| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Circles |
| Type | Perpendicular bisector of chord |
| Difficulty | Moderate -0.3 This is a standard C2 circle geometry question requiring perpendicular chord bisector property, finding line equations, and circle equations. The steps are routine: find gradient of AM, use perpendicular gradient for line l, substitute x=6 to find y, then use distance formula for radius. Slightly easier than average due to clear structure and straightforward calculations. |
| 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 |
\includegraphics{figure_3}
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 $AB$ has coordinates $(3, 1)$. The line $l$ passes through the points $M$ and $P$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l$.
[4]
\end{enumerate}
Given that the $x$-coordinate of $P$ is 6,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item use your answer to part (a) to show that the $y$-coordinate of $P$ is $-1$.
[1]
\item find an equation for the circle.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [9]}}