Question 9 - A-Level Maths

AQA Paper 1 2023 June — Question 9 9 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2023
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circles
Type	Circle through three points using perpendicular bisectors
Difficulty	Moderate -0.3 This is a multi-part question requiring standard techniques: midpoint formula, perpendicular bisector (negative reciprocal gradient), and finding a circle's center as the intersection of two lines. While it has multiple steps (7+ marks total), each individual technique is routine A-level content with no novel problem-solving required. Slightly easier than average due to straightforward application of formulas.
Spec	1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 1.03b Straight lines: parallel and perpendicular relationships 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 1.03e Complete the square: find centre and radius of circle

9 The points $P$ and $Q$ have coordinates ( $- 6,15$ ) and (12, 19) respectively. 9

1. Find the coordinates of the midpoint of $P Q$ 9
  1. (ii) Find the equation of the perpendicular bisector of $P Q$ Give your answer in the form $a x + b y = c$ where $a , b$ and $c$ are integers.
    [0pt] [4 marks]
    9
2. 1. A circle passes through the points $P$ and $Q$ The centre of the circle lies on the line with equation $2 x - 5 y = - 30$ Find the equation of the circle. 9
3. (ii) The circle intersects the coordinate axes at $n$ points.
  State the value of $n$ $$y = \sin x ^ { \circ }$$ for $- 360 \leq x \leq 360$ is shown below. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-12_613_1552_532_246}

Show mark scheme Show mark scheme source

Question 9:

9(a)(i):

Answer	Marks	Guidance
$(3, 17)$	B1	Condone position vectors, missing brackets or $x=3$ and $y=17$

9(a)(ii):

Answer	Marks	Guidance
$m_{PQ} = \frac{19-15}{12--6} = \frac{4}{18}$	B1	Obtains gradient of PQ; PI correct gradient used in equation of perpendicular bisector
$y - 17 = -\frac{9}{2}(x-3)$	M1	Forms an equation of a line using the negative reciprocal of their gradient or their midpoint
$2y - 34 = -9x + 27$	M1	Forms an equation of a line using the negative reciprocal of their gradient and their midpoint
$9x + 2y = 61$	A1	Obtains $9x + 2y = 61$; OE in required form

9(b)(i):

Answer	Marks	Guidance
Centre $(5, 8)$; $(x-5)^2 + (y-8)^2 = r^2$; $(12-5)^2 + (19-8)^2 = 170$	M1	Solves simultaneously using $9x+2y=61$ from (a)(ii) with $2x-5y=-30$ to obtain centre; PI by $(5,8)$ or $x=5$, $y=8$
$(x-5)^2 + (y-8)^2 = 170$	M1	Uses $P$ or $Q$ and their centre to find radius or radius$^2$
$(x-5)^2 + (y-8)^2 = 170$	A1	ACF e.g. $x^2 - 10x + y^2 - 16y = 81$

9(b)(ii):

Answer	Marks	Guidance
$4$	R1	States 4; must have correct centre and correct radius or radius$^2$

## Question 9:

### 9(a)(i):
$(3, 17)$ | B1 | Condone position vectors, missing brackets or $x=3$ and $y=17$

### 9(a)(ii):
$m_{PQ} = \frac{19-15}{12--6} = \frac{4}{18}$ | B1 | Obtains gradient of PQ; PI correct gradient used in equation of perpendicular bisector

$y - 17 = -\frac{9}{2}(x-3)$ | M1 | Forms an equation of a line using the negative reciprocal of their gradient **or** their midpoint

$2y - 34 = -9x + 27$ | M1 | Forms an equation of a line using the negative reciprocal of their gradient **and** their midpoint

$9x + 2y = 61$ | A1 | Obtains $9x + 2y = 61$; OE in required form

### 9(b)(i):
Centre $(5, 8)$; $(x-5)^2 + (y-8)^2 = r^2$; $(12-5)^2 + (19-8)^2 = 170$ | M1 | Solves simultaneously using $9x+2y=61$ from (a)(ii) with $2x-5y=-30$ to obtain centre; PI by $(5,8)$ or $x=5$, $y=8$

$(x-5)^2 + (y-8)^2 = 170$ | M1 | Uses $P$ or $Q$ and their centre to find radius or radius$^2$

$(x-5)^2 + (y-8)^2 = 170$ | A1 | ACF e.g. $x^2 - 10x + y^2 - 16y = 81$

### 9(b)(ii):
$4$ | R1 | States 4; must have correct centre and correct radius or radius$^2$

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Show LaTeX source

9 The points $P$ and $Q$ have coordinates ( $- 6,15$ ) and (12, 19) respectively.

9
\begin{enumerate}[label=(\alph*)]
\item (i) Find the coordinates of the midpoint of $P Q$

9 (a) (ii) Find the equation of the perpendicular bisector of $P Q$\\
Give your answer in the form $a x + b y = c$ where $a , b$ and $c$ are integers.\\[0pt]
[4 marks]\\

9
\item (i) A circle passes through the points $P$ and $Q$

The centre of the circle lies on the line with equation $2 x - 5 y = - 30$\\
Find the equation of the circle.

9 (b) (ii) The circle intersects the coordinate axes at $n$ points.\\
State the value of $n$

$$y = \sin x ^ { \circ }$$

for $- 360 \leq x \leq 360$ is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-12_613_1552_532_246}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2023 Q9 [9]}}

This paper (16 questions)

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Q1 1 Q2 1 Q3 1 Q4 1 Q5 4 Q6 5 Q7 4 Q8 6 Q9 9 Q10 8 Q11 9 Q12 8 Q13 9 Q14 13 Q15 9 Q16 14

Answer	Marks	Guidance
\(m_{PQ} = \frac{19-15}{12--6} = \frac{4}{18}\)	B1	Obtains gradient of PQ; PI correct gradient used in equation of perpendicular bisector
\(y - 17 = -\frac{9}{2}(x-3)\)	M1	Forms an equation of a line using the negative reciprocal of their gradient or their midpoint
\(2y - 34 = -9x + 27\)	M1	Forms an equation of a line using the negative reciprocal of their gradient and their midpoint
\(9x + 2y = 61\)	A1	Obtains \(9x + 2y = 61\); OE in required form

Answer	Marks	Guidance
Centre \((5, 8)\); \((x-5)^2 + (y-8)^2 = r^2\); \((12-5)^2 + (19-8)^2 = 170\)	M1	Solves simultaneously using \(9x+2y=61\) from (a)(ii) with \(2x-5y=-30\) to obtain centre; PI by \((5,8)\) or \(x=5\), \(y=8\)
\((x-5)^2 + (y-8)^2 = 170\)	M1	Uses \(P\) or \(Q\) and their centre to find radius or radius\(^2\)
\((x-5)^2 + (y-8)^2 = 170\)	A1	ACF e.g. \(x^2 - 10x + y^2 - 16y = 81\)