| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Perpendicular bisector of chord |
| Difficulty | Moderate -0.5 This is a straightforward C2 circle question testing standard properties: perpendicular bisector of a chord passes through the centre. Part (a) requires finding the gradient of AM then the perpendicular gradient (routine), part (b) is simple substitution, and part (c) uses the standard circle equation formula. All steps are mechanical applications of well-practiced techniques with no problem-solving 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^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(AM\): \(\frac{1-(-2)}{3-1} = \frac{3}{2}\) or \(\frac{-3}{-2}\) | B1 | |
| Gradient of \(l = -\frac{2}{3}\) | M1 | Use of \(m_1 m_2 = -1\) or equiv. |
| \(y - 1 = -\frac{2}{3}(x-3)\) or \(\frac{y-1}{x-3} = -\frac{2}{3}\), i.e. \(3y = -2x + 9\) | M1 A1 | Any equiv. form |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 6\): \(3y = -12 + 9 = -3\), \(y = -1\) (or show that for \(y = -1\), \(x = 6\)) | B1 | A conclusion is not required |
| Answer | Marks | Guidance |
|---|---|---|
| \(r^2 = (6-1)^2 + (-1-(-2))^2\) | M1 A1 | Attempt \(r^2\) or \(r\); simplification not required |
| \((x \pm 6)^2 + (y \pm 1)^2 = k\), \(k \neq 0\) | M1 | Value for \(k\) not needed, could be \(r^2\) or \(r\) |
| \((x-6)^2 + (y+1)^2 = 26\) | A1 | Allow \((\sqrt{26})^2\) or other exact equivalents for 26 |
## Question 7:
### Part (a):
| Gradient of $AM$: $\frac{1-(-2)}{3-1} = \frac{3}{2}$ or $\frac{-3}{-2}$ | B1 | |
| Gradient of $l = -\frac{2}{3}$ | M1 | Use of $m_1 m_2 = -1$ or equiv. |
| $y - 1 = -\frac{2}{3}(x-3)$ or $\frac{y-1}{x-3} = -\frac{2}{3}$, i.e. $3y = -2x + 9$ | M1 A1 | Any equiv. form |
### Part (b):
| $x = 6$: $3y = -12 + 9 = -3$, $y = -1$ (or show that for $y = -1$, $x = 6$) | B1 | A conclusion is not required |
### Part (c):
| $r^2 = (6-1)^2 + (-1-(-2))^2$ | M1 A1 | Attempt $r^2$ or $r$; simplification not required |
| $(x \pm 6)^2 + (y \pm 1)^2 = k$, $k \neq 0$ | M1 | Value for $k$ not needed, could be $r^2$ or $r$ |
| $(x-6)^2 + (y+1)^2 = 26$ | A1 | Allow $(\sqrt{26})^2$ or other exact equivalents for 26 |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{22ebc302-765c-4734-b312-b286ccb20be9-09_778_988_223_500}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The points $A$ and $B$ lie on a circle with centre $P$, as shown in Figure 3.\\
The point $A$ has coordinates $( 1 , - 2 )$ and the mid-point $M$ of $A B$ has coordinates $( 3,1 )$. The line $l$ passes through the points $M$ and $P$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l$.
Given that the $x$-coordinate of $P$ is 6 ,
\item use your answer to part (a) to show that the $y$-coordinate of $P$ is - 1 ,
\item find an equation for the circle.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2007 Q7 [9]}}