| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Circles |
| Type | Area of region bounded by circle and line |
| Difficulty | Standard +0.3 This question requires completing the square to find centre and radius (standard technique), then using perpendicular tangent properties and coordinate geometry to find point D, followed by a triangle area calculation. While multi-step, each component uses routine A-level methods without requiring novel insight or particularly challenging problem-solving. |
| 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 circle |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)^2 - 9 + \left(y + \dfrac{9}{2}\right)^2 - \dfrac{81}{4} + 19 = 0\) | M1 (AO 1.1) | \(\left(x \pm 3\right)^2 \pm \ldots + \left(y \pm \dfrac{9}{2}\right)^2 \pm \ldots + 19 = 0\) |
| \(C\!\left(3, -\dfrac{9}{2}\right)\) | A1 (AO 1.1) | cao |
| Radius is \(\dfrac{\sqrt{41}}{2}\) | A1 (AO 1.1) | cao (oe) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \dfrac{11}{\frac{55}{4}}x - 11 \left(\Rightarrow y = \dfrac{4}{5}x - 11\right)\) | B1 (AO 2.1) | Equation of line \(AB\) (any equivalent form) – allow unsimplified |
| \(x^2 + \left(\dfrac{4}{5}x - 11\right)^2 - 6x + 9\!\left(\dfrac{4}{5}x - 11\right) + 19 = 0\) | M1* (AO 3.1a) | Substitute equation of line into equation of circle; or \(m_{CD} = -\dfrac{5}{4}\) (use of \(m_1 m_2 = -1\) with their gradient of \(AB\)); equation of \(CD\) is \(y + \dfrac{9}{2} = -\dfrac{5}{4}(x-3)\) (using their \(C\) from (a)) |
| \(\dfrac{41}{25}x^2 - \dfrac{82}{5}x + 41 = 0\) | M1dep* (AO 1.1) | Simplify to three-term quadratic in \(x\) (or \(y\)) |
| \(x\)-coordinate of \(D\) is \(5\) (or \(y\)-coordinate of \(D\) is \(-7\)) | A1 (AO 1.1) | BC |
| Area of \(OBD = \dfrac{1}{2}(11)({'}5{'})\) | M1 (AO 3.2a) | \(\dfrac{1}{2}(11)(x\text{-coordinate of }D)\) or other complete method; dependent on both previous M marks |
| \(= 27.5\) | A1 (AO 1.1) |
## Question 7(a):
$x^2 + y^2 - 6x + 9y + 19 = 0$
$(x-3)^2 - 9 + \left(y + \dfrac{9}{2}\right)^2 - \dfrac{81}{4} + 19 = 0$ | M1 (AO 1.1) | $\left(x \pm 3\right)^2 \pm \ldots + \left(y \pm \dfrac{9}{2}\right)^2 \pm \ldots + 19 = 0$
$C\!\left(3, -\dfrac{9}{2}\right)$ | A1 (AO 1.1) | cao
Radius is $\dfrac{\sqrt{41}}{2}$ | A1 (AO 1.1) | cao (oe)
**Total: [3]**
## Question 7(b):
$y = \dfrac{11}{\frac{55}{4}}x - 11 \left(\Rightarrow y = \dfrac{4}{5}x - 11\right)$ | B1 (AO 2.1) | Equation of line $AB$ (any equivalent form) – allow unsimplified
$x^2 + \left(\dfrac{4}{5}x - 11\right)^2 - 6x + 9\!\left(\dfrac{4}{5}x - 11\right) + 19 = 0$ | M1* (AO 3.1a) | Substitute equation of line into equation of circle; or $m_{CD} = -\dfrac{5}{4}$ (use of $m_1 m_2 = -1$ with their gradient of $AB$); equation of $CD$ is $y + \dfrac{9}{2} = -\dfrac{5}{4}(x-3)$ (using their $C$ from **(a)**)
$\dfrac{41}{25}x^2 - \dfrac{82}{5}x + 41 = 0$ | M1dep* (AO 1.1) | Simplify to three-term quadratic in $x$ (or $y$)
$x$-coordinate of $D$ is $5$ (or $y$-coordinate of $D$ is $-7$) | A1 (AO 1.1) | BC
Area of $OBD = \dfrac{1}{2}(11)({'}5{'})$ | M1 (AO 3.2a) | $\dfrac{1}{2}(11)(x\text{-coordinate of }D)$ or other complete method; dependent on both previous M marks
$= 27.5$ | A1 (AO 1.1) |
**Total: [6]**7\\
\includegraphics[max width=\textwidth, alt={}, center]{31b0d5b6-1593-489b-bbcd-486e7c96ff18-05_848_1049_260_242}
The diagram shows the circle with equation $x ^ { 2 } + y ^ { 2 } - 6 x + 9 y + 19 = 0$ and centre $C$.
\begin{enumerate}[label=(\alph*)]
\item Find the following.
\begin{itemize}
\item The coordinates of $C$.
\item The exact radius of the circle.
\end{itemize}
The tangent to the circle at $D$ meets the $x$-axis at the point $A \left( \frac { 55 } { 4 } , 0 \right)$ and the $y$-axis at the point $B ( 0 , - 11 )$.
\item Determine the area of triangle $O B D$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q7 [9]}}