Circle theorems · past-paper pack
Real exam questions, remixed. Every question is labelled with the paper it came from.
1
AQA Jun 2023 Paper 2 (H) · Q20
(a)
$P$, $Q$ and $R$ lie on a circle.
Point $S$ sits inside triangle $PQR$.
Point $S$ sits inside triangle $PQR$.
Assume that $S$ is the centre of this circle.
Work out the size of angle $x$.
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.
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
$AC = CD$
$AB$ 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 $y$.
Give a reason for your answer.
Angle $ADC = 76^\circ$ (alternate segment theorem)
Therefore $y = 76^\circ$ (angles in an isosceles triangle)
Is he right?Therefore $y = 76^\circ$ (angles in an isosceles triangle)
Give a reason for your answer.
[1]
2
AQA Nov 2023 Paper 1 (H) · Q22
The diagram shows a cyclic quadrilateral.
$a : b : c \ = \ 8 : 10 : 7$
$a : b : c \ = \ 8 : 10 : 7$
d = °[3]
3
AQA Nov 2023 Paper 3 (H) · Q19
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 $BD$.
$E$ also bisects the chord $BD$, so $E$ is the midpoint of $BD$.
Prove that triangles $ABE$ and $CDE$ are congruent.
[4]
4
Edexcel Jun 2022 Paper 2 (H) · Q20
$A$, $B$, $C$ and $D$ all lie on the circumference of a circle with centre $O$.
$ADE$ and $BCE$ are both straight lines.
$ADE$ and $BCE$ are both straight lines.
Work out the size of angle $CDE$.
Give a reason for each step of your working.
Give a reason for each step of your working.
°[4]
5
Edexcel Nov 2023 Paper 3 (H) · Q21
$A$, $B$, $C$ and $D$ all lie on the circumference of a circle with centre $O$.
$AC$ is a diameter.
The lines $ADE$ and $BCE$ are straight.
$AC$ is a diameter.
The lines $ADE$ and $BCE$ are straight.
Work out the size of angle $BDC$.
Write down any circle theorems that you use.
Write down any circle theorems that you use.
[4]
Mark scheme
1
AQA Jun 2023 Paper 2 (H) · Q20
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.
1
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.
2
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.
⚠️ 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.
1
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 $PR$.
2
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 $54°$.
⚠️ 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.
1
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.
2
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 $2y + 76 = 180$.
⚠️ 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
AQA Nov 2023 Paper 1 (H) · Q22
| 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).
1
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.
2
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: $10 \times 12 = 120$.
3
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 $d = 60^\circ$.
⚠️ 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
AQA Nov 2023 Paper 3 (H) · Q19
| 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.
1
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.
2
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.
3
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 $DE$.
4
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.
⚠️ 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
Edexcel Jun 2022 Paper 2 (H) · Q20
| 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.
1
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 $120^\circ$.
2
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 $180^\circ$.
3
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 $BCE$.
4
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.
⚠️ 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
Edexcel Nov 2023 Paper 3 (H) · Q21
| M1 | one relevant angle, eg angle $ADC = 90°$ (semicircle) or angle $DBC = 32°$ (same segment) |
| M1 | a second angle, eg angle $DCE = 54°$ or angle $DCB = 126°$ |
| A1 | angle $BDC = 22$ |
| C1 | (dep M1) states a correct circle theorem used, eg "angle in a semicircle is 90°" or "angles in the same segment are equal" |
Worked solution
Method Chase the angle round the figure one theorem at a time. "Write down any circle theorems you use" means every new angle must be justified by a named theorem, quoted in full.
1
Use the diameter to get a right angle at $D$.
$AC$ is a diameter, so $\angle ADC = 90^\circ$.
The angle in a semicircle is $90^\circ$: the diameter $AC$ subtends a right angle at any point on the circle, and $D$ is such a point.
2
Use the straight line $ADE$.
$ADE$ is straight, so $\angle CDE = 180^\circ - 90^\circ = 90^\circ$.
Angles on a straight line sum to $180^\circ$. $\angle ADC$ and $\angle CDE$ sit side by side on the straight line $ADE$, and $\angle ADC = 90^\circ$.
3
Finish triangle $CDE$ to find $\angle DCE$.
$\angle DCE = 180^\circ - 90^\circ - 36^\circ = 54^\circ$.
Angles in a triangle sum to $180^\circ$. Triangle $CDE$ has $90^\circ$ at $D$ and $36^\circ$ at $E$, so the angle at $C$ is what is left.
4
Use the straight line $BCE$ to turn round to $\angle DCB$.
$BCE$ is straight, so $\angle DCB = 180^\circ - 54^\circ = 126^\circ$.
Angles on a straight line sum to $180^\circ$: $\angle DCE$ and $\angle DCB$ are next to each other on the straight line $BCE$.
5
Bring in $\angle DBC$, then finish triangle $BDC$.
$\angle DBC = 32^\circ$ (same segment as $\angle DAC$), so $\angle BDC = 180^\circ - 32^\circ - 126^\circ = 22^\circ$.
Angles in the same segment are equal: $\angle DBC$ and $\angle DAC = 32^\circ$ both stand on arc $DC$. Then angles in triangle $BDC$ sum to $180^\circ$, giving $\angle BDC$.
⚠️ Watch out Vague reasons score nothing: quote each theorem exactly ("angle in a semicircle is $90^\circ$", "angles in the same segment are equal"). The right angle comes from the DIAMETER $AC$, and $\angle DBC$ equals $\angle DAC$ (same segment), not $\angle E$.
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