| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find centre and radius from equation |
| Difficulty | Moderate -0.8 This is a standard C2 circle question requiring completion of the square to find centre and radius, then substituting x=0 for y-intercepts. All techniques are routine and well-practiced at this level, making it easier than average but not trivial due to the algebraic manipulation required. |
| 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 |
|---|---|---|
| Working | Mark | Guidance |
| \((x+2)^2+(y-1)^2=16\), centre \((-2,1)\), radius \(r=4\) | — | — |
| \(d_1 = \sqrt{4^2-2^2}=\sqrt{12}\) | M1 | Applying \(\sqrt{\text{their }r^2 - |
| A1 aef | \(\sqrt{12}\) | |
| Hence \(y = 1 \pm \sqrt{12}\) | M1 | Applies \(y = \text{their }y_i \pm \text{their }d\) |
| So \(y = 1 \pm 2\sqrt{3}\) | A1 cao cso | \(1 \pm 2\sqrt{3}\) |
## Question 4(c) — Aliter Way 2:
| Working | Mark | Guidance |
|---|---|---|
| $(x+2)^2+(y-1)^2=16$, centre $(-2,1)$, radius $r=4$ | — | — |
| $d_1 = \sqrt{4^2-2^2}=\sqrt{12}$ | M1 | Applying $\sqrt{\text{their }r^2 - |\text{their }x_i|^2}$ |
| | A1 aef | $\sqrt{12}$ |
| Hence $y = 1 \pm \sqrt{12}$ | M1 | Applies $y = \text{their }y_i \pm \text{their }d$ |
| So $y = 1 \pm 2\sqrt{3}$ | A1 cao cso | $1 \pm 2\sqrt{3}$ |
**Special Case:** SC: M1A1M1A0 if candidate achieves any one of $y=1+2\sqrt{3}$, $y=1-2\sqrt{3}$, $y=1+\sqrt{12}$, or $y=1-\sqrt{12}$
**[4 marks]**
---4. The circle $C$ has equation
$$x ^ { 2 } + y ^ { 2 } + 4 x - 2 y - 11 = 0$$
Find
\begin{enumerate}[label=(\alph*)]
\item the coordinates of the centre of $C$,
\item the radius of $C$,
\item the coordinates of the points where $C$ crosses the $y$-axis, giving your answers as simplified surds.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2011 Q4 [8]}}