| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| 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 standard circle geometry question requiring knowledge that radius ⊥ tangent, followed by solving a quadratic and finding a circle equation. While multi-step, each component is routine A-level technique with clear scaffolding through parts (a) and (b). |
| 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 |
|---|---|---|
| Answer | Mark | Guidance |
| tangent \(= -2 \rightarrow\) normal \(= \frac{1}{2}\) | B1 | Deduces gradient between \(P\) and \(Q\) is \(\frac{1}{2}\). May be implied by further work or seen in an equation |
| \(\frac{k^2-2k-k-8}{3--1}\) | M1 | Attempts expression for gradient of \(PQ\) in terms of \(k\). Condone one sign slip. May be seen in \(y_2-y_1=m(x_2-x_1)\) |
| \(\frac{1}{2}=\frac{k^2-2k-k-8}{3--1} \Rightarrow \frac{1}{2}=\frac{k^2}{4}-\frac{2k}{4}-\frac{k}{4}-2\) | dM1 | Sets gradient of \(PQ\) equal to \(\frac{1}{2}\) (or negative reciprocal of \(PQ\) set equal to \(-2\)) and proceeds to a quadratic with fraction split into separate terms. e.g. \(k^2-2k-k-8=2\). Dependent on first M1 |
| \(\Rightarrow k^2-3k-10=0\) * | A1* | Achieves given answer with no errors seen. Must see \(=0\). Must achieve a quadratic which is not the given answer before proceeding |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k=5\) | B1 | Deduces \(k=5\) only. If both roots found then 5 must be selected or negative root rejected. Note B0B1ftM1A0 is possible |
| \(y\) coordinate of \(P\) is \(13\) | B1ft | Follow through on positive value of \(k\) (add 8 to their \(k\)). Check on diagram or implied by further work |
| Attempts \(PQ\left(=\sqrt{(3--1)^2+("25-15-8")^2}\right)\) (or \(PQ^2\)) | M1 | Finds distance \(PQ\) (or \(PQ^2\)) using \(y\) coordinates from chosen value of \(k\). Value of \(k\) must be consistently used to find \(P\) and \(Q\) |
| \((x+1)^2+(y-13)^2=20\) | A1 | Only or equivalent e.g. \(x^2+y^2+2x-26y+150=0\) |
## Question 10:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| tangent $= -2 \rightarrow$ normal $= \frac{1}{2}$ | B1 | Deduces gradient between $P$ and $Q$ is $\frac{1}{2}$. May be implied by further work or seen in an equation |
| $\frac{k^2-2k-k-8}{3--1}$ | M1 | Attempts expression for gradient of $PQ$ in terms of $k$. Condone one sign slip. May be seen in $y_2-y_1=m(x_2-x_1)$ |
| $\frac{1}{2}=\frac{k^2-2k-k-8}{3--1} \Rightarrow \frac{1}{2}=\frac{k^2}{4}-\frac{2k}{4}-\frac{k}{4}-2$ | dM1 | Sets gradient of $PQ$ equal to $\frac{1}{2}$ (or negative reciprocal of $PQ$ set equal to $-2$) and proceeds to a quadratic with fraction split into separate terms. e.g. $k^2-2k-k-8=2$. Dependent on first M1 |
| $\Rightarrow k^2-3k-10=0$ * | A1* | Achieves given answer with no errors seen. Must see $=0$. Must achieve a quadratic which is not the given answer before proceeding |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k=5$ | B1 | Deduces $k=5$ only. If both roots found then 5 must be selected or negative root rejected. Note B0B1ftM1A0 is possible |
| $y$ coordinate of $P$ is $13$ | B1ft | Follow through on positive value of $k$ (add 8 to their $k$). Check on diagram or implied by further work |
| Attempts $PQ\left(=\sqrt{(3--1)^2+("25-15-8")^2}\right)$ (or $PQ^2$) | M1 | Finds distance $PQ$ (or $PQ^2$) using $y$ coordinates from chosen value of $k$. Value of $k$ must be consistently used to find $P$ and $Q$ |
| $(x+1)^2+(y-13)^2=20$ | A1 | Only or equivalent e.g. $x^2+y^2+2x-26y+150=0$ |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{23689deb-7eed-4022-848f-1278231a4056-26_748_764_296_646}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the circle $C$
\begin{itemize}
\item the point $P ( - 1 , k + 8 )$ is the centre of $C$
\item the point $Q \left( 3 , k ^ { 2 } - 2 k \right)$ lies on $C$
\item $k$ is a positive constant
\item the line $l$ is the tangent to $C$ at $Q$
\end{itemize}
Given that the gradient of $l$ is - 2
\begin{enumerate}[label=(\alph*)]
\item show that
$$k ^ { 2 } - 3 k - 10 = 0$$
\item Hence find an equation for $C$
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2024 Q10 [8]}}