| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle geometry question requiring standard techniques: midpoint formula, circle equation from diameter endpoints, and finding intersection points. Part (b) involves some algebraic manipulation but follows a routine method. Part (c) requires finding x-intercepts and calculating triangle area, all using familiar AS-level methods with no novel insight needed. Slightly easier than average due to its predictable structure and standard techniques. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | Finds correct midpoint of AB | 1.1b |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 6(b) | Calculates length of radius, AC, |
| Answer | Marks | Guidance |
|---|---|---|
| centre. | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| radius or square of the radius. | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone sign error in brackets. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 6(c) | Substitutes y = 0 into the given |
| Answer | Marks | Guidance |
|---|---|---|
| or EM | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains DM or EM = √17 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI | 2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| CAO | 1.1b | A1 |
| Subtotal | 4 | |
| Question 6 Total | 9 | |
| Q | Marking instructions | AO |
Question 6:
--- 6(a) ---
6(a) | Finds correct midpoint of AB | 1.1b | B1 | (4,1)
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b) ---
6(b) | Calculates length of radius, AC,
BC or half AB using ‘their’
centre. | 3.1a | M1 | r = √(4 – 1)2 + (1 – 4)2
=
(x – 4)2 + √(y1 –8 1)2 = 18
2 2
𝑥𝑥 −8𝑥𝑥+16+𝑦𝑦 −2𝑦𝑦+1 = 18
2 2
𝑥𝑥 +𝑦𝑦 −8𝑥𝑥−2𝑦𝑦 = 1
Obtains correct value for the
radius or square of the radius. | 1.1b | A1
Derives circle equation in any
form using their centre and
radius.
Condone sign error in brackets.
Or
Completes the square on given
equation to obtain centre and
radius.
Condone sign error in brackets. | 1.1a | M1
Completes reasoned argument
to obtain equation in given form
Or
Justifies that the centre and
radius obtained from completing
the square on the given
equation corresponds to the
midpoint from part (a) and the
radius from part (b)
AG | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 6(c) ---
6(c) | Substitutes y = 0 into the given
formula and solves the quadratic
equation.
Or
Uses Pythagoras on the triangle
CDM where M(4 , 0) to find DM
or EM | 3.1a | M1 | 2
𝑥𝑥 −8𝑥𝑥−1 = 0
x = 4 + √17 and 4 – √17
DE = 2√17
Area = ½ × 1 × 2√17
=√17
Obtains 2 correct values for x
Condone decimal equivalents
AWRT 8.1 and – 0.1
Or
Obtains DM or EM = √17 | 1.1b | A1
Deduces length of DE as the
difference between their two
values of x
Or
Deduces the length of DE as
twice the length of DM or EM
PI | 2.2a | M1
Obtains √17
CAO | 1.1b | A1
Subtotal | 4
Question 6 Total | 9
Q | Marking instructions | AO | Marks | Typical solution$AB$ is a diameter of a circle where $A$ is $(1, 4)$ and $B$ is $(7, -2)$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the midpoint of $AB$.
[1 mark]
\item Show that the equation of the circle may be written as
$$x^2 + y^2 - 8x - 2y = 1$$
[4 marks]
\item The circle has centre $C$ and crosses the $x$-axis at points $D$ and $E$.
Find the exact area of triangle $DEC$.
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2022 Q6 [9]}}