| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.8 This is a straightforward C2 circle question requiring basic application of two standard results: (1) diameter subtends a right angle, giving a⊥ equation to find a=13, and (2) finding circle equation from diameter endpoints using midpoint as centre. Both are routine textbook exercises with clear signposting and minimal problem-solving required. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(PQ:\ m_1=\frac{10-2}{9-(-3)}\left(=\frac{2}{3}\right)\) and \(QR:\ m_2=\frac{10-4}{9-a}\) | M1 | Must be \(y\)-difference/\(x\)-difference |
| \(m_1 m_2=-1:\ \frac{8}{12}\times\frac{6}{9-a}=-1 \Rightarrow a=13\) | M1 A1 | M1: substitute gradients into product \(=-1\); A1: \(a=13\) with no errors |
| Total: [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Centre at \((5,3)\) | B1 | Can be implied by use in part (b) |
| \(r^2=(10-3)^2+(9-5)^2\) or equiv. | M1 A1 | M1: attempt to find \(r^2\), allow one slip in a bracket |
| \((x-5)^2+(y-3)^2=65\) or \(x^2+y^2-10x-6y-31=0\) | M1 A1 | M1: for \((x\pm5)^2+(y\pm3)^2=k^2\); A1: rhs must be 65 |
| Total: [5] |
## Question 5:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $PQ:\ m_1=\frac{10-2}{9-(-3)}\left(=\frac{2}{3}\right)$ and $QR:\ m_2=\frac{10-4}{9-a}$ | M1 | Must be $y$-difference/$x$-difference |
| $m_1 m_2=-1:\ \frac{8}{12}\times\frac{6}{9-a}=-1 \Rightarrow a=13$ | M1 A1 | M1: substitute gradients into product $=-1$; A1: $a=13$ with no errors |
| **Total: [3]** | | |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Centre at $(5,3)$ | B1 | Can be implied by use in part (b) |
| $r^2=(10-3)^2+(9-5)^2$ or equiv. | M1 A1 | M1: attempt to find $r^2$, allow one slip in a bracket |
| $(x-5)^2+(y-3)^2=65$ **or** $x^2+y^2-10x-6y-31=0$ | M1 A1 | M1: for $(x\pm5)^2+(y\pm3)^2=k^2$; A1: rhs must be 65 |
| **Total: [5]** | | |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{12e54724-64a3-4dc0-b7d5-6ef6cc04124c-06_828_956_244_457}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The points $P ( - 3,2 ) , Q ( 9,10 )$ and $R ( a , 4 )$ lie on the circle $C$, as shown in Figure 2. Given that $P R$ is a diameter of $C$,
\begin{enumerate}[label=(\alph*)]
\item show that $a = 13$,
\item find an equation for $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2009 Q5 [8]}}