| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Perpendicular bisector of chord |
| Difficulty | Standard +0.3 Part (a) is a standard coordinate geometry exercise requiring midpoint, gradient, and perpendicular gradient—routine AS-level skills. Part (b) adds a mild problem-solving element (recognizing the perpendicular bisector must pass through the center on the x-axis), but the overall question remains straightforward with well-signposted steps and familiar techniques, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | Obtains midpoint (–2, 3) | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| rule to their gradient | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ISW ACF | 1.1b | A1 |
| Subtotal | 5 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(b) | Uses x-intercept of their |
| Answer | Marks | Guidance |
|---|---|---|
| (x − a)2 + y 2 = r2 | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| find their r | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct answer | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 1.1b | A1 |
| Subtotal | 4 | |
| Question 7 Total | 9 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Obtains midpoint (–2, 3) | 1.1b | B1 | Midpoint of AB is (–2, 3)
5 − 1 1
Gradient of AB = = –
− 8 − 4 3
Gradient of perpendicular to AB = 3
Equation is y – 3 = 3(x + 2)
y = 3x + 9
Finds the gradient of the line AB
Using 5 with 1 divided by
−8 with 4
1
PI by Gradient =
3 | 1.1a | M1
1
Obtains gradient of AB = –
3
OE | 1.1b | A1
Applies perpendicular gradient
rule to their gradient | 1.1a | M1
Obtains correct equation of
perpendicular bisector
ISW ACF | 1.1b | A1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b) ---
7(b) | Uses x-intercept of their
equation to find their centre of
the circle
Or
Forms two equations of the form
(x − a)2 + y 2 = r2 | 3.1a | M1 | When y = 0, x = –3
(–3, 0) to (4, 1) = 7 2 + 1 2
radius = 50
(x + 3)2 + y 2 = 50
Uses the distance formula
between two points at least one
of which must be A(4 , 1) or
B(−8 , 5)
Or
Solves their two equations to
find their a and uses this a to
find their r | 1.1a | M1
Obtains r = 5 0 or r2 = 50
PI by correct answer | 1.1b | A1
Obtains (x + 3)2 + y 2 = 50
OE | 1.1b | A1
Subtotal | 4
Question 7 Total | 9
Q | Marking instructions | AO | Marks | Typical solutionPoint $A$ has coordinates $(4, 1)$ and point $B$ has coordinates $(-8, 5)$
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the perpendicular bisector of $AB$
[5 marks]
\item A circle passes through the points $A$ and $B$
A diameter of the circle lies along the $x$-axis.
Find the equation of the circle.
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2024 Q7 [9]}}