| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using perpendicular bisectors |
| Difficulty | Moderate -0.3 This is a standard AS-level circle geometry question requiring perpendicular gradients and finding a circle equation from center and radius. Part (a) is routine (gradient of L1 is -1/7, so perpendicular is 7, then point-slope form). Part (b) requires finding where two radii intersect to get the center, then using distance formula. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks |
|---|---|
| 11(a) | Obtains correct gradient of L |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 11(b) | Obtains correct equation for | |
| radius through (0, 3) | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| intersection of the radii | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation of C | 1.1a | M1 |
| Obtains correct equation for C | 2.1 | R1 |
| Subtotal | 4 | |
| Question 11 Total | 7 | |
| Q | Marking instructions | AO |
Question 11:
--- 11(a) ---
11(a) | Obtains correct gradient of L
1
PI by fully rearranged equation
of L
1
41
Condone slip in term
7 | 1.1b | B1 | 1 41
y = - x +
7 7
−1
Gradient of L is
1
7
Gradient of radius is 7
Equation of radius is
(y – 5) = 7(x – 6)
Equation of radius is y = 7x – 37
Uses perpendicular gradients
rule on their gradient of L
1 | 1.1a | M1
Completes reasoned argument
to obtain given equation
AG | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 11(b) ---
11(b) | Obtains correct equation for
radius through (0, 3) | 1.1b | B1 | Gradient of L is 1
2
Gradient of radius is –1
Equation of radius is y = –x + 3
Radii intersect when
7x – 37 = –x + 3
x = 5, y = –2
Distance from (5, –2) to (0, 3) is
52 +52 = √50
Equation of C is
(x – 5)2 + (y + 2)2 = 50
Solves for y = 7x – 37 and their
linear radius equation for L to
2
find the coordinates of their
intersection of the radii | 3.1a | M1
Calculates distance from their
centre to either contact point
PI by √50 or 5√2 or 50 seen in
equation of C | 1.1a | M1
Obtains correct equation for C | 2.1 | R1
Subtotal | 4
Question 11 Total | 7
Q | Marking instructions | AO | Marks | Typical solutionThe line $L_1$ has equation $x + 7y - 41 = 0$
$L_1$ is a tangent to the circle C at the point P(6, 5)
The line $L_2$ has equation $y = x + 3$
$L_2$ is a tangent to the circle C at the point Q(0, 3)
The lines $L_1$ and $L_2$ and the circle C are shown in the diagram below.
\includegraphics{figure_11}
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the radius of the circle through P is $y = 7x - 37$ [3 marks]
\item Find the equation of C [4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2023 Q11 [7]}}