Circle theorems · past-paper pack

Exam-style questions written to the standard of the published papers. Each is labelled with the tier it is pitched at.

5 questions · 18 marks · about 21 minutes

1 Higher calculator · ~3 min

(a)

$P$, $Q$ and $R$ lie on a circle.
Point $S$ sits inside triangle $PQR$.

Assume that $S$ is the centre of this circle.
Work out the size of angle $x$.

x = °[1]

(b)

In fact the centre of the circle lies on $PS$ but not at $S$.
What does this mean for angle $x$?
Tick one box.

The same as your part (a) answer

Greater than your part (a) answer

Smaller than your part (a) answer

Impossible to tell

[1]

(c)

For a different circle,
   $AB$ is a tangent to the circle at $A$
   $C$ and $D$ lie on the circumference
   $AC = CD$

Simon writes out the method below to find the size of angle $y$.

Angle $ADC = 76^\circ$ (alternate segment theorem)
Therefore $y = 76^\circ$ (angles in an isosceles triangle)

Is he right?
Give a reason for your answer.

[1]

2 Higher non-calculator · AO1 · ~3 min

The diagram shows a cyclic quadrilateral.
$a : b : c \ = \ 8 : 10 : 7$

d = °[3]

3 Higher calculator · AO2 · ~5 min SYNOPTIC

In the circle shown, $E$ is the centre and $AC$ passes through it as a diameter.
$E$ also bisects the chord $BD$, so $E$ is the midpoint of $BD$.

Prove that triangles $ABE$ and $CDE$ are congruent.

[4]

4 Higher calculator · AO2 · ~5 min

$A$, $B$, $C$ and $D$ all lie on the circumference of a circle with centre $O$.
$ADE$ and $BCE$ are both straight lines.

Work out the size of angle $CDE$.
Give a reason for each step of your working.

°[4]

5 Higher calculator · AO2 · ~5 min SYNOPTIC

$O$ is the centre of a circle, and $A$, $B$ are two points on its circumference.
$PA$ and $PB$ are tangents to the circle, drawn from the external point $P$.

Prove that $AP = BP$

[4]

Mark scheme

1 Higher 3 marks

Worked solution below (no per-mark scheme recorded for this item).

Worked solution

(a)

Method You are told to ASSUME $S$ is the centre, so the $108°$ at $S$ is the angle at the centre standing on chord $PR$, and $x$ at $Q$ is the angle at the circumference standing on the SAME chord. The angle at the centre is twice the angle at the circumference, so halve it.

Match the two angles to the same chord $PR$.

angle at the centre $= 108°$ (vertex $S$); angle at the circumference $= x$ (vertex $Q$)

Both marked angles stand on the same arc $PR$: one with its point at the centre $S$, one with its point on the circle at $Q$. That is exactly the pairing the centre theorem needs.

Halve the central angle to get $x$.

$x = \dfrac{108}{2} = 54°$

The angle at the centre is twice the angle at the circumference on the same arc, so the circumference angle is half of the centre angle.

Answer x = 54°

⚠️ Watch out Halve, do not double: $x$ is the SMALLER one, at the circumference. Doubling to $216°$ is the classic slip.

(b)

Method In part (a) you HALVED the angle at $S$. That only works if $S$ really is the centre. The genuine centre lies on $PS$ but nearer to the chord $PR$, and a point nearer a chord sees a WIDER angle: so the true $x$ is bigger.

Recall what part (a) assumed.

assuming $S$ is the centre gave $x = \dfrac{108}{2} = 54°$

That value used the centre theorem with $S$ playing the part of the centre. If $S$ is NOT the centre, the $108°$ is no longer the true angle at the centre on $PR$.

Compare the real centre with $S$.

the real centre is on $PS$ but closer to $PR$ than $S$, so it subtends a WIDER angle on $PR$; half of a wider angle is bigger, so $x$ is greater

As a vertex moves toward a chord, the angle it makes on that chord grows. The true centre is nearer $PR$ than $S$, so the true angle at the centre is more than $108°$, making $x$ more than $54°$.

Answer Greater

⚠️ Watch out The answer is a comparison, not a number: it is GREATER than your part (a) answer. Do not try to compute a new exact value, and do not assume it stays the same just because the picture barely changes.

(c)

Method The alternate segment theorem matches the tangent-chord angle to the angle in the OTHER segment on the SAME chord. Find which angle that really is, then finish with the isosceles triangle.

