| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 5 |
| Topic | Circles |
| Type | Circle equation from centre and radius |
| Difficulty | Easy -1.8 This is a straightforward circle geometry question requiring only basic recall: finding radius from center to origin using distance formula, writing standard circle equation, then substituting a point to solve for 'a'. The mechanics is routine with no problem-solving insight needed, making it significantly easier than average A-level questions. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| \((x - a)^2 + (y + a)^2 = K\) | B1 | Correct LHS (accept if expanded: \(x^2 + y^2 - 2ax + 2ay + 2a^2\)) |
| \(K = 2a^2\) | B1 | Correct RHS. Allow full marks for any equivalent form, e.g. \(x^2 + y^2 - 2ax + 2ay = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((1 - a)^2 + (-5 + a)^2 = 2a^2\) | M1 | Substitute (1, -5) into their circle equation |
| \(a = \frac{13}{6} \Rightarrow \text{Area} = \pi \times 2(\frac{13}{6})^2\) | M1 | Solve for \(a\) and substitute into \(\pi r^2\) with their \(r^2\) |
| \(= \frac{169}{18}\pi\) | A1 |
## Part (a)
$(x - a)^2 + (y + a)^2 = K$ | B1 | Correct LHS (accept if expanded: $x^2 + y^2 - 2ax + 2ay + 2a^2$)
$K = 2a^2$ | B1 | Correct RHS. Allow full marks for any equivalent form, e.g. $x^2 + y^2 - 2ax + 2ay = 0$
## Part (b)
$(1 - a)^2 + (-5 + a)^2 = 2a^2$ | M1 | Substitute (1, -5) into their circle equation
$a = \frac{13}{6} \Rightarrow \text{Area} = \pi \times 2(\frac{13}{6})^2$ | M1 | Solve for $a$ and substitute into $\pi r^2$ with their $r^2$
$= \frac{169}{18}\pi$ | A1 |\includegraphics{figure_3}
The diagram shows a circle with centre $(a, -a)$ that passes through the origin.
\begin{enumerate}[label=(\alph*)]
\item Write down an equation for the circle in terms of $a$. [2]
\item Given that the point $(1, -5)$ lies on the circle, find the exact area of the circle. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q3 [5]}}