| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Sector and arc length |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard C2 circle techniques: completing the square to find centre and radius, finding intersection points by substitution, and applying the sector perimeter formula (2r + rθ). All steps are routine with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation needed. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Obtain \((x \pm 10)^2\) and \((y \pm 8)^2\) | M1 | As in scheme, can be implied by \((\pm 10, \pm 8)\) |
| Obtain \((x-10)^2\) and \((y-8)^2\) | A1 | |
| Centre is \((10, 8)\) | A1 | May be indicated on diagram only. Correct centre (10,8) implies M1A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| See \((x \pm 10)^2 + (y \pm 8)^2 = 25\) \((= r^2)\) or \((r^2 =)\) "100"+"64"\(-\)139 | M1 | Correct method leading to \(r=\ldots\), or \(r^2 =\) "100"+"64"\(-\)139 |
| \(r = 5\) | A1 | Printed answer so need one of the above two reasons. NB \(r=5\) should follow \(k^2=25\) or \(r^2=100+64-139\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use \(x=13\) in either form of equation of circle and solve resulting quadratic to give \(y=\) | M1 | |
| e.g. \(x=13 \Rightarrow (13-10)^2+(y-8)^2=25 \Rightarrow (y-8)^2=16\) so \(y=\) | ||
| or \(13^2+y^2-20\times13-16y+139=0 \Rightarrow y^2-16y+48=0\) so \(y=\) | ||
| \(y=4\) or \(y=12\) | A1, A1 | On EPEN mark one correct value as A1A0 and both correct as A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of \(r\theta\) with \(r=5\) and \(\theta = 1.855\) (may be implied by 9.275) | M1 | |
| Perimeter \(PTQ = 2r +\) their arc \(PQ\) (Finding perimeter of triangle is M0 here) | M1 | |
| \(= 19.275\) or \(19.28\) or \(19.3\) | A1 | Full marks available for calculation using major sector: \(r=5\), \(\theta=4.428\) leading to perimeter of 32.14 |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Obtain $(x \pm 10)^2$ **and** $(y \pm 8)^2$ | M1 | As in scheme, can be implied by $(\pm 10, \pm 8)$ |
| Obtain $(x-10)^2$ **and** $(y-8)^2$ | A1 | |
| Centre is $(10, 8)$ | A1 | May be indicated on diagram only. **Correct centre (10,8) implies M1A1A1** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| See $(x \pm 10)^2 + (y \pm 8)^2 = 25$ $(= r^2)$ or $(r^2 =)$ "100"+"64"$-$139 | M1 | Correct method leading to $r=\ldots$, or $r^2 =$ "100"+"64"$-$139 |
| $r = 5$ | A1 | Printed answer so need one of the above two reasons. **NB** $r=5$ should follow $k^2=25$ or $r^2=100+64-139$ |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $x=13$ in either form of equation of circle and solve resulting quadratic to give $y=$ | M1 | |
| e.g. $x=13 \Rightarrow (13-10)^2+(y-8)^2=25 \Rightarrow (y-8)^2=16$ so $y=$ | | |
| or $13^2+y^2-20\times13-16y+139=0 \Rightarrow y^2-16y+48=0$ so $y=$ | | |
| $y=4$ or $y=12$ | A1, A1 | On EPEN mark one correct value as A1A0 and both correct as A1 A1 |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $r\theta$ with $r=5$ and $\theta = 1.855$ (may be implied by 9.275) | M1 | |
| Perimeter $PTQ = 2r +$ their **arc** $PQ$ (Finding perimeter of triangle is M0 here) | M1 | |
| $= 19.275$ or $19.28$ or $19.3$ | A1 | Full marks available for calculation using major sector: $r=5$, $\theta=4.428$ leading to perimeter of 32.14 |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1ef99f0-4ad4-49d8-bee7-d5bb9cc84660-04_734_1262_237_315}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The circle $C$ with centre $T$ and radius $r$ has equation
$$x ^ { 2 } + y ^ { 2 } - 20 x - 16 y + 139 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre of $C$.
\item Show that $r = 5$
The line $L$ has equation $x = 13$ and crosses $C$ at the points $P$ and $Q$ as shown in Figure 1.
\item Find the $y$ coordinate of $P$ and the $y$ coordinate of $Q$.
Given that, to 3 decimal places, the angle $P T Q$ is 1.855 radians,
\item find the perimeter of the sector $P T Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2012 Q3 [11]}}