Maths · 5 Geometry

5.6 Circle theorems I

Before you start

Angles on a straight line sum to $180 °$ , angles at a point sum to $360 °$ , and vertically opposite angles are equal.
The angles in a triangle sum to $180 °$ and the angles in a quadrilateral sum to $360 °$ .
An isosceles triangle has two equal sides, and the two angles opposite them (the base angles) are equal.
Three-letter angle notation: angle $A B C$ is the angle at $B$ , between the lines $B A$ and $B C$ .
The circle words centre, radius, diameter, chord, tangent, arc and segment.
Forming and solving a linear equation such as $5 x + 5 = 180$ .

Syllabus objectives covered (1)

Calculate unknown angles and give explanations using the following geometrical properties of circles: • angle in a semicircle = 90° • angle between tangent and radius = 90° • angle at the centre is twice the angle at the circumference • angles in the same segment are equal • opposite angles of a cyclic quadrilateral sum to 180° (supplementary) • alternate segment theorem.

Part 1 of 8

Circle words, and what a reason has to say

A chord is a straight line joining two points on a circle. A diameter is a chord that passes through the centre. A tangent is a straight line that touches the circle at exactly one point, called the point of contact.
graph diagram — not drawn yet
A circle with centre O at the origin and radius 5. A chord AB from A(-3.83, -3.21) to B(3.83, -3.21); the small region below the chord shaded and labelled 'minor segment', the large region above it labelled 'major segment', and the arc below the chord labelled 'minor arc AB'. A diameter CD from C(-5, 0) to D(5, 0) through O. A horizontal tangent line touching the circle at T(0, 5), with T labelled 'point of contact', and the radius OT drawn.
A chord splits the circle into two segments. A tangent meets the circle at one point only.
A chord cuts the circle into two segments. The smaller one is the minor segment and the larger one is the major segment. In the same way, the ends of the chord split the circumference into a minor arc and a major arc.
When the two arms of an angle run to the two ends of a chord $A B$ , the angle is subtended by $A B$ : it stands on the arc $A B$ . Angle $A P B$ , with $P$ on the circle, is an angle at the circumference. Angle $A O B$ , with $O$ the centre, is the angle at the centre.
A cyclic quadrilateral is a quadrilateral with all four vertices on the circle. A vertex at the centre does not count, because the centre is not on the circle.
Every radius of a circle has the same length. So a triangle with two radii as sides, such as $O A B$ with $A$ and $B$ on the circle, is isosceles, and its base angles $O A B$ and $O B A$ are equal. You will need this alongside the circle properties.
The objective asks you to give explanations, and the syllabus expects your reasons to use the properties it lists. So a reason is the property written out in words: 'angles in the same segment are equal'. 'Circle theorem' on its own does not say which property, and 'from the diagram' is not a property at all.
Set out each step on its own line: the angle in three-letter notation, its value, then the reason in brackets. For example: angle $A C B = 58 °$ (angles in the same segment are equal).

Part 2 of 8

The angle at the centre is twice the angle at the circumference

Let $A$ , $B$ and $P$ be points on a circle with centre $O$ , with $P$ on the major arc $A B$ . The angle at the centre, $A O B$ , and the angle at the circumference, $A P B$ , both stand on the minor arc $A B$ . The angle at the centre is twice the angle at the circumference.
$∠ A O B = 2 \times ∠ A P B$
Here is why. Join $P$ to $O$ and continue the line to meet the circle again at $D$ . $O A = O P$ (radii), so triangle $O A P$ is isosceles. Call each of its base angles $a$ . Its third angle, $A O P$ , is then $180 ° - 2 a$ .
graph diagram — not drawn yet
A circle with centre O at the origin and radius 5. P at the top, (0, 5); A on the lower left at (-3.83, -3.21); B on the lower right at (4.33, -2.50). Chords PA and PB and radii OA and OB drawn solid. A dashed line from P through O to D, the lowest point of the circle, (0, -5). In triangle OAP the base angles at A and at P are each marked a; in triangle OBP the base angles at B and at P are each marked b. Angle AOD is marked 2a and angle BOD is marked 2b.
Each isosceles triangle turns its base angle into twice that angle at the centre, so angle AOB = 2a + 2b while angle APB = a + b.
$P O D$ is a straight line, so angle $A O D = 180 ° - (180 ° - 2 a) = 2 a$ (angles on a straight line). The same argument in triangle $O B P$ , with base angles $b$ , gives angle $B O D = 2 b$ .
Add the two parts. Angle $A O B$ is $2 a + 2 b$ , and angle $A P B$ is $a + b$ .
$∠ A O B = 2 a + 2 b = 2 (a + b) = 2 \times ∠ A P B$
Use it in either direction. Halve to go from the centre to the circumference; double to go from the circumference to the centre. If angle $A P B = 47 °$ , then angle $A O B = 94 °$ .
Before you double or halve, check that both angles stand on the same arc: the arms of both must end at the same two points, $A$ and $B$ . The property holds wherever $P$ is on the major arc, whether $O$ is inside triangle $A P B$ or outside it, so check the end points of the arms, not the look of the picture.
If $P$ is on the minor arc instead, angle $A P B$ is obtuse and stands on the major arc. The angle at the centre on the major arc is the reflex angle $A O B$ , so reflex angle $A O B = 2 \times$ angle $A P B$ . With angle $A P B = 115 °$ , reflex angle $A O B = 230 °$ , and the angle $A O B$ that is less than $180 °$ is $360 ° - 230 ° = 130 °$ .

