Moderate -0.3 This is a standard circle geometry question requiring completing the square (routine A-level technique), reading off centre coordinates, finding intersection points by substitution, and using basic trigonometry with Pythagoras. All parts follow predictable patterns with no novel insight required, making it slightly easier than average but not trivial due to the multi-step nature and final trigonometric calculation.
Figure 1 below shows a circle.
**Figure 1**
\includegraphics{figure_9}
The centre of the circle is \(P\) and the circle intersects the \(y\)-axis at \(Q\) as shown in Figure 1.
The equation of the circle is
$$x^2 + y^2 = 12y - 8x - 27$$
\begin{enumerate}[label=(\alph*)]
\item Express the equation of the circle in the form
$$(x - a)^2 + (y - b)^2 = k$$
where \(a\), \(b\) and \(k\) are constants to be found.
[3 marks]
\item State the coordinates of \(P\)
[1 mark]
\item Find the \(y\)-coordinate of \(Q\)
[2 marks]
\item The line segment \(QR\) is a tangent to the circle as shown in Figure 2 below.
**Figure 2**
\includegraphics{figure_9d}
The point \(R\) has coordinates \((9, -3)\).
Find the angle \(QPR\)
Give your answer in radians to three significant figures.
[3 marks]
Question 9:
--- 9(a) ---
9(a) | Obtains ( x4 )2 ...or ( y6 )2 ... | 1.1a | M1 | x 2 + y 2 = 1 2 y − 8 x − 2 7
x 2 + 8 x + y 2 − 1 2 y = − 2 7
( x + 4 2 ) − 1 6 + ( y − 6 ) 2 − 3 6 = − 2 7
( x + 4 2 ) + ( y − 6 ) 2 = 2 5
Obtains ( x + 4 ) 2 ... + ( y − 6 ) 2 ... | 1.1a | M1
Obtains ( x + 4 ) 2 + ( y − 6 ) 2 = 2 5
Accept 52 for 25
Condone ( x − − 4 ) 2 . . . throughout | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | States (–4, 6)
Follow through their squared
brackets from part 9(a) | 1.2 | B1F | (–4, 6)
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c) ---
9(c) | Substitutes x = 0 correctly into
the given or their equation of
circle to form a quadratic
equation in y
or
uses their right-angled triangle
correctly to find vertical distance
of Q above P
PI by y = 9 or (0, 9)
May be seen on diagram | 3.1a | M1 | 4 2 + ( y − 6 ) 2 = 2 5
y 2 − 1 2 + y 2 7 = 0
y = 3 o r y = 9
y = 9
Obtains 9
Accept (0, 9)
Must come from centre (–4, 6) | 2.2a | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 9(d) ---
9(d) | Uses distance formula to find
QR or PR
PI by their 15 for QR or their
5 10for PR OE
PI by correct answer or
AWFW [71.5, 72]° | 1.1a | M1 | PR= ( 9−−4 )2 +(−3−6 )2
=5 10
5
cosQPR=
5 10
QPR=1.25
Substitutes all their relevant
lengths correctly into an
appropriate trigonometric
equation
PI by correct answer or
AWFW [71.5, 72]° | 3.1a | M1
Obtains AWFW [1.249, 1.25] | 1.1b | A1
Subtotal | 3
Question 9 Total | 9
Q | Marking instructions | AO | Marks | Typical solution
Figure 1 below shows a circle.
**Figure 1**
\includegraphics{figure_9}
The centre of the circle is $P$ and the circle intersects the $y$-axis at $Q$ as shown in Figure 1.
The equation of the circle is
$$x^2 + y^2 = 12y - 8x - 27$$
\begin{enumerate}[label=(\alph*)]
\item Express the equation of the circle in the form
$$(x - a)^2 + (y - b)^2 = k$$
where $a$, $b$ and $k$ are constants to be found.
[3 marks]
\item State the coordinates of $P$
[1 mark]
\item Find the $y$-coordinate of $Q$
[2 marks]
\item The line segment $QR$ is a tangent to the circle as shown in Figure 2 below.
**Figure 2**
\includegraphics{figure_9d}
The point $R$ has coordinates $(9, -3)$.
Find the angle $QPR$
Give your answer in radians to three significant figures.
[3 marks]
</end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2024 Q9 [9]}}