| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.3 Part (a) is a routine completion of the square exercise to find the centre. Part (b) requires finding the gradient of the radius, then the perpendicular gradient for the tangent, and writing the equation—all standard techniques for AS level. The 'hence' structure guides students through the method, making this slightly easier than average but still requiring multiple steps and careful algebra. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03f Circle properties: angles, chords, tangents |
| Answer | Marks |
|---|---|
| 6(a) | Completes a square correctly once |
| Answer | Marks | Guidance |
|---|---|---|
| centre) | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains correct centre | 1.1b | A1 |
| Subtotal | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | Finds correct gradient from ‘their’ | |
| centre from (a) to the point (1, 10) | 1.1b | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| property for ‘their’ gradient | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| for M1 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑎𝑎𝑑𝑑+𝑏𝑏𝑑𝑑+𝑐𝑐 = 0 | 1.1b | A1F |
| Subtotal | 4 | |
| Question Total | 6 | |
| Q | Marking Instructions | AO |
Question 6:
--- 6(a) ---
6(a) | Completes a square correctly once
OE
(PI by one correct coordinate of the
centre) | 1.1a | M1 | x2 +10x + 25 + y2 – 4y + 4 – 100 = 0
(x + 5)2 + (y – 2)2 = 100
Centre = (–5, 2)
Obtains correct centre | 1.1b | A1
Subtotal | 2
--- 6(b) ---
6(b) | Finds correct gradient from ‘their’
centre from (a) to the point (1, 10) | 1.1b | B1F | Gradient from (–5, 2) to (1, 10) is
8
6
So perpendicular gradient is –
3
y – 10 = – (x – 1)
4
3
4y – 40 = –43 x + 3
3x + 4y – 43 = 0
Uses perpendicular gradient
property for ‘their’ gradient | 1.1a | M1
Finds ‘their’ equation of the line
based on ‘their’ perpendicular
gradient.
Finding c = 10.75 (OE) is sufficient
for M1 | 1.1a | M1
Rearranges “their” equation of the
line into form
(FT ‘their’ centre of the circle only)
𝑎𝑎𝑑𝑑+𝑏𝑏𝑑𝑑+𝑐𝑐 = 0 | 1.1b | A1F
Subtotal | 4
Question Total | 6
Q | Marking Instructions | AO | Marks | Typical SolutionA circle has equation
$$x^2 + y^2 + 10x - 4y - 71 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the centre of the circle.
[2 marks]
\item Hence, find the equation of the tangent to the circle at the point $(1, 10)$, giving your answer in the form $ax + by + c = 0$ where $a$, $b$ and $c$ are integers.
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2020 Q6 [6]}}