How many circle theorems do I need to know?

About eight: centre, semicircle, same segment, cyclic quadrilateral, tangent and radius, two tangents, alternate segment, and the chord bisector.

What counts as a good reason in the exam?

Use the full name of the theorem or its statement, such as 'angle at the centre is twice the angle at the circumference', not just 'circle rule'.

How do I start a circle theorem question?

Look for radii and diameters, because they create isosceles triangles and right angles. Mark each angle on the diagram as you find it.

Is this on the Foundation paper?

No. Circle theorems and their proofs are Higher tier.

Year 11 circle theorems revision notes to download as PDF

This Year 11 circle theorems revision notes is written for pupils aged 15-16 and follows the National Curriculum. In short, it shows you how to use circle theorems to find angles and give reasons. Before you start, it helps to revise sine rule, cosine rule and area of a triangle and similarity and congruence, because this lesson builds on them.

Year 11 circle theorems revision notes: what you will learn

This lesson is Higher tier and rewards clear reasoning. In the exam you must find the angle and give a reason using the correct theorem name. Draw the circle and write each angle you find on the diagram.

Starter activity

Draw a circle with a compass and mark two points \(A\) and \(B\) on it. Now mark three different points \(C\), \(D\), \(E\) on the major arc and measure angle \(ACB\), \(ADB\) and \(AEB\) with a protractor. What do you notice? Next, join \(A\) and \(B\) to the centre and measure angle \(AOB\). How does it compare with your first three angles? Write down your conjecture.

Parts of a circle

Definition: centre, radius, chord, diameter and tangent

The centre is the middle point and any line from the centre to the circle is a radius. A chord joins two points on the circle. A diameter is a chord through the centre. A tangent touches the circle at one point.

Definition: arc and segment

An arc is part of the circumference, and a segment is the region between a chord and an arc.

Key fact: isosceles triangles

A triangle formed by two radii and a chord is isosceles, because the two radii are equal, so its base angles are equal.

The theorems

Theorem: angle at the centre

The angle at the centre is twice the angle at the circumference when both stand on the same arc. If the arc subtends \(80^\circ\) at the centre it subtends \(40^\circ\) at the circumference.

Theorem: angle in a semicircle

The angle in a semicircle is \(90^\circ\). If \(AB\) is a diameter and \(C\) is on the circle, then angle \(ACB=90^\circ\).

Theorem: angles in the same segment

Angles at the circumference standing on the same arc are equal.

Theorem: cyclic quadrilateral

Opposite angles of a cyclic quadrilateral (all four vertices on the circle) add up to \(180^\circ\). An exterior angle equals the interior opposite angle.

Theorem: tangent and radius

A tangent is perpendicular to the radius at the point of contact.

Theorem: two tangents

Tangents drawn to a circle from the same point are equal in length, so the triangle formed with the chord is isosceles.

Theorem: alternate segment

The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the other side of the chord).

Theorem: chord and perpendicular

The perpendicular from the centre to a chord bisects the chord.

Worked examples

Worked example 1: angle at the centre

The angle at the circumference is \(64^\circ\) on arc \(AB\). Find angle \(AOB\) and angle \(OAB\).

The angle at the centre is \(128^\circ\). Triangle \(OAB\) is isosceles (\(OA=OB\)), so \(\angle OAB=\dfrac{180-128}{2}=26^\circ\).

Worked example 2: cyclic quadrilateral with algebra

Opposite angles of a cyclic quadrilateral are \((2x+10)^\circ\) and \((3x-5)^\circ\). Find \(x\).

\(2x+10+3x-5=180\Rightarrow5x=175\Rightarrow x=35\). The angles are \(80^\circ\) and \(100^\circ\).

Worked example 3: tangents

\(TA\) and \(TB\) are tangents from \(T\) and angle \(ATB=50^\circ\). Find \(\angle TAB\), \(\angle OAB\) and \(\angle AOB\).

