AQA Paper 1 Specimen — Question 11 8 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeChord length calculation
DifficultyModerate -0.3 Part (a) is a routine completing-the-square exercise to find centre and radius. Part (b) requires applying the perpendicular-from-centre-to-chord theorem and Pythagoras, which is standard circle geometry but involves multiple steps and algebraic manipulation. Overall slightly easier than average due to being a familiar multi-part question with well-signposted techniques.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle

A circle with centre \(C\) has equation \(x^2 + y^2 + 8x - 12y = 12\)
  1. Find the coordinates of \(C\) and the radius of the circle. [3 marks]
  2. The points \(P\) and \(Q\) lie on the circle. The origin is the midpoint of the chord \(PQ\). Show that \(PQ\) has length \(n\sqrt{3}\), where \(n\) is an integer. [5 marks]

Question 11:

AnswerMarks Guidance
11(a)Completes the square twice or
applies standard formulaAO1.1a M1
(x4)2 (y6)2= 64
Centre (–4, 6)
Radius = 8
AnswerMarks Guidance
Obtains correct equationAO1.1b A1
Obtains correct radius and correct
coordinates of C
AnswerMarks Guidance
Follow through ‘their’ equationAO1.1b A1F
(b)Demonstrates a method to find the
length OP or OQ (or their squares),
or the coordinates P or Q using
AnswerMarks Guidance
‘their’ values from part (a)AO3.1a M1
OP2  r2 OC2
= 64 – 52 = 12
PQ = 2OP
2 12 4 3
Uses a circle property that may lead
to a solution, eg radius and chord
meet at right-angles (evidence for
this could be the use of Pythagoras
AnswerMarks Guidance
or perpendicular gradients)AO3.1a M1
Finds OP or OQ or coordinates of P
AnswerMarks Guidance
or Q CAOAO1.1b A1
Obtains length of PQ
Follow through from ‘their’
coordinate of P and Q
(Does not need to be in the required
AnswerMarks Guidance
form)AO1.1b A1F
Sets out a well-constructed
mathematical argument, using
precise statements and correct use
of symbols throughout to show the
correct required result in required
form
Only award if they have a
completely correct solution, which is
clear, easy to follow and contains no
AnswerMarks Guidance
slipsAO2.1 R1
Total8
QMarking Instructions AO
Question 11:
--- 11(a) ---
11(a) | Completes the square twice or
applies standard formula | AO1.1a | M1 | (x4)2(y6)2 163612
(x4)2 (y6)2= 64
Centre (–4, 6)
Radius = 8
Obtains correct equation | AO1.1b | A1
Obtains correct radius and correct
coordinates of C
Follow through ‘their’ equation | AO1.1b | A1F
(b) | Demonstrates a method to find the
length OP or OQ (or their squares),
or the coordinates P or Q using
‘their’ values from part (a) | AO3.1a | M1 | OC2 42 62= 52
OP2  r2 OC2
= 64 – 52 = 12
PQ = 2OP
2 12 4 3
Uses a circle property that may lead
to a solution, eg radius and chord
meet at right-angles (evidence for
this could be the use of Pythagoras
or perpendicular gradients) | AO3.1a | M1
Finds OP or OQ or coordinates of P
or Q CAO | AO1.1b | A1
Obtains length of PQ
Follow through from ‘their’
coordinate of P and Q
(Does not need to be in the required
form) | AO1.1b | A1F
Sets out a well-constructed
mathematical argument, using
precise statements and correct use
of symbols throughout to show the
correct required result in required
form
Only award if they have a
completely correct solution, which is
clear, easy to follow and contains no
slips | AO2.1 | R1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
A circle with centre $C$ has equation $x^2 + y^2 + 8x - 12y = 12$

\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $C$ and the radius of the circle.
[3 marks]
\item The points $P$ and $Q$ lie on the circle.

The origin is the midpoint of the chord $PQ$.

Show that $PQ$ has length $n\sqrt{3}$, where $n$ is an integer.
[5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1  Q11 [8]}}