| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Area of region bounded by circle and line |
| Difficulty | Standard +0.3 Part (a) is direct substitution into the circle equation formula (trivial). Part (b) requires recognizing the right angle in a radius-tangent configuration and using inverse trig, which is standard C2 material. Part (c) involves finding a sector area and subtracting a triangle, requiring careful geometric decomposition but using routine formulas. This is a well-structured multi-part question slightly easier than average due to its guided nature and standard techniques. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks |
|---|---|
| (a) \((x - 6)^2 + (y - 4)^2 = ; 3^2\) | B1; B1 (2) |
| (b) Complete method for \(MP\): \(= \sqrt{(12-6)^2 + (6-4)^2}\) | M1 |
| \[= \sqrt{40}\] \((= 6.325)\) | A1 |
| Answer | Marks |
|---|---|
| Complete method for \(\cos\theta\), \(\sin\theta\) or \(\tan\theta\) | M1 |
| e.g. \(\cos\theta = \frac{MT}{MP} = \frac{3}{\text{candidate's } s\sqrt{40}}\) (= 0.4743) (\(\theta = 61.6835°\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = 1.0766\) rad | AG; A1 (4) | |
| (c) Complete method for area \(TMP\); e.g., \(= \frac{1}{2} \times 3 \times \sqrt{40}\sin\theta\) | M1 | |
| \[= \frac{3}{2}\sqrt{31}\] \((= 8.3516...)\) allow awrt 8.35 | A1 | |
| Area (sector) \(MTQ = 0.5 \times 3^2 \times 1.0766\) | M1 | (= 4.8446...) |
| Area \(TPQ =\) candidate's \((8.3516... - 4.8446...)\) | M1 | |
| \[= 3.507\] (awrt) | A1 (5) [11] |
**(a)** $(x - 6)^2 + (y - 4)^2 = ; 3^2$ | B1; B1 (2) |
**(b)** Complete method for $MP$: $= \sqrt{(12-6)^2 + (6-4)^2}$ | M1 |
$$= \sqrt{40}$$ $(= 6.325)$ | A1 |
[These first two marks can be scored if seen as part of solution for (c)]
Complete method for $\cos\theta$, $\sin\theta$ or $\tan\theta$ | M1 |
e.g. $\cos\theta = \frac{MT}{MP} = \frac{3}{\text{candidate's } s\sqrt{40}}$ (= 0.4743) ($\theta = 61.6835°$) | |
[If TP = 6 is used, then M0]
$\theta = 1.0766$ rad | AG; A1 (4) |
**(c)** Complete method for area $TMP$; e.g., $= \frac{1}{2} \times 3 \times \sqrt{40}\sin\theta$ | M1 |
$$= \frac{3}{2}\sqrt{31}$$ $(= 8.3516...)$ allow awrt 8.35 | A1 |
Area (sector) $MTQ = 0.5 \times 3^2 \times 1.0766$ | M1 | (= 4.8446...)
Area $TPQ =$ candidate's $(8.3516... - 4.8446...)$ | M1 |
$$= 3.507$$ (awrt) | A1 (5) [11] |
[Note: 3.51 is A0]
**Notes:**
**(a)** Allow 9 for $3^2$.
**(b)** First M1 can be implied by $\sqrt{40}$
For second M1:
May find $TP = \sqrt{(\sqrt{40})^2 - 3^2} = \sqrt{31}$, then either:
$$\sin\theta = \frac{TP}{MP} = \frac{\sqrt{31}}{\sqrt{40}} (= 0.8803...)$$ or $$\tan\theta = \frac{\sqrt{31}}{3} (= 1.8859...)$$ or cos rule
**NB. Answer is given, but allow final A1 if all previous work is correct.**
**(c)** First M1: (alternative) $\frac{1}{2} \times 3 \times \sqrt{40 - 9}$
---A circle $C$ has centre $M$ $(6, 4)$ and radius 3.
\begin{enumerate}[label=(\alph*)]
\item Write down the equation of the circle in the form
$$(x - a)^2 + (y - b)^2 = r^2.$$
[2]
\end{enumerate}
\includegraphics{figure_3}
Figure 3 shows the circle $C$. The point $T$ lies on the circle and the tangent at $T$ passes through the point $P$ $(12, 6)$. The line $MP$ cuts the circle at $Q$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the angle $TMQ$ is 1.0766 radians to 4 decimal places. [4]
\end{enumerate}
The shaded region $TPQ$ is bounded by the straight lines $TP$, $QP$ and the arc $TQ$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the area of the shaded region $TPQ$. Give your answer to 3 decimal places. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q8 [11]}}