| Exam Board | SPS |
|---|---|
| Module | SPS FM (SPS FM) |
| Year | 2024 |
| Session | October |
| Marks | 9 |
| Topic | Circles |
| Type | Chord length calculation |
| Difficulty | Standard +0.8 This is a multi-step coordinate geometry problem requiring finding the perpendicular from centre to chord, using the midpoint property, calculating distances, and deriving an area formula with algebraic manipulation. While the techniques are standard A-level (perpendicular bisector of chord, distance formula, triangle area), the question requires careful coordination of several concepts and algebraic fluency to reach the given answer, placing it moderately above average difficulty. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.08e Area between curve and x-axis: using definite integrals |
A circle has centre $C$ which lies on the $x$-axis, as shown in the diagram. The line $y = x$ meets the circle at $A$ and $B$. The midpoint of $AB$ is $M$.
\includegraphics{figure_9}
The equation of the circle is $x^2 - 6x + y^2 + a = 0$, where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item In this question you must show detailed reasoning.
Find the $x$-coordinate of $M$ and hence show that the area of triangle $ABC$ is $\frac{3}{2}\sqrt{9 - 2a}$.
[6]
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $a$ when the area of triangle $ABC$ is zero.
[1]
\item Give a geometrical interpretation of the case in part (b)(i).
[1]
\end{enumerate}
\item Give a geometrical interpretation of the case where $a = 5$.
[1]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM 2024 Q9 [9]}}