Part 3 of 8

The angle in a semicircle is 90°

If $A B$ is a diameter and $C$ is any other point on the circle, then angle $A C B = 90 °$ . The angle is at the circumference and its arms go to the two ends of the diameter.
graph diagram — not drawn yet
A circle with centre O at the origin and radius 5. A horizontal diameter AB from A(-5, 0) to B(5, 0), passing through O. C on the upper arc at (2.94, 4.05). Chords AC and BC drawn, a right-angle box at C, angle CAB marked 27° and angle ABC marked 63°.
The arms of the angle at C run to the ends of a diameter, so angle ACB is 90° wherever C is on the circle.
This is the centre property with a straight angle at the centre. $A O B$ is a straight line, so angle $A O B = 180 °$ , and half of $180 °$ is $90 °$ .
So triangle $A C B$ is right-angled at $C$ , and its other two angles sum to $90 °$ . If angle $C A B = 27 °$ , then angle $A B C = 180 ° - 90 ° - 27 ° = 63 °$ (angle sum of a triangle).
The chord must pass through the centre. If a chord does not go through $O$ , the angle it subtends at the circumference is not $90 °$ , however close to a diameter it looks. Look for $O$ marked on the line, or the word diameter, before you write $90 °$ .

Part 4 of 8

Angles in the same segment are equal

Let $C$ and $D$ be two points on the circle on the same side of the chord $A B$ , so they are in the same segment. Then angle $A C B$ = angle $A D B$ .
graph diagram — not drawn yet
A circle with centre O at the origin and radius 5. Chord AB from A(-4.70, -1.71) to B(4.70, -1.71). C at (1.71, 4.70) and D at (-2.50, 4.33), both on the major arc above AB. Lines AC, BC, AD and BD drawn, so that AC and BD cross inside the circle. Angles ACB and ADB each marked 70°. E on the minor arc at the bottom, (0, -5), with AE and BE drawn dashed and angle AEB marked 110°.
C and D are in the same segment, so their angles are equal. E is in the other segment, and its angle is 180° minus theirs.
Why: both angles stand on the same arc $A B$ , so each one is half of angle $A O B$ . Two halves of the same angle are equal.
To spot it, look for two triangles that share the chord $A B$ as a base and have their third vertices on the same arc. Joined up, two of their sides cross, making a shape like a bow tie.
The trap is a point on the other side of the chord. $E$ , in the minor segment, does not give the same angle. $A E B C$ is a cyclic quadrilateral, so angle $A E B = 180 ° -$ angle $A C B$ . In the figure that is $180 ° - 70 ° = 110 °$ .
Name the chord before you use this property. Angle $A C B$ stands on chord $A B$ ; angle $C A D$ stands on chord $C D$ . Equal angles come only from the same chord and the same segment.

