| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Two circles intersection or tangency |
| Difficulty | Standard +0.3 This is a straightforward multi-part circle question requiring standard techniques: midpoint formula for the centre, distance formula for radius, and understanding that circles touching means the distance between centres equals sum or difference of radii. All steps are routine applications of Core 1/2 formulas with no novel problem-solving required, 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 circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{1+7}{2}, \frac{4+8}{2}\right) = (4, 6)\) | M1A1 | M1 for attempt at midpoint; A1 for \((4,6)\); no working required, correct answer scores both marks; condone lack of brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sqrt{(7-1)^2 + (8-4)^2}}{2}\) or \(\sqrt{(4-1)^2 + (6-4)^2}\) or \(\sqrt{(7-4)^2 + (8-6)^2}\) | M1 | Scored for using Pythagoras to find distance between centre and a point; if original coordinates used there must be some attempt to halve |
| Radius \(= \sqrt{13}\) | A1 | Correct answer scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of \(C_2\) is \(x^2 + y^2 = r^2\) | M1 | For stating equation of \(C_2\) as \(x^2 + y^2 = r^2\) or \((x-0)^2 + (y-0)^2 = r^2\) for any \(r\); accept \(x^2 + y^2 = k\) if \(k\) is positive |
| Attempts either value of \(r\) as \(\left(\sqrt{4^2 + 6^2} \pm \text{their } r\right)\) | M1 | Look for \(r = \frac{\sqrt{4^2+6^2}}{2} \pm\) their \(r\); accept \(r = \frac{\sqrt{4^2+6^2}}{2}\) |
| When \(r = \sqrt{52} - \sqrt{13} = \sqrt{13} \Rightarrow x^2 + y^2 = 13\) | A1 | Either \(x^2 + y^2 = 13\) or \(x^2 + y^2 = 117\); allow \((x-0)^2 + (y-0)^2 = \sqrt{13}^2\) |
| When \(r = \sqrt{52} + \sqrt{13} = 3\sqrt{13} \Rightarrow x^2 + y^2 = 117\) | A1 | Both \(x^2 + y^2 = 13\) and \(x^2 + y^2 = 117\); equations must be simplified; any one correct equation implies first two M marks |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1+7}{2}, \frac{4+8}{2}\right) = (4, 6)$ | M1A1 | M1 for attempt at midpoint; A1 for $(4,6)$; no working required, correct answer scores both marks; condone lack of brackets |
**(2 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sqrt{(7-1)^2 + (8-4)^2}}{2}$ or $\sqrt{(4-1)^2 + (6-4)^2}$ or $\sqrt{(7-4)^2 + (8-6)^2}$ | M1 | Scored for using Pythagoras to find distance between centre and a point; if original coordinates used there must be some attempt to halve |
| Radius $= \sqrt{13}$ | A1 | Correct answer scores both marks |
**(2 marks)**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of $C_2$ is $x^2 + y^2 = r^2$ | M1 | For stating equation of $C_2$ as $x^2 + y^2 = r^2$ or $(x-0)^2 + (y-0)^2 = r^2$ for any $r$; accept $x^2 + y^2 = k$ if $k$ is positive |
| Attempts either value of $r$ as $\left(\sqrt{4^2 + 6^2} \pm \text{their } r\right)$ | M1 | Look for $r = \frac{\sqrt{4^2+6^2}}{2} \pm$ their $r$; accept $r = \frac{\sqrt{4^2+6^2}}{2}$ |
| When $r = \sqrt{52} - \sqrt{13} = \sqrt{13} \Rightarrow x^2 + y^2 = 13$ | A1 | Either $x^2 + y^2 = 13$ or $x^2 + y^2 = 117$; allow $(x-0)^2 + (y-0)^2 = \sqrt{13}^2$ |
| When $r = \sqrt{52} + \sqrt{13} = 3\sqrt{13} \Rightarrow x^2 + y^2 = 117$ | A1 | Both $x^2 + y^2 = 13$ and $x^2 + y^2 = 117$; equations must be simplified; any one correct equation implies first two M marks |
**(4 marks) — Total: 8 marks**
---14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08b1be3e-2d9a-4832-b230-d5519540f494-48_771_812_237_575}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a sketch of the circle $C _ { 1 }$\\
The points $A ( 1,4 )$ and $B ( 7,8 )$ lie on $C _ { 1 }$\\
Given that $A B$ is a diameter of the circle $C _ { 1 }$
\begin{enumerate}[label=(\alph*)]
\item find the coordinates for the centre of $C _ { 1 }$
\item find the exact radius of $C _ { 1 }$ simplifying your answer.
Two distinct circles $C _ { 2 }$ and $C _ { 3 }$ each have centre $( 0,0 )$.\\
Given that each of these circles touch circle $C _ { 1 }$
\item find the equation of circle $C _ { 2 }$ and the equation of circle $C _ { 3 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q14 [8]}}