| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - find equation |
| Difficulty | Moderate -0.3 This is a standard circle geometry question testing routine techniques: reading centre/radius from equation, finding tangent using perpendicular gradient, and applying Pythagoras to a tangent-radius configuration. Part (a) is trivial recall, part (b) is a standard tangent calculation, and part (c) requires recognizing the right-angled triangle QTC but is a well-practiced technique. Slightly easier than average due to straightforward setup and standard methods throughout. |
| 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 | Guidance |
|---|---|---|
| 11(a)(i) | States correct radius CAO | AO1.2 |
| (a)(ii) | States correct centre CAO | AO1.2 |
| (b) | Finds gradient of the line |
| Answer | Marks | Guidance |
|---|---|---|
| Condone one sign error | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| obtained from ‘their’ CP gradient | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ‘their’ tangent gradient | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| part (a) | AO1.1b | A1F |
| (c) | Identifies QTC as a right-angled | |
| triangle PI | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| FT ‘their’ C found in part (a) | AO1.1b | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| correctly for ‘their’ triangle | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| found in part (a) | AO1.1b | A1F |
| Total | 10 | |
| Q | Marking Instructions | AO |
Question 11:
--- 11(a)(i) ---
11(a)(i) | States correct radius CAO | AO1.2 | B1 | Radius = 5
(a)(ii) | States correct centre CAO | AO1.2 | B1 | C is (7, –2)
(b) | Finds gradient of the line
through the points P and ‘their’
C (as found in part (a))
Condone one sign error | AO3.1a | M1 | −1−(−2) 1
Gradient CP = = −
5−7 2
So tangent gradient = 2
y – (–1) = 2(x – 5)
y = 2x – 11
Correct tangent gradient
obtained from ‘their’ CP gradient | AO3.1a | M1
Uses a correct form for the
equation of a straight line with
correct coordinates of P and
‘their’ tangent gradient | AO1.1a | M1
States correct final answer in
required form (y = mx + c)
FT from ‘their’ C found in
part (a) | AO1.1b | A1F
(c) | Identifies QTC as a right-angled
triangle PI | AO3.1a | M1 | QTC is a right-angled triangle so we
can use Pythagoras
QC2 = (7 – 3)2+ (–2–3)2
42 + 52 = ( 5 )2 + QT 2
QT 2 = 36 so QT = 6
2
Finds QC or QC
FT ‘their’ C found in part (a) | AO1.1b | B1F
Uses Pythagoras’ theorem
correctly for ‘their’ triangle | AO1.1a | M1
Correct evaluation of length of
QT
FT ‘their’ QC and ‘their’ radius
found in part (a) | AO1.1b | A1F
Total | 10
Q | Marking Instructions | AO | Marks | Typical SolutionThe circle with equation $(x - 7)^2 + (y + 2)^2 = 5$ has centre C.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the radius of the circle. [1 mark]
\item Write down the coordinates of C. [1 mark]
\end{enumerate}
\item The point $P(5, -1)$ lies on the circle.
Find the equation of the tangent to the circle at $P$, giving your answer in the form $y = mx + c$ [4 marks]
\item The point Q(3, 3) lies outside the circle and the point T lies on the circle such that QT is a tangent to the circle. Find the length of QT. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 Q11 [10]}}