Part 5 of 8

Opposite angles of a cyclic quadrilateral sum to 180°

In a cyclic quadrilateral $A B C D$ , the angles at opposite corners add up to $180 °$ : angle $D A B$ + angle $B C D = 180 °$ , and angle $A B C$ + angle $C D A = 180 °$ . Two angles that add up to $180 °$ are supplementary.
$∠ D A B + ∠ B C D = 180^{\circ}$
graph diagram — not drawn yet
A circle with centre O at the origin and radius 5. Quadrilateral ABCD with A(-4.33, 2.50), B(-3.83, -3.21), C(3.83, -3.21) and D(1.71, 4.70) on the circle, in that order. Angle DAB marked 105°, angle BCD marked 75°, angle ABC marked 95° and angle CDA marked 85°. Side BC continued beyond C as a dashed line to E(6.89, -3.21), and angle DCE marked 105°.
Opposite angles add up to 180°, and the exterior angle at C equals the interior angle at A.
Why: angle $D A B$ stands on the arc $B C D$ , and angle $B C D$ stands on the arc $D A B$ . Together those two arcs make the whole circle, so the two angles at the centre on them add up to $360 °$ . Each angle of the quadrilateral is half of its angle at the centre, so the two halves add up to $180 °$ .
All four vertices must be on the circle. A quadrilateral such as $O A B C$ , with one corner at the centre $O$ , is not a cyclic quadrilateral of that circle, and its opposite angles need not sum to $180 °$ .
Extend one side and you get an exterior angle. With $B C$ extended to $E$ , angle $D C E = 180 ° -$ angle $B C D$ (angles on a straight line), and angle $D A B = 180 ° -$ angle $B C D$ (opposite angles of a cyclic quadrilateral sum to $180 °$ ). So angle $D C E$ = angle $D A B$ . Give both reasons, because the result comes from the two properties together.
When the angles are given in terms of $x$ , the property gives you the equation. If angle $D A B = (5 x + 4) °$ and angle $B C D = (3 x - 8) °$ , then $(5 x + 4) + (3 x - 8) = 180$ , so $8 x - 4 = 180$ and $x = 23$ . Substitute back: the angles are $119 °$ and $61 °$ , and $119 ° + 61 ° = 180 °$ checks it.

Part 6 of 8

A tangent meets the radius at 90°

A tangent is perpendicular to the radius at the point of contact. If the tangent touches the circle at $T$ , then angle $O T P = 90 °$ for any point $P$ on the tangent.
graph diagram — not drawn yet
Two diagrams side by side. Left: a circle with centre O and a vertical tangent line touching it at T on the right-hand side, the radius OT drawn, a right-angle box at T, and a point P on the tangent above T. Right: a circle with centre O(0, 0) and radius 3, an external point P(7.38, 0), and two tangents from P touching the circle at A(1.22, 2.74) and B(1.22, -2.74); radii OA and OB drawn, right-angle boxes at A and B, angle APB marked 48° and angle AOB marked 132°.
The radius to the point of contact is at right angles to the tangent. With two tangents, the two right angles leave 180° for angles AOB and APB together.
Why it is true: every point of the tangent other than $T$ lies outside the circle, so it is further than one radius from $O$ . That makes $O T$ the shortest line from $O$ to the tangent, and the shortest line from a point to a straight line is the perpendicular one.
A tangent therefore gives you a right-angled triangle $O T P$ , and its other two angles sum to $90 °$ . If angle $O P T = 22 °$ , then angle $T O P = 180 ° - 90 ° - 22 ° = 68 °$ .
Two tangents from the same outside point $P$ , touching at $A$ and $B$ , make the quadrilateral $O A P B$ with right angles at $A$ and $B$ . The angles of a quadrilateral sum to $360 °$ , so angle $A O B$ + angle $A P B = 360 ° - 90 ° - 90 ° = 180 °$ .
$∠ A O B + ∠ A P B = 180^{\circ}$
Give the reasons as 'angle between tangent and radius $= 90 °$ ' and 'angle sum of a quadrilateral'. Do not quote the cyclic quadrilateral property here: $O$ and $P$ are not on the circle.
The fact that the two tangents $P A$ and $P B$ are equal in length is a separate property, in Circle theorems II. You do not need it here: the two right angles are enough.
If a chord joins the two points of contact, triangle $O A B$ is isosceles because $O A = O B$ (radii). Once you know angle $A O B$ , its base angles give you the angle between the chord and each radius.

