| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using right angle in semicircle |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear guidance ('show that', 'hence'). Part (a) uses straightforward vector dot product or gradient multiplication. Part (b) exploits the right angle to identify the diameter (a key geometric insight, but explicitly prompted). Part (c) requires finding the point diametrically opposite Q and then the tangent equation. While it involves several steps and coordinate geometry techniques, the question scaffolds the solution path clearly, making it slightly easier than average for A-level. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m_{PQ} = \frac{14+30}{23-15},\ m_{QR} = \frac{-30+26}{15+7} \Rightarrow m_{PQ} \times m_{QR} = \ldots\) or \(PQ^2 = 8^2 + 44^2,\ QR^2 = 22^2 + 4^2,\ PR^2 = 30^2 + 40^2 \Rightarrow PQ^2 + QR^2 = \ldots\) | M1 | Correct strategy to show \(\angle PQR = 90°\); attempts gradient of \(PQ\) and \(QR\) and product, or finds side lengths and attempts Pythagoras |
| \(m_{PQ} \times m_{QR} = \frac{11}{2} \times \left(-\frac{2}{11}\right) = -1 \Rightarrow \angle PQR = 90°\) or \(PQ^2 = 2000,\ QR^2 = 500,\ PR^2 = 2500;\ 2000 + 500 = 2500 \Rightarrow \angle PQR = 90°\) | A1 | Correct proof and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Centre is \((8, -6)\) | B1 | Correct coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = \sqrt{(23-8)^2 + (14+6)^2}\) or \(r = \frac{1}{2}\sqrt{(23+7)^2 + (14+26)^2}\) | M1 | Fully correct method for the radius |
| \(r = 25\) | A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S\) is \((1, 18)\) or \(m_T = \frac{7}{24}\) | B1 | Correct coordinates for \(S\) or correct gradient for the tangent |
| \(m_N = \frac{18+6}{1-8} \Rightarrow m_T = \frac{8-1}{18+6} \Rightarrow y - 18 = \frac{7}{24}(x-1)\) | M1 | Uses a correct straight line method for the tangent using their \(S\) and the negative reciprocal of the radius gradient |
| \(7x - 24y + 425 = 0\) | A1 | Allow any integer multiple |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m_{PQ} = \frac{14+30}{23-15},\ m_{QR} = \frac{-30+26}{15+7} \Rightarrow m_{PQ} \times m_{QR} = \ldots$ **or** $PQ^2 = 8^2 + 44^2,\ QR^2 = 22^2 + 4^2,\ PR^2 = 30^2 + 40^2 \Rightarrow PQ^2 + QR^2 = \ldots$ | M1 | Correct strategy to show $\angle PQR = 90°$; attempts gradient of $PQ$ and $QR$ and product, or finds side lengths and attempts Pythagoras |
| $m_{PQ} \times m_{QR} = \frac{11}{2} \times \left(-\frac{2}{11}\right) = -1 \Rightarrow \angle PQR = 90°$ **or** $PQ^2 = 2000,\ QR^2 = 500,\ PR^2 = 2500;\ 2000 + 500 = 2500 \Rightarrow \angle PQR = 90°$ | A1 | Correct proof and conclusion |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Centre is $(8, -6)$ | B1 | Correct coordinates |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \sqrt{(23-8)^2 + (14+6)^2}$ or $r = \frac{1}{2}\sqrt{(23+7)^2 + (14+26)^2}$ | M1 | Fully correct method for the radius |
| $r = 25$ | A1 | Cao |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S$ is $(1, 18)$ **or** $m_T = \frac{7}{24}$ | B1 | Correct coordinates for $S$ or correct gradient for the tangent |
| $m_N = \frac{18+6}{1-8} \Rightarrow m_T = \frac{8-1}{18+6} \Rightarrow y - 18 = \frac{7}{24}(x-1)$ | M1 | Uses a correct straight line method for the tangent using their $S$ and the negative reciprocal of the radius gradient |
| $7x - 24y + 425 = 0$ | A1 | Allow any integer multiple |
**Total: 8 marks**6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{59c9f675-e7eb-47b9-b233-dfbe1844f792-18_579_620_219_667}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The points $P ( 23,14 ) , Q ( 15 , - 30 )$ and $R ( - 7 , - 26 )$ lie on the circle $C$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that angle $P Q R = 90 ^ { \circ }$
\item Hence, or otherwise, find
\begin{enumerate}[label=(\roman*)]
\item the centre of $C$,
\item the radius of $C$.
Given that the point $S$ lies on $C$ such that the distance $Q S$ is greatest,
\end{enumerate}\item find an equation of the tangent to $C$ at $S$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2022 Q6 [8]}}