Since \(TA=TB\), the triangle \(TAB\) is isosceles and \(\angle TAB=\dfrac{180-50}{2}=65^\circ\). Also \(\angle OAT=90^\circ\), so \(\angle OAB=90-65=25^\circ\). In the kite \(OATB\), \(\angle AOB=360-90-90-50=130^\circ\).

Worked example 4: chord and radius

A circle has radius 13 cm and a chord of length 24 cm. How far is the chord from the centre?

The perpendicular from the centre bisects the chord, giving a right-angled triangle with hypotenuse 13 and one side 12. The distance is \(\sqrt{13^2-12^2}=5\) cm.

Worked example 5: alternate segment

A tangent at \(A\) makes angles of \(70^\circ\) with chord \(AB\) and \(50^\circ\) with chord \(AC\) on the other side. Find the angles of triangle \(ABC\).

By the alternate segment theorem \(\angle ACB=70^\circ\) and \(\angle ABC=50^\circ\). The angles at \(A\) on the straight line are \(70+x+50=180\), so \(\angle BAC=60^\circ\), and \(70+50+60=180\) checks the triangle.

Proof and method

Proof: angle in a semicircle

Let \(AB\) be a diameter with centre \(O\) and let \(C\) be on the circle. Join \(OC\). Then \(OA=OB=OC\) (radii), so triangles \(OAC\) and \(OBC\) are isosceles. Let \(\angle OAC=\angle OCA=x\) and \(\angle OBC=\angle OCB=y\). The angles of triangle \(ABC\) add up to \(180^\circ\): \(x+y+(x+y)=180\), so \(x+y=90\). Therefore angle \(ACB=x+y=90^\circ\).

Method: circle problems

Look for radii (isosceles triangles), diameters and tangents first.
Mark every angle you can find on the diagram.
Write each statement as "angle = value, because ...".
Check that angles in a triangle sum to \(180^\circ\) and angles on a line sum to \(180^\circ\).

Watch out: common mistakes

Using the centre theorem when the angle is not at the centre.
Forgetting the reflex angle at the centre: a reflex angle \(220^\circ\) gives \(110^\circ\) at the circumference on the minor arc.
Giving a vague reason like "circle rule".
Assuming the quadrilateral is cyclic when its vertices are not all on the circle.

Worked example 6: Using the tangent-radius angle

\(TA\) is a tangent at \(A\) to a circle with centre \(O\). Angle \(OTA=24^\circ\). Find angle \(AOT\).

The radius \(OA\) meets the tangent at \(90^\circ\), so angle \(OAT=90^\circ\). The angles in triangle \(OAT\) add to \(180^\circ\): \(AOT=180-90-24=66^\circ\). Always write the reason "angle between tangent and radius is \(90^\circ\)" next to your working.

Worked example 7: Two tangents and a kite

\(TA\) and \(TB\) are tangents to a circle with centre \(O\). Angle \(AOB=116^\circ\). Find angle \(ATB\).

\(OATB\) is a kite with right angles at \(A\) and \(B\). The angles of a quadrilateral add to \(360^\circ\): \(ATB=360-90-90-116=64^\circ\). Alternatively, triangle \(TAB\) is isosceles and angle \(TAB=\tfrac12(180-64)=58^\circ\), which equals \(90^\circ-32^\circ\), where \(32^\circ\) is angle \(OAB\) from the isosceles triangle \(OAB\) (\(180-116=64\), \(64\div2=32\)).

Worked example 8: Angles in the same segment

\(A\), \(B\), \(C\), \(D\) lie on a circle. Angle \(ADB=48^\circ\) and angle \(DBC=30^\circ\). Find angle \(ACB\) and angle \(ACD\) if \(AC\) is a diameter.

\(ACB\) and \(ADB\) are in the same segment on chord \(AB\), so \(ACB=48^\circ\). Since \(AC\) is a diameter, angle \(ADC=90^\circ\) and angle \(ABC=90^\circ\). In triangle \(ABC\) the third angle is \(BAC=180-90-48=42^\circ\). This is a typical exam chain: one equal-angle step, then a right angle step.