Part 7 of 8

The alternate segment theorem

Draw a tangent at $T$ and a chord $T B$ from the point of contact. The alternate segment theorem says: the angle between the tangent and the chord equals the angle that the chord subtends at any point on the circle in the segment on the other side of the chord, the alternate segment.
graph diagram — not drawn yet
A circle with centre O at the origin and radius 5. A horizontal tangent touching the circle at its lowest point T(0, -5), with S on the tangent to the left of T and U on it to the right. A chord TB to B(4.33, 2.50). A point A on the upper left of the circle at (-4.33, 2.50), with chords TA and AB drawn. Angle UTB, between the tangent and the chord on the right, marked 60°; angle TAB marked 60°. The small region between chord TB and the arc to its right lightly shaded.
Angle UTB is on the right of the chord TB. The equal angle is at A, in the segment on the other side of the chord.
Find the matching angle in three moves. One: find the chord that starts at the point of contact. Two: note which side of that chord the tangent angle is on. Three: go to the other side of the chord and find a triangle with the chord as its base and its third vertex on the circle. The angle at that third vertex is the equal one.
In the figure, angle $U T B = 60 °$ is on the right of $T B$ . $A$ is on the left of $T B$ , so angle $T A B = 60 °$ (alternate segment theorem).
The tangent makes two angles with the chord, one on each side, and they add up to $180 °$ : here angle $S T B = 120 °$ . Each one pairs with the segment opposite it. A point $Q$ on the minor arc $T B$ , on the right of the chord, gives angle $T Q B = 120 °$ , not $60 °$ .
Why it is true, using three properties you already have. Draw the diameter $T D$ . Angle $D T U = 90 °$ (angle between tangent and radius), so angle $D T B = 90 ° - 60 ° = 30 °$ . Angle $T B D = 90 °$ (angle in a semicircle), so angle $T D B = 180 ° - 90 ° - 30 ° = 60 °$ . $A$ and $D$ are in the same segment of chord $T B$ , so angle $T A B$ = angle $T D B = 60 °$ (angles in the same segment are equal). The same steps work for any tangent angle, not just $60 °$ .
The reason to write is 'alternate segment theorem'. It names the property, so you do not write out the proof.

Part 8 of 8

Choosing the property and chaining the steps

A question can need several properties in a row, each answer feeding the next. Start by marking every given angle on the diagram. Then look for the features that switch each property on.
A diameter, or a chord through $O$ , with an angle at the circumference on it: angle in a semicircle. A tangent with a radius to its point of contact: $90 °$ . A tangent with a chord from its point of contact: alternate segment theorem.
Lines from $O$ to two points on the circle: the angle at the centre, and an isosceles triangle from the two radii. Two angles on the same chord and on the same side of it: same segment. Four points on the circle joined in order: cyclic quadrilateral.
Write one angle per line: name, value, reason. An angle you have found becomes a given for the next line. Mark it on the diagram as you go, so you can see what it unlocks.
When you cannot see the next step, find any angle you can, even one you were not asked for. The angles next to the unknown one can lead you to it.
Diagrams in these questions may not be drawn to scale, so do not measure with a protractor. A measured angle is not a reason, and it may be wrong.
Check as you go: the angles of every triangle add up to $180 °$ , and if two routes lead to the same angle, they must give the same value.

Worked examples

Example 1. Points $A$ , $B$ , $C$ and $D$ lie on a circle with centre $O$ . $C$ and $D$ are on the major arc $A B$ , and angle $A O B = 116 °$ . Find, giving a reason for each answer, (a) angle $A C B$ (b) angle $A D B$ (c) angle $O A B$ .

Show the working

(a) Angle $A O B$ is at the centre and angle $A C B$ is at the circumference. The arms of both end at $A$ and $B$ , and $C$ is on the major arc, so both stand on the minor arc $A B$ . Halve (angle at the centre is twice the angle at the circumference).
$∠ A C B = 116^{\circ} \div 2 = 58^{\circ}$
(b) $D$ is also on the major arc, so $C$ and $D$ are in the same segment of the chord $A B$ . Angle $A D B$ = angle $A C B = 58 °$ (angles in the same segment are equal). Halving $116 °$ again, with the centre property as the reason, is also correct.
(c) $O A$ and $O B$ are radii, so they are equal and triangle $O A B$ is isosceles, with base angles $O A B$ and $O B A$ equal. The two base angles share what is left of $180 °$ .
$∠ O A B = \frac{180^{\circ} - 116^{\circ}}{2} = 32^{\circ}$
Check: in triangle $O A B$ , $116 ° + 32 ° + 32 ° = 180 °$ .

