| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Circles |
| Type | Chord length calculation |
| Difficulty | Standard +0.3 This is a multi-part coordinate geometry question requiring standard techniques: finding circle center, using perpendicular chord properties, calculating distances, and applying triangle area formula. While it has 7 marks for part (a) and requires careful algebraic manipulation, the methods are all standard A-level fare with no novel insights needed. The geometric interpretations in parts (b) and (c) are straightforward once the algebra is complete. Slightly above average due to the multi-step nature and algebraic complexity, but well within typical A-level scope. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01d Proof by contradiction |
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_6}
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.
Show that the area of triangle $ABC$ is $\frac{5}{2}\sqrt{9 - 2a}$. [7]
\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{OCR H240/02 2023 Q6 [10]}}