| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: finding perpendicular gradient (tangent ⊥ radius), writing a line equation, solving simultaneous equations, and finding circle equation from centre and radius. All steps are routine applications of formulae with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient \(XN = -\dfrac{2}{5}\) | B1 | Cannot be scored from candidates working backwards from given answer |
| Uses \((4,-3)\) with gradient \(\pm\dfrac{2}{5}\) to form equation: \(y+3 = -\dfrac{2}{5}(x-4)\) | M1 | |
| \(2x+5y+7=0\) * | A1* | Sufficient working must be shown between \(y+3=-\dfrac{2}{5}(x-4)\) and \(2x+5y+7=0\) e.g. \(5y+15=-2x+8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solves \(2x+5y+7=0\) and \(y=\dfrac{5}{2}x-\dfrac{55}{2}\) | M1 | Allow values just written down e.g. via calculator |
| \(N=(9,-5)\) | A1 | May be written \(x=9, y=-5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(("9"-4)^2+("-5"+3)^2\) | M1 | Condone slips on lhs; must attempt difference for both components before squaring |
| Correct form of equation \((x-4)^2+(y+3)^2 = ("9"-4)^2+("-5"+3)^2\) | M1 | Condone slip on one component when finding \(r^2\) |
| \((x-4)^2+(y+3)^2=29\) | A1 | Constant terms must be collected; ISW after correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States equation of circle is \(x^2+y^2-8x+6y+c=0\) | M1 | |
| Substitutes \(N=(9,-5)\) into \(x^2+y^2-8x+6y+c=0\) and finds \(c\) | M1 | |
| \(x^2+y^2-8x+6y-4=0\) | A1 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient $XN = -\dfrac{2}{5}$ | B1 | Cannot be scored from candidates working backwards from given answer |
| Uses $(4,-3)$ with gradient $\pm\dfrac{2}{5}$ to form equation: $y+3 = -\dfrac{2}{5}(x-4)$ | M1 | |
| $2x+5y+7=0$ * | A1* | Sufficient working must be shown between $y+3=-\dfrac{2}{5}(x-4)$ and $2x+5y+7=0$ e.g. $5y+15=-2x+8$ |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $2x+5y+7=0$ and $y=\dfrac{5}{2}x-\dfrac{55}{2}$ | M1 | Allow values just written down e.g. via calculator |
| $N=(9,-5)$ | A1 | May be written $x=9, y=-5$ |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $("9"-4)^2+("-5"+3)^2$ | M1 | Condone slips on lhs; must attempt difference for both components before squaring |
| Correct form of equation $(x-4)^2+(y+3)^2 = ("9"-4)^2+("-5"+3)^2$ | M1 | Condone slip on one component when finding $r^2$ |
| $(x-4)^2+(y+3)^2=29$ | A1 | Constant terms must be collected; ISW after correct answer |
**Alternative for (b)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| States equation of circle is $x^2+y^2-8x+6y+c=0$ | M1 | |
| Substitutes $N=(9,-5)$ into $x^2+y^2-8x+6y+c=0$ and finds $c$ | M1 | |
| $x^2+y^2-8x+6y-4=0$ | A1 | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{66abdef1-072e-41eb-a933-dd51a96330ff-16_949_940_246_566}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of
\begin{itemize}
\item the circle $C$ with centre $X ( 4 , - 3 )$
\item the line $l$ with equation $y = \frac { 5 } { 2 } x - \frac { 55 } { 2 }$
\end{itemize}
Given that $l$ is the tangent to $C$ at the point $N$,
\begin{enumerate}[label=(\alph*)]
\item show that an equation for the straight line passing through $X$ and $N$ is
$$2 x + 5 y + 7 = 0$$
\item Hence find
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $N$,
\item an equation for $C$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2023 Q7 [8]}}