Answer: (a) $58 °$ (angle at the centre is twice the angle at the circumference) (b) $58 °$ (angles in the same segment are equal) (c) $32 °$ ( $O A = O B$ are radii, so triangle $O A B$ is isosceles; angle sum of a triangle)

Example 2. $A B$ is a diameter of a circle with centre $O$ , and $C$ is another point on the circle. $T A$ is the tangent to the circle at $A$ , with $T$ and $C$ on the same side of $A B$ . Angle $T A C = 54 °$ . Find angle $B A C$ , angle $A C B$ and angle $A B C$ , giving reasons.

Show the working

$O A$ is a radius, and the tangent at $A$ is perpendicular to it. The diameter $A B$ runs along that radius, so angle $T A B = 90 °$ (angle between tangent and radius $= 90 °$ ).
graph diagram — not drawn yet
A circle with centre O at the origin and radius 5. A vertical diameter AB with A at the bottom (0, -5) and B at the top (0, 5). A horizontal tangent through A, with T on it to the left at (-7, -5). C on the left of the circle at (-4.76, 1.55). Chords AC and BC drawn, and angle TAC marked 54°.
The tangent at A is at right angles to the diameter AB, and the chord AC lies between them.
$C$ lies inside the right angle $T A B$ , so angle $B A C = 90 ° - 54 ° = 36 °$ .
$A B$ is a diameter, so angle $A C B = 90 °$ (angle in a semicircle).
In triangle $A B C$ : angle $A B C = 180 ° - 90 ° - 36 ° = 54 °$ (angle sum of a triangle).
Check with the alternate segment theorem. The tangent $T A$ and the chord $A C$ make angle $T A C = 54 °$ , and $B$ is on the other side of $A C$ from $T$ , so angle $A B C$ should be $54 °$ . It is.

Answer: Angle $B A C = 36 °$ , angle $A C B = 90 °$ , angle $A B C = 54 °$ .

Example 3. $A B C D$ is a cyclic quadrilateral. Angle $D A B = (3 x + 10) °$ , angle $B C D = (2 x - 5) °$ and angle $A B C = 98 °$ . The side $B C$ is extended beyond $C$ to $E$ . (a) Find $x$ . (b) Find angle $A D C$ . (c) Find angle $D C E$ .

Show the working

(a) $A$ and $C$ are opposite corners, so their angles sum to $180 °$ (opposite angles of a cyclic quadrilateral sum to $180 °$ ). Write that as an equation.
$(3 x + 10) + (2 x - 5) = 180$
Collect the terms: $5 x + 5 = 180$ , so $5 x = 175$ and $x = 35$ .
Substitute back: angle $D A B = 3 (35) + 10 = 115 °$ and angle $B C D = 2 (35) - 5 = 65 °$ . Check: $115 ° + 65 ° = 180 °$ .
(b) $D$ is opposite $B$ , so angle $A D C = 180 ° - 98 ° = 82 °$ (opposite angles of a cyclic quadrilateral sum to $180 °$ ).
(c) $B C E$ is a straight line, so angle $D C E = 180 ° - 65 ° = 115 °$ (angles on a straight line). It equals angle $D A B$ : the exterior angle of a cyclic quadrilateral equals the interior opposite angle.

Answer: (a) $x = 35$ (b) $82 °$ (c) $115 °$

Example 4. $P A$ and $P B$ are tangents to a circle with centre $O$ , touching it at $A$ and $B$ . Angle $A P B = 48 °$ . $C$ is a point on the major arc $A B$ . Find, giving reasons, (a) angle $A O B$ (b) angle $O A B$ (c) angle $P A B$ (d) angle $A C B$ .