Apply the alternate segment theorem to the RIGHT angle.

the tangent-chord angle is $76°$ between tangent $AB$ and chord $AD$; its alternate-segment partner is the angle $AD$ makes at $C$, so angle $ACD = 76°$ (NOT angle $ADC$)

The tangent-chord angle pairs with the inscribed angle on the SAME chord from the other segment. Chord $AD$ makes angle $ACD$ at $C$, so that is the equal angle. Simon matched it to $ADC$ instead, which is the error.

Use the isosceles triangle $ACD$ to find $y$.

$$\begin{aligned} 2y &= 180 - 76 && \textcolor{#6b7280}{\small\text{apex } ACD = 76°,\ \text{base angles equal ( } AC = CD \text{ )}} \\ 2y &= 104 \\ y &= 104 \div 2 = 52° \end{aligned}$$

$AC = CD$ makes triangle $ACD$ isosceles, so its base angles $CAD$ and $ADC$ are equal ($= y$). With the apex $ACD = 76°$, the three angles give $2y + 76 = 180$.

Answer No

⚠️ Watch out So Simon is wrong: angle $ADC = 52°$, not $76°$. The alternate segment theorem gives the angle at $C$ (chord $AD$), not the nearest-looking angle at $D$.

2 Higher 3 marks

M1	$\dfrac{180}{8+7}$ or $12$, or $7\times12$ or $84$ (angle c), or $8\times12$ or $96$ (angle a)
M1dep	$10\times12$ or $120$ (angle b), or via opposite angles
A1	$60$

Worked solution

Method In a cyclic quadrilateral, opposite angles add up to $180^\circ$. Angles $a$ and $c$ are an opposite pair, so split $180^\circ$ in the ratio $8 : 7$ to find one "part", scale up to $b$, then take $b$ off $180^\circ$ to reach $d$ (its opposite).

Use the opposite pair $a$ and $c$ to find the size of one ratio part.

$a + c = 180^\circ$, and $a : c = 8 : 7$, so one part $= \dfrac{180}{8 + 7} = \dfrac{180}{15} = 12^\circ$

$a$ and $c$ are opposite corners of the cyclic quadrilateral, so together they make $180^\circ$. The ratio $8 : 7$ has $15$ equal parts, and $180 \div 15 = 12$ gives the value of one part.

Scale the "$b$" share up to find angle $b$.

$b = 10 \times 12 = 120^\circ$

The ratio $a : b : c = 8 : 10 : 7$ uses the same part throughout, so $b$ is $10$ parts: $10 \times 12 = 120$.

Use the OTHER opposite pair, $b$ and $d$, to find $d$.

$d = 180 - 120 = 60^\circ$

$b$ and $d$ are also opposite corners, so $b + d = 180^\circ$. Subtracting the $120^\circ$ you found leaves $d = 60^\circ$.

Answer d = 60°

⚠️ Watch out Pair the OPPOSITE corners, not adjacent ones: $a$ pairs with $c$ and $b$ pairs with $d$. A common slip is adding $a$ and $b$ (which are next to each other) to $180^\circ$ - that is not a cyclic-quadrilateral rule.

3 Higher 4 marks

M1	$AE = CE$ (radii)
M1	angle $AEB$ = angle $CED$ (vertically opposite angles)
M1	$BE = DE$ ($E$ is the midpoint of $BD$, given)
A1	$AE=CE$ and radii AND angle $AEB$=angle $CED$ and (vertically) opposite AND $BE=DE$ and $E$ is the midpoint AND SAS

Worked solution

Method To prove two triangles congruent by SAS, find two pairs of equal sides with the INCLUDED angle between them equal too. Triangle $ABE$ and triangle $CDE$ share the vertex $E$, so look at what the circle and the midpoint give you there.

Show one pair of equal sides: $AE$ and $CE$.

$AE = CE$ (radii of the same circle, centre $E$)

$A$ and $C$ are both on the circle and $E$ is the centre, so $AE$ and $CE$ are both radii, and all radii of one circle are equal.

Show the angle BETWEEN those sides is equal: angle $AEB$ and angle $CED$.

angle $AEB =$ angle $CED$ (vertically opposite angles)

$AC$ and $BD$ are two straight lines crossing at $E$, so angle $AEB$ and angle $CED$ are vertically opposite, and vertically opposite angles are always equal. This is the INCLUDED angle for SAS; it sits between the two sides just shown.

Show the second pair of equal sides: $BE$ and $DE$.

$BE = DE$ ($E$ is the midpoint of $BD$, given)

The question states $E$ is the midpoint of $BD$, so it splits $BD$ into two equal halves, $BE$ and $DE$.

Put the three facts together and name the SAS rule.

$AE = CE$, angle $AEB =$ angle $CED$, $BE = DE$; so triangle $ABE$ is congruent to triangle $CDE$ (SAS).

