angle stands on?

Follow the two arms of the angle to the circle. The arc between those two points, on the far side from the vertex, is the arc the angle stands on.

Do I need to give reasons in circle geometry?

Yes. Each step should name the theorem or fact used, such as angles on the same arc, radii, or angle sum of a triangle. Marks are given for the reasons as well as the answer.

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