Show the working

(a) Angle $O A P$ = angle $O B P = 90 °$ (angle between tangent and radius $= 90 °$ ). Then use the angle sum of quadrilateral $O A P B$ .
$∠ A O B = 360^{\circ} - 90^{\circ} - 90^{\circ} - 48^{\circ} = 132^{\circ}$
graph diagram — not drawn yet
A circle with centre O(0, 0) and radius 3, an external point P(7.38, 0) on the right, and tangents PA and PB touching the circle at A(1.22, 2.74) and B(1.22, -2.74). Radii OA and OB, chord AB, and a point C on the major arc at (-3, 0) with chords CA and CB drawn. Angle APB marked 48°.
Quadrilateral OAPB has right angles at A and B. C is in the major segment of the chord AB.
(b) $O A = O B$ (radii), so triangle $O A B$ is isosceles and its base angles are equal.
$∠ O A B = \frac{180^{\circ} - 132^{\circ}}{2} = 24^{\circ}$
(c) Angle $P A B$ is what is left of the right angle $O A P$ after angle $O A B$ : $90 ° - 24 ° = 66 °$ .
(d) Angle $A C B$ stands on the minor arc $A B$ , the same arc as angle $A O B$ , so angle $A C B = 132 ° \div 2 = 66 °$ (angle at the centre is twice the angle at the circumference).
Check with the alternate segment theorem. Angle $P A B = 66 °$ is between the tangent $P A$ and the chord $A B$ , and $C$ is in the other segment, so angle $A C B$ must also be $66 °$ . The two routes agree.

Answer: (a) $132 °$ (b) $24 °$ (c) $66 °$ (d) $66 °$

Where people go wrong

Mistake: Doubling or halving between an angle at the centre and an angle at the circumference that stand on different arcs.

Instead: Check that the arms of both angles end at the same two points, and that the point on the circumference is on the opposite arc. If that point is on the minor arc, pair its angle with the reflex angle at the centre.

The property links an angle at the circumference to the angle at the centre on the same arc. With $P$ on the minor arc and angle $A P B = 115 °$ , the matching angle at the centre is reflex angle $A O B = 230 °$ . Halving the $130 °$ angle instead gives $65 °$ , which is the angle at a point on the major arc, not at $P$ .

Mistake: Using 'opposite angles sum to $180 °$ ' in a quadrilateral that has a corner at the centre $O$ .

Instead: Use the cyclic quadrilateral property only when all four corners are on the circle. For a shape like $O A P B$ or $O A C B$ , use the angle sum of a quadrilateral or the angle at the centre.

The centre is not on the circle, so the property does not apply. In $O A C B$ with $C$ on the major arc and angle $A C B = 58 °$ , angle $A O B$ is $2 \times 58 ° = 116 °$ , not $180 ° - 58 ° = 122 °$ .

Mistake: Taking two angles on the same chord to be equal when they are on opposite sides of it.

Instead: They are equal only in the same segment. On opposite sides of the chord, the two points and the ends of the chord make a cyclic quadrilateral, so the two angles add up to $180 °$ .

Angles in the same segment are equal because both are half of the same angle at the centre. A point on the other side stands on the other arc: with angle $A C B = 70 °$ , a point $E$ across the chord gives angle $A E B = 110 °$ .

Mistake: In the alternate segment theorem, matching the tangent–chord angle with an angle in the segment on the same side of the chord.

Instead: The equal angle is in the alternate segment: on the other side of the chord from the tangent angle.

A point on the same side as the tangent angle gives $180 °$ minus that angle. With a tangent–chord angle of $60 °$ , an angle at a point on that same side is $120 °$ , not $60 °$ .

Mistake: Writing $90 °$ (angle in a semicircle) for a chord that does not pass through the centre.

Instead: Write $90 °$ only when the chord is a diameter: $O$ is on it, or the question says it is a diameter.

The $90 °$ comes from the straight angle of $180 °$ at the centre. A chord that misses $O$ subtends an angle other than $180 °$ at the centre, so the angle at the circumference is not $90 °$ .

