| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Moderate -0.3 This is a standard P2 circle question requiring knowledge that radius is perpendicular to tangent (giving m=-2 for PN), finding point P by solving simultaneous equations, then using distance formula for radius. All steps are routine applications of well-practiced techniques 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^21.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts line with gradient \(-2\) and point \((4,-1)\): \(y+1=-2(x-4)\) | M1 | Condone a slip on one of the signs of \((4,-1)\). If \(y=mx+c\) form used, must proceed to \(c=\ldots\) |
| \(y=-2x+7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=\frac{1}{2}x\) meets \(y=-2x+7\): \(\frac{1}{2}x=-2x+7 \Rightarrow x=\frac{14}{5}, y=\frac{7}{5}\) | M1 A1 | Expect candidates to proceed to \(x=\ldots\) |
| Attempts \(r^2=\left(4-\frac{14}{5}\right)^2+\left(-1-\frac{7}{5}\right)^2=\frac{36}{5}\) | dM1 A1 | Dependent on previous M1. Attempt to subtract coordinates, square and add. Condone slip on signs. |
| \((x-4)^2+(y+1)^2=\frac{36}{5}\) | A1 | Also accept \((x-4)^2+(y+1)^2=7.2\) or \(5(x-4)^2+5(y+1)^2=36\) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts line with gradient $-2$ and point $(4,-1)$: $y+1=-2(x-4)$ | M1 | Condone a slip on one of the signs of $(4,-1)$. If $y=mx+c$ form used, must proceed to $c=\ldots$ |
| $y=-2x+7$ | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\frac{1}{2}x$ meets $y=-2x+7$: $\frac{1}{2}x=-2x+7 \Rightarrow x=\frac{14}{5}, y=\frac{7}{5}$ | M1 A1 | Expect candidates to proceed to $x=\ldots$ |
| Attempts $r^2=\left(4-\frac{14}{5}\right)^2+\left(-1-\frac{7}{5}\right)^2=\frac{36}{5}$ | dM1 A1 | Dependent on previous M1. Attempt to subtract coordinates, square and add. Condone slip on signs. |
| $(x-4)^2+(y+1)^2=\frac{36}{5}$ | A1 | Also accept $(x-4)^2+(y+1)^2=7.2$ or $5(x-4)^2+5(y+1)^2=36$ |
**Alt (b):** Substitutes $y=\frac{1}{2}x$ into $(x-4)^2+(y+1)^2=r^2$, giving $5x^2-28x+68-4r^2=0$ or $5y^2-14y+17-r^2$; uses discriminant $b^2-4ac=0$ to find $r$ or $r^2$.
**Note:** Candidates who substitute $y=-2x+\text{"7"}$ into $(x-4)^2+(y+1)^2=r^2$ and attempt the discriminant approach will score **0 marks in (b)**
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bfeb1724-9a00-4a36-9606-520395792b2b-16_677_826_258_559}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a circle $C$ with centre $N ( 4 , - 1 )$.
The line $l$ with equation $y = \frac { 1 } { 2 } x$ is a tangent to $C$ at the point $P$.
Find
\begin{enumerate}[label=(\alph*)]
\item the equation of line $P N$ in the form $y = m x + c$, where $m$ and $c$ are constants,
\item the equation of $C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{bfeb1724-9a00-4a36-9606-520395792b2b-16_2256_52_311_1978}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q6 [7]}}