Two sides ($AE$/$CE$ and $BE$/$DE$) with the angle BETWEEN them (angle $AEB$/angle $CED$) all equal is exactly the SAS condition, so the triangles must be congruent.

Answer triangle ABE is congruent to triangle CDE (SAS)

⚠️ Watch out All three reasons must be named IN FULL: "radii", "vertically opposite angles" and "$E$ is the midpoint of $BD$"; a correct-looking equality with no reason loses the mark. Also write the full three-letter angle names (angle $AEB$, angle $CED$), never just "angle $E$": $E$ has several different angles meeting there.

4 Higher 4 marks

M1	for $BAD = 120 \div 2 \; (= 60)$
M1	e.g. $BCD = 180 - 60 \; (= 120)$ or $ABE = 180 - 60 - 20 \; (= 100)$
A1	for finding $CDE = 100$
C1	(dep on at least M2) for one relevant circle theorem stated, e.g. "the angle at the centre is double the angle at the circumference" or "opposite angles in a cyclic quadrilateral sum to 180"

Worked solution

Method Chase the angle from the centre out to triangle $CDE$, one theorem per step. "Give a reason for each stage" means every new angle is written down WITH the theorem that gives it, quoted in full.

Use the angle at the centre to find $\angle BAD$ at the circumference.

$\angle BAD = 120^\circ \div 2 = 60^\circ$

The angle at the centre is twice the angle at the circumference when both stand on the same arc. $\angle BOD$ (at $O$) and $\angle BAD$ (at $A$) both stand on arc $BD$, so the angle at $A$ is half of $120^\circ$.

Move to the opposite corner of cyclic quadrilateral $ABCD$.

$\angle BCD = 180^\circ - 60^\circ = 120^\circ$

Opposite angles of a cyclic quadrilateral sum to $180^\circ$. All four corners $A$, $B$, $C$, $D$ sit on the circle. $A$ and $C$ are opposite corners, so their two angles add to $180^\circ$.

Step outside the circle using the straight line $BCE$.

$\angle DCE = 180^\circ - 120^\circ = 60^\circ$

Angles on a straight line sum to $180^\circ$: $\angle BCD$ (red, from step 2) and $\angle DCE$ sit side by side on the straight line $BCE$.

Finish inside triangle $CDE$.

$\angle CDE = 180^\circ - 60^\circ - 20^\circ = 100^\circ$

Angles in a triangle sum to $180^\circ$. Triangle $CDE$ already has $60^\circ$ at $C$ (from step 3) and $20^\circ$ at $E$, so the angle at $D$ is what remains.

Answer 100°

⚠️ Watch out Vague reasons lose the communication mark: "angles in a circle" or "opposite angles are equal" score nothing. Quote each theorem precisely, e.g. "the angle at the centre is double the angle at the circumference". And halve $120^\circ$ (the centre angle), not $20^\circ$: the doubling theorem links the CENTRE to the circumference, nothing else.

5 Higher 4 marks

C1	$\angle OAP = \angle OBP = 90^{\circ}$ (angle between a radius and a tangent is $90^{\circ}$)
C1	$OA = OB$ (both radii)
C1	$OP$ is common
C1	(dep on previous 3) complete proof with all reasons given, e.g. triangles $OAP$ and $OBP$ congruent RHS so $AP = BP$

Worked solution

Method Build the two triangles $OAP$ and $OBP$ that share the tangent point $P$, and show they are congruent: congruent triangles have matching sides, so the two tangent lengths must be equal.

Mark the two right angles.

$\angle OAP = \angle OBP = 90^{\circ}$

A tangent meets its radius at a right angle, at BOTH points of contact, so this holds at $A$ and at $B$.

Compare the two triangles' other sides.

$OA = OB$
$OP$ is common to both triangles

$OA$ and $OB$ are both radii of the same circle, so they are equal. $OP$ belongs to both triangles $OAP$ and $OBP$, so it is shared, not just equal.

State the congruence and conclude.

Triangles $OAP$ and $OBP$ are congruent (RHS: right angle, hypotenuse $OP$, side $OA = OB$).
So $AP = BP$.

RHS needs a right angle, the hypotenuse, and one other matching side: all three are shown above. Congruent triangles have every pair of corresponding sides equal, and $AP$, $BP$ are the corresponding sides opposite the equal angles at $O$.

Answer AP = BP

⚠️ Watch out Every reason must be named, not just the equal facts: "$OA = OB$" alone is not a mark unless it is identified as "both radii", and the final line needs the congruence condition (RHS) stated, not just "the triangles are congruent".

Circle theorems · past-paper pack

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