| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Pulley, cord, and tangent applications |
| Difficulty | Challenging +1.2 This is a multi-part circle geometry question requiring completing the square, finding a radius, using tangent-radius perpendicularity to find an angle via trigonometry, and calculating a composite area (sector minus triangle). While it involves several steps and techniques, each component is a standard application of Core 1/2 circle and trigonometry methods with no novel insight required. The multi-step nature and sector area calculation elevate it slightly above average difficulty. |
| 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtains \((x \pm 3)^2\) and \((y \pm 1)^2\) | M1 | Could be implied by writing centre \((\pm3, \pm1)\) |
| Obtains \((x-3)^2\) and \((y+1)^2\) | A1 | |
| Centre \(= (3, -1)\) | A1 | Accept without working for M1A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r^2 = {'}3{'}^2 + {'-1'}^2 - 5 = 5 \Rightarrow r = \sqrt{5}\) | M1A1 | \(r^2\) must be positive. Do not accept 2.24 on its own |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(TQ = \sqrt{(8-3)^2 + (4--1)^2} = \sqrt{50}\) | M1A1 | Must attempt difference of coordinates. Accept \(5\sqrt{2}\), awrt 7.07 |
| \(\cos\theta = \dfrac{\sqrt{5}}{\sqrt{50}} \Rightarrow \theta = 1.249...\) | M1A1 | Accept \(\theta\) = awrt 1.25, 71.6° or \(\theta = \arccos\!\left(\dfrac{\sqrt{5}}{\sqrt{50}}\right)\) |
| Angle \(MQN = 2.498\) radians | A1* | cso. Must follow accuracy of at least 1.249 for half angle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of sector \(= \frac{1}{2}r^2\theta = \frac{1}{2}\times(\sqrt{5})^2 \times 2.498\) (= awrt 6.24/6.25) | M1A1 | Formula must be correct if quoted. Accept \(\frac{1}{2}\times 5 \times 2.498\) or awrt 6.24/6.25 |
| Area of triangle \(= \frac{1}{2}ab\sin C = \frac{1}{2}\times\sqrt{5}\times\sqrt{50}\times\sin 1.249 = (7.50)\) | M1 | Accept use of \(\frac{1}{2}ab\sin C\) with their \(r\), \(TQ\) and 1.249 |
| Shaded Area \(= 15.0 - 6.245 = 8.76\) or \(8.75\) | dM1, A1 | Dependent on both previous M marks. Accept awrt 8.76 or 8.75 |
# Question 11:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtains $(x \pm 3)^2$ and $(y \pm 1)^2$ | M1 | Could be implied by writing centre $(\pm3, \pm1)$ |
| Obtains $(x-3)^2$ and $(y+1)^2$ | A1 | |
| Centre $= (3, -1)$ | A1 | Accept without working for M1A1A1 |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r^2 = {'}3{'}^2 + {'-1'}^2 - 5 = 5 \Rightarrow r = \sqrt{5}$ | M1A1 | $r^2$ must be positive. Do not accept 2.24 on its own |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $TQ = \sqrt{(8-3)^2 + (4--1)^2} = \sqrt{50}$ | M1A1 | Must attempt difference of coordinates. Accept $5\sqrt{2}$, awrt 7.07 |
| $\cos\theta = \dfrac{\sqrt{5}}{\sqrt{50}} \Rightarrow \theta = 1.249...$ | M1A1 | Accept $\theta$ = awrt 1.25, 71.6° or $\theta = \arccos\!\left(\dfrac{\sqrt{5}}{\sqrt{50}}\right)$ |
| Angle $MQN = 2.498$ radians | A1* | cso. Must follow accuracy of at least 1.249 for half angle |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of sector $= \frac{1}{2}r^2\theta = \frac{1}{2}\times(\sqrt{5})^2 \times 2.498$ (= awrt 6.24/6.25) | M1A1 | Formula must be correct if quoted. Accept $\frac{1}{2}\times 5 \times 2.498$ or awrt 6.24/6.25 |
| Area of triangle $= \frac{1}{2}ab\sin C = \frac{1}{2}\times\sqrt{5}\times\sqrt{50}\times\sin 1.249 = (7.50)$ | M1 | Accept use of $\frac{1}{2}ab\sin C$ with their $r$, $TQ$ and 1.249 |
| Shaded Area $= 15.0 - 6.245 = 8.76$ or $8.75$ | dM1, A1 | Dependent on both previous M marks. Accept awrt 8.76 or 8.75 |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b85872d4-00b2-499b-9765-f7559d3de66a-17_1000_956_264_500}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the circle $C$ with centre $Q$ and equation
$$x ^ { 2 } + y ^ { 2 } - 6 x + 2 y + 5 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $Q$,
\item the exact value of the radius of $C$.
The tangents to $C$ from the point $T ( 8,4 )$ meet $C$ at the points $M$ and $N$, as shown in Figure 4.
\end{enumerate}\item Show that the obtuse angle $M Q N$ is 2.498 radians to 3 decimal places.
The region $R$, shown shaded in Figure 4, is bounded by the tangent $T N$, the minor arc $N M$, and the tangent $M T$.
\item Find the area of region $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q11 [15]}}