| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Area of region bounded by circle and line |
| Difficulty | Standard +0.3 This is a standard C2 circle question requiring completing the square (routine), finding y-intercepts (substitution), applying cosine rule (given formula), and calculating a segment area (standard formula application). All techniques are textbook exercises with clear scaffolding across 14 marks, making it slightly easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.05b Sine and cosine rules: including ambiguous case1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \((x-4)^2 - 16 + (y-5)^2 - 25 + 16 = 0\) | M1 | |
| \((x-4)^2 + (y-5)^2 = 25\) | ||
| \(\therefore \text{centre } (4, 5), \text{ radius } = 5\) | A2 | |
| (b) \(x = 0\) \(\therefore y^2 - 10y + 16 = 0\) | M1 | |
| \((y-2)(y-8) = 0\) | M1 | |
| \(y = 2, 8\) \(\therefore (0,2), (0,8)\) | A1 | |
| (c) \(6^2 = 5^2 + 5^2 - (2 \times 5 \times 5 \times \cos\theta)\) | M2 A1 | |
| \(\cos\theta = \frac{25 + 25 - 36}{50} = \frac{7}{25}\) | A1 | |
| (d) \(\theta = \cos^{-1}\frac{7}{25} = 1.287\), \(\sin\theta = 0.96\) | M2 A1 | |
| \(\text{area} = \frac{1}{2} \times 5^2 \times \theta - \frac{1}{2} \times 5^2 \times \sin\theta = \frac{25}{2}(1.287 - 0.96)\) | M2 A1 | |
| \(= 4.09 (3\text{sf})\) | A1 | (14) |
**(a)** $(x-4)^2 - 16 + (y-5)^2 - 25 + 16 = 0$ | M1 |
$(x-4)^2 + (y-5)^2 = 25$ | |
$\therefore \text{centre } (4, 5), \text{ radius } = 5$ | A2 |
**(b)** $x = 0$ $\therefore y^2 - 10y + 16 = 0$ | M1 |
$(y-2)(y-8) = 0$ | M1 |
$y = 2, 8$ $\therefore (0,2), (0,8)$ | A1 |
**(c)** $6^2 = 5^2 + 5^2 - (2 \times 5 \times 5 \times \cos\theta)$ | M2 A1 |
$\cos\theta = \frac{25 + 25 - 36}{50} = \frac{7}{25}$ | A1 |
**(d)** $\theta = \cos^{-1}\frac{7}{25} = 1.287$, $\sin\theta = 0.96$ | M2 A1 |
$\text{area} = \frac{1}{2} \times 5^2 \times \theta - \frac{1}{2} \times 5^2 \times \sin\theta = \frac{25}{2}(1.287 - 0.96)$ | M2 A1 |
$= 4.09 (3\text{sf})$ | A1 | (14)
**Total: (75)**\includegraphics{figure_3}
Figure 3 shows the circle $C$ with equation
$$x^2 + y^2 - 8x - 10y + 16 = 0.$$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre and the radius of $C$. [3]
\end{enumerate}
$C$ crosses the $y$-axis at the points $P$ and $Q$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the coordinates of $P$ and $Q$. [3]
\end{enumerate}
The chord $PQ$ subtends an angle of $\theta$ at the centre of $C$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Using the cosine rule, show that $\cos \theta = \frac{7}{25}$. [4]
\item Find the area of the shaded minor segment bounded by $C$ and the chord $PQ$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [14]}}