Worked example 9: A circle theorem with algebra

In a cyclic quadrilateral \(ABCD\), angle \(A=(3x+10)^\circ\), angle \(B=(2x+30)^\circ\) and angle \(C=(5x-10)^\circ\). Find \(x\) and angle \(D\).

Opposite angles \(A\) and \(C\) add to \(180^\circ\): \(3x+10+5x-10=180\), so \(8x=180\) and \(x=22.5\). Then \(A=77.5^\circ\), \(B=75^\circ\) and \(C=102.5^\circ\). Angle \(D=180-B=105^\circ\). Check: \(77.5+75+102.5+105=360\).

Worked example 10: Perpendicular from the centre to a chord

A chord \(AB\) of length 10 cm is drawn in a circle of radius 13 cm. How far is the chord from the centre?

The perpendicular from the centre bisects the chord, so half the chord is 5 cm. By Pythagoras the distance is \(\sqrt{13^2-5^2}=\sqrt{144}=12\) cm. Notice the 5, 12, 13 Pythagorean triple.

Worked example 11: Finding the angle at the centre in a quadrilateral

\(A\), \(B\), \(C\) are on a circle, centre \(O\), and angle \(ABC=70^\circ\). Find the reflex angle \(AOC\) and angle \(OAC\).

The angle at the centre standing on the same arc as \(B\) is \(2\times70=140^\circ\), so the non-reflex angle \(AOC=140^\circ\) and the reflex angle is \(360-140=220^\circ\). Triangle \(OAC\) is isosceles with \(OA=OC\), so \(OAC=(180-140)\div2=20^\circ\).

Worked example 12: Alternate segment with a tangent from an external point

\(TA\) and \(TB\) are tangents to a circle. \(C\) is on the major arc \(AB\). Angle \(ATB=48^\circ\). Find angle \(ACB\) and give reasons.

Triangle \(TAB\) is isosceles because \(TA=TB\) (tangents from a point), so angle \(TAB=(180-48)\div2=66^\circ\). By the alternate segment theorem the angle between the tangent \(TA\) and the chord \(AB\) equals the angle in the alternate segment, so angle \(ACB=66^\circ\). A second method uses the centre: \(AOB=180-48=132^\circ\) from the kite, and \(ACB=132\div2=66^\circ\). Both methods agree, which is a good way to check your answer.

Note: how to write a reason

In circle theorem questions the marks are for the reason as much as for the angle. Use short standard phrases: "angle at the centre is twice the angle at the circumference", "angle in a semicircle is \(90^\circ\)", "angles in the same segment are equal", "opposite angles of a cyclic quadrilateral add to \(180^\circ\)", "angle between tangent and radius is \(90^\circ\)", "tangents from a point are equal in length" and "alternate segment theorem". Write the reason next to the angle, and add an isosceles triangle reason ("radii are equal") whenever you use the centre.

Watch out: spotting the wrong theorem

The angle at the centre rule only works for angles standing on the same arc. Check both angles are made from the same chord.
A cyclic quadrilateral has all four vertices on the circle; opposite angles add to \(180^\circ\), not adjacent angles.
The alternate segment angle is on the other side of the chord from the angle with the tangent.
Do not assume that a line through the centre is a diameter unless both ends touch the circle.

Did you know?

Circle theorems were already known to the ancient Greeks. Thales of Miletus, who lived about 600 BC, is said to have proved that the angle in a semicircle is a right angle, and Euclid collected the other theorems in his book Elements around 300 BC. Craftspeople still use the semicircle theorem: a carpenter checks that a corner is square by placing a set square so that its corner touches a circle drawn on its hypotenuse.

Otto’s tip

When you see a circle, ask "where is the centre, and which radii can I draw?" Equal radii give isosceles triangles, and these are the key to most circle theorem questions.

How to use this Year 11 circle theorems revision notes

First, read the key points so that you know where the Year 11 circle theorems revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the circle theorems worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the circle theorems test and mark it with the test answers.

Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 11 circle theorems revision notes gives you everything you need for the chapter.