| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Two circles intersection or tangency |
| Difficulty | Standard +0.8 This question requires finding the angle at the origin using the cosine rule with circle radii and distance between centres, then calculating arc lengths using the sector formula. While the individual techniques are standard A-level content, the multi-step geometric reasoning (finding intersection geometry, then applying it to calculate a perimeter involving two arcs) and the need to carefully identify which arcs bound the shaded region elevate this above a routine exercise. |
| 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 | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solves \(x^2 + y^2 = 100\) and \((x-15)^2 + y^2 = 40\) simultaneously to find \(x\) or \(y\). E.g. \((x-15)^2 + 100 - x^2 = 40 \Rightarrow x = ...\) | M1 | Key step: set up equation in single variable leading to value for \(x\) or \(y\) |
| \(\Rightarrow -30x + 325 = 40 \Rightarrow x = 9.5\), or \(y = \frac{\sqrt{39}}{2} = \text{awrt} \pm 3.12\) | A1 | |
| Attempts to find angle \(AOB\) in circle \(C_1\). E.g. attempts \(\cos\alpha = \frac{"9.5"}{10}\) to find \(\alpha\) then \(\times 2\) | M1 | Uses radius of circle and correct trig to find angle \(AOB\) in \(C_1\) |
| Angle \(AOB = 2\times\arccos\left(\frac{9.5}{10}\right) = 0.635\) rads (3sf) | A1* | Correct and careful work leading to given answer. Condone truncated intermediate values provided trig equation is correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(10\times(2\pi - 0.635) = 56.48\) | M1 | Attempts to use \(s = r\theta\) with \(r=10\) and \(\theta = 2\pi - 0.635\) |
| Attempts to find angle \(AXB\) or \(AXO\) in circle \(C_2\). E.g. \(\cos\beta = \frac{15-"9.5"}{\sqrt{40}} \Rightarrow \beta = ...\) (Note \(AXB = 1.03\) rads) | M1 | Correct method to find angle \(AXB\) or \(AXO\) in circle \(C_2\) |
| Attempts \(10\times(2\pi - 0.635) + \sqrt{40}\times(2\pi - 2\beta)\) | dM1 | Full and complete attempt to find perimeter. Dependent on both M marks |
| \(= 89.7\) | A1 | awrt 89.7 |
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $x^2 + y^2 = 100$ and $(x-15)^2 + y^2 = 40$ simultaneously to find $x$ or $y$. E.g. $(x-15)^2 + 100 - x^2 = 40 \Rightarrow x = ...$ | M1 | Key step: set up equation in single variable leading to value for $x$ or $y$ |
| $\Rightarrow -30x + 325 = 40 \Rightarrow x = 9.5$, or $y = \frac{\sqrt{39}}{2} = \text{awrt} \pm 3.12$ | A1 | |
| Attempts to find angle $AOB$ in circle $C_1$. E.g. attempts $\cos\alpha = \frac{"9.5"}{10}$ to find $\alpha$ then $\times 2$ | M1 | Uses radius of circle and correct trig to find angle $AOB$ in $C_1$ |
| Angle $AOB = 2\times\arccos\left(\frac{9.5}{10}\right) = 0.635$ rads (3sf) | A1* | Correct and careful work leading to given answer. Condone truncated intermediate values provided trig equation is correct |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $10\times(2\pi - 0.635) = 56.48$ | M1 | Attempts to use $s = r\theta$ with $r=10$ and $\theta = 2\pi - 0.635$ |
| Attempts to find angle $AXB$ or $AXO$ in circle $C_2$. E.g. $\cos\beta = \frac{15-"9.5"}{\sqrt{40}} \Rightarrow \beta = ...$ (Note $AXB = 1.03$ rads) | M1 | Correct method to find angle $AXB$ or $AXO$ in circle $C_2$ |
| Attempts $10\times(2\pi - 0.635) + \sqrt{40}\times(2\pi - 2\beta)$ | dM1 | Full and complete attempt to find perimeter. Dependent on both M marks |
| $= 89.7$ | A1 | awrt 89.7 |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dc0ac5df-24a7-41b5-8410-f0e9b332ba64-30_738_837_242_614}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Circle $C _ { 1 }$ has equation $x ^ { 2 } + y ^ { 2 } = 100$\\
Circle $C _ { 2 }$ has equation $( x - 15 ) ^ { 2 } + y ^ { 2 } = 40$\\
The circles meet at points $A$ and $B$ as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that angle $A O B = 0.635$ radians to 3 significant figures, where $O$ is the origin.
The region shown shaded in Figure 3 is bounded by $C _ { 1 }$ and $C _ { 2 }$
\item Find the perimeter of the shaded region, giving your answer to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q11 [8]}}