| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find centre and radius from equation |
| Difficulty | Moderate -0.8 This is a straightforward C2 question requiring completing the square to find centre and radius, then using Pythagoras to find a chord length. All steps are routine applications of standard techniques with no problem-solving insight required. The multi-part structure and final exact form answer add minimal difficulty beyond basic recall. |
| 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 |
|---|---|
| \((x - 6)^2 - 36 + (y + 4)^2 - 16 + 16 = 0\) | M1 |
| \(\therefore\) centre (6, -4) | A1 |
| Answer | Marks |
|---|---|
| \((x - 6)^2 + (y + 4)^2 = 36\) | M1 |
| \(\therefore\) radius = 6 | A1 |
| Answer | Marks |
|---|---|
| [Diagram shown with circle centered at (6, -4) with radius 6] | B2 |
| Answer | Marks |
|---|---|
| \(y = 0 \Rightarrow (x - 6)^2 + 16 = 36\) | M1 |
| \(x = 6 \pm \sqrt{20} = 6 \pm 2\sqrt{5}\) | A1 |
| \(AB = 6 + 2\sqrt{5} - (6 - 2\sqrt{5}) = 4\sqrt{5}\) | M1 A1 |
**Part (a)**
$(x - 6)^2 - 36 + (y + 4)^2 - 16 + 16 = 0$ | M1 |
$\therefore$ centre (6, -4) | A1 |
**Part (b)**
$(x - 6)^2 + (y + 4)^2 = 36$ | M1 |
$\therefore$ radius = 6 | A1 |
**Part (c)**
[Diagram shown with circle centered at (6, -4) with radius 6] | B2 |
**Part (d)**
$y = 0 \Rightarrow (x - 6)^2 + 16 = 36$ | M1 |
$x = 6 \pm \sqrt{20} = 6 \pm 2\sqrt{5}$ | A1 |
$AB = 6 + 2\sqrt{5} - (6 - 2\sqrt{5}) = 4\sqrt{5}$ | M1 A1 |
---6. The circle $C$ has the equation
$$x ^ { 2 } + y ^ { 2 } - 12 x + 8 y + 16 = 0 .$$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre of $C$.
\item Find the radius of $C$.
\item Sketch C.
Given that $C$ crosses the $x$-axis at the points $A$ and $B$,
\item find the length $A B$, giving your answer in the form $k \sqrt { 5 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [10]}}