Check yourself

$A$ , $B$ and $C$ lie on a circle with centre $O$ , with $C$ on the major arc $A B$ . Angle $A O B = 146 °$ . Find angle $A C B$ , giving a reason.
Answer
$73 °$ (angle at the centre is twice the angle at the circumference).
$P$ is on the minor arc $A B$ of a circle with centre $O$ , and angle $A P B = 115 °$ . Find the reflex angle $A O B$ , and the angle $A O B$ that is less than $180 °$ .
Answer
Reflex angle $A O B = 2 \times 115 ° = 230 °$ (angle at the centre is twice the angle at the circumference), so angle $A O B = 360 ° - 230 ° = 130 °$ .
$A B$ is a diameter of a circle and $C$ is another point on the circle. Angle $C A B = 38 °$ . Find angle $A B C$ .
Answer
$52 °$ . Angle $A C B = 90 °$ (angle in a semicircle), so angle $A B C = 180 ° - 90 ° - 38 ° = 52 °$ .
$D$ and $E$ are points on a circle, on opposite sides of the chord $A B$ . Angle $A D B = 41 °$ . Find angle $A E B$ .
Answer
$139 °$ . $A D B E$ is a cyclic quadrilateral, so angle $A E B = 180 ° - 41 ° = 139 °$ (opposite angles of a cyclic quadrilateral sum to $180 °$ ). It is not $41 °$ , because $E$ is not in the same segment as $D$ .
$P T$ is a tangent to a circle with centre $O$ , touching it at $T$ . Angle $O P T = 31 °$ . Find angle $T O P$ .
Answer
$59 °$ . Angle $O T P = 90 °$ (angle between tangent and radius $= 90 °$ ), so angle $T O P = 180 ° - 90 ° - 31 ° = 59 °$ .
The tangent at $A$ makes an angle of $72 °$ with the chord $A B$ . $C$ is a point on the circle on the other side of $A B$ from that angle. Find angle $A C B$ , giving a reason.
Answer
$72 °$ (alternate segment theorem).
$P Q R S$ is a cyclic quadrilateral with angle $P = 2 y °$ and angle $R = (y + 30) °$ . Find $y$ .
Answer
$P$ and $R$ are opposite, so $2 y + y + 30 = 180$ , giving $3 y = 150$ and $y = 50$ . Check: angle $P = 100 °$ and angle $R = 80 °$ , which sum to $180 °$ .
$O$ is the centre of a circle and $A$ , $B$ , $C$ are points on it. A student writes: 'In quadrilateral $O A B C$ , angle $A O C$ + angle $A B C = 180 °$ , because opposite angles of a cyclic quadrilateral sum to $180 °$ .' Explain what is wrong.
Answer
$O A B C$ is not a cyclic quadrilateral of this circle, because $O$ is the centre and is not on the circle. The opposite-angle property needs all four vertices on the circle. The property that links them is the angle at the centre: the angle at $O$ that stands on the same arc as angle $A B C$ is twice angle $A B C$ . The angle of quadrilateral $O A B C$ at $O$ is the other angle $A O C$ , the one on the arc through $B$ , so it equals $360 ° - 2 \times$ angle $A B C$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Points $A$ , $B$ , $C$ and $D$ lie on a circle with centre $O$ , in that order around the circle. $A C$ is a diameter. Angle $B A C = 28 °$ and angle $C A D = 41 °$ . The diagram is not drawn to scale.

(a) Write down the size of angle $A B C$ . Give a reason for your answer. 2 marks No calculator

Mark scheme

B1 90
B1 angle in a semicircle [= 90°] oe

(b) Find angle $B D C$ . Give a reason for your answer. 2 marks No calculator

Mark scheme

B1 28
B1 angles in the same segment [are equal] oe

(c) Find angle $B C D$ . 2 marks No calculator

Mark scheme

M1 180 − (28 + 41) oe, or 62 + 49 from angles BCA and ACD
A1 111 cao

(d) Find angle $A O D$ . 2 marks No calculator

Mark scheme

M1 180 − 2 × 41 (triangle AOD isosceles, OA = OD), or 2 × 49 using angle ACD = 49 oe
A1 98 cao

Question 2. $T A S$ is a straight line that is the tangent to a circle at $A$ . $O$ is the centre of the circle. $B$ and $C$ are points on the circle, with $C$ on the major arc $A B$ . Angle $T A B = 64 °$ . The diagram is not drawn to scale.

