Edexcel C12 2016 June — Question 12 5 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2016
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeCircle touching axes
DifficultyModerate -0.5 This is a straightforward circle question requiring understanding that a circle touching the y-axis has radius equal to its x-coordinate, then solving a simple equation. Part (a) is direct recall of circle properties, part (b) involves substituting a point and solving a quadratic equation. Slightly easier than average due to minimal problem-solving required.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^2

12. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-18_636_887_274_534} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a circle \(C\) \(C\) touches the \(y\)-axis and has centre at the point ( \(a , 0\) ) where \(a\) is a positive constant.
  1. Write down an equation for \(C\) in terms of \(a\) Given that the point \(P ( 4 , - 3 )\) lies on \(C\),
  2. find the value of \(a\)

Question 12:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Writes \(C\) as \((x-a)^2+(y-0)^2=a^2\)M1A1 M1: attempts equation with centre \((a,0)\) radius \(a\); accept \((x\pm a)^2+y^2=a^2\). A1: \((x-a)^2+(y-0)^2=a^2\) or equivalent \(x^2+y^2-2ax=0\)
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Subs \((4,-3)\): \((4-a)^2+(-3-0)^2=a^2\)M1 Subs \(x=4\), \(y=-3\) into circle equation of form \((x\pm a)^2+(y\pm0)^2=a^2\)
\(\Rightarrow 16-8a+a^2+9=a^2 \Rightarrow 25=8a\)dM1 Proceeds to linear equation in \(a\); condone numerical slips
\(\Rightarrow a=\frac{25}{8}\)A1 Accept exact alternatives
## Question 12:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Writes $C$ as $(x-a)^2+(y-0)^2=a^2$ | M1A1 | M1: attempts equation with centre $(a,0)$ radius $a$; accept $(x\pm a)^2+y^2=a^2$. A1: $(x-a)^2+(y-0)^2=a^2$ or equivalent $x^2+y^2-2ax=0$ |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Subs $(4,-3)$: $(4-a)^2+(-3-0)^2=a^2$ | M1 | Subs $x=4$, $y=-3$ into circle equation of form $(x\pm a)^2+(y\pm0)^2=a^2$ |
| $\Rightarrow 16-8a+a^2+9=a^2 \Rightarrow 25=8a$ | dM1 | Proceeds to linear equation in $a$; condone numerical slips |
| $\Rightarrow a=\frac{25}{8}$ | A1 | Accept exact alternatives |
12.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-18_636_887_274_534}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

Figure 3 shows a circle $C$\\
$C$ touches the $y$-axis and has centre at the point ( $a , 0$ ) where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Write down an equation for $C$ in terms of $a$

Given that the point $P ( 4 , - 3 )$ lies on $C$,
\item find the value of $a$
\end{enumerate}

\hfill \mbox{\textit{Edexcel C12 2016 Q12 [5]}}