(a) Find angle $A C B$ . Give a reason for your answer. 2 marks No calculator

Mark scheme

B1 64
B1 alternate segment theorem oe

(b) Find angle $O A B$ . Give a reason for your answer. 2 marks No calculator

Mark scheme

B1 26
B1 angle between tangent and radius = 90° oe

(c) Find angle $A O B$ . 2 marks No calculator

Mark scheme

M1 180 − 2 × their 26, or 2 × their 64
A1 128 cao

(d) $D$ is a point on the minor arc $A B$ . Show that angle $A D B = 116 °$ . 2 marks No calculator

Mark scheme

B1 180 − 64 [= 116] seen, or (360 − 128) ÷ 2 [= 116] seen oe
B1 reason: opposite angles of a cyclic quadrilateral (ADBC) sum to 180° oe, or angle SAB = 116° with alternate segment theorem, or reflex angle AOB = 232° with angle at the centre is twice the angle at the circumference

Question 3. $P Q R S$ is a cyclic quadrilateral. Angle $S P Q = (4 x - 7) °$ and angle $Q R S = (2 x + 25) °$ . The side $S R$ is extended beyond $R$ to the point $U$ .

(a) Find the value of $x$ . 2 marks No calculator

Mark scheme

M1 (4x − 7) + (2x + 25) = 180 oe
A1 x = 27

(b) Calculate angle $Q R U$ . Give a reason for your answer. 2 marks No calculator

Mark scheme

B1 101, FT 180 − (2 × their x + 25)
B1 angles on a straight line with angle QRS = 79°, or the exterior angle of a cyclic quadrilateral equals the interior opposite angle SPQ oe

How this is examined

Chaining several circle properties to reach an angle, with a reason at every step: A diagram of points on a circle with centre $O$ , sometimes with a tangent, followed by a run of Find parts worth 1 or 2 marks each, some of them ending 'Give a reason for your answer' In each part, the correct angle; and for each reason mark, the property in the syllabus words, such as 'angles in the same segment are equal' or 'angle between tangent and radius = 90°'. The syllabus expects the listed properties as reasons, so 'circle theorem' or 'from the diagram' does not state one.
Turning a tangent into a right angle, then using the isosceles triangle made by two radii: Find parts with a tangent at $A$ and a chord $A B$ , asking for angle $O A B$ and then angle $A O B$ , about 1 to 2 marks each Angle $O A T = 90 °$ with the reason 'angle between tangent and radius = 90°', then $90 °$ minus the tangent–chord angle for angle $O A B$ ; for angle $A O B$ , the working $180 ° - 2 \times$ angle $O A B$ with ' $O A = O B$ (radii)' or 'isosceles triangle'.
A cyclic quadrilateral with two opposite angles given in terms of $x$ : A Find $x$ part with two opposite angles written as expressions, then a part that uses the value of $x$ to find an angle The two opposite angles added and set equal to $180$ , e.g. $(4 x - 7) + (2 x + 25) = 180$ , is the method; the value of $x$ is the answer to that part; then substitute back for the angle. Adding two neighbouring angles to make $180$ is not the property.
The angle at the centre when the point on the circumference is on the minor arc: A 2-mark Find angle $A O B$ or reflex angle $A O B$ , with the circumference point between $A$ and $B$ on the short side Doubling to the reflex angle ( $2 \times 115 ° = 230 °$ ), then $360 ° - 230 ° = 130 °$ if the angle less than $180 °$ is asked for, with the reason 'angle at the centre is twice the angle at the circumference'.
Finding the angle in the alternate segment: A 2-mark Find part with a tangent and a chord from its point of contact: one mark for the angle, one for the reason The angle at a point on the other side of the chord, equal to the tangent–chord angle, with the reason 'alternate segment theorem'. A point on the same side as the tangent angle gives $180 °$ minus it.
The right angle at the circumference from a diameter: A 2-mark Write down or Find part for an angle whose arms reach the two ends of a diameter $90 °$ with the reason 'angle in a semicircle'; that right angle then feeds the angle sum of the triangle for the next angle.

Words to know

Chord: A straight line joining two points on a circle. A diameter is a chord through the centre.
Tangent: A straight line that touches a circle at exactly one point, the point of contact.
Segment: One of the two regions a chord cuts a circle into: the smaller is the minor segment, the larger the major segment.
Arc: Part of the circumference. The ends of a chord split it into a minor arc and a major arc.
Subtended: An angle is subtended by a chord or arc when its two arms run to the two ends of that chord or arc.
Cyclic quadrilateral: A quadrilateral with all four vertices on a circle.
Alternate segment: For a tangent and a chord from its point of contact, the segment on the other side of the chord from the angle between them.
Supplementary: Two angles that add up to $180 °$ .
Reflex angle: An angle greater than $180 °$ and less than $360 °$ .

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