| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle touching axes |
| Difficulty | Moderate -0.3 This is a straightforward C2 circle question requiring completing the square to find the centre (standard technique) and using the tangency condition (radius equals perpendicular distance). While it involves two parts and some algebraic manipulation, both are routine applications of well-practiced methods with no conceptual challenges, 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 |
|---|---|---|
| \((x + 4)^2 - 16 + (y - 2)^2 - 4 + k = 0\) | M1 | |
| \(\therefore\) centre \((-4, 2)\) | A1 | |
| For x-axis to be tangent, radius must be 2 | B1 | |
| \((x + 4)^2 + (y - 2)^2 = 20 - k\) | ||
| \(\therefore 20 - k = 2^2\) | M1 | |
| \(k = 16\) | A1 | (5) |
$(x + 4)^2 - 16 + (y - 2)^2 - 4 + k = 0$ | M1 |
$\therefore$ centre $(-4, 2)$ | A1 |
For x-axis to be tangent, radius must be 2 | B1 |
$(x + 4)^2 + (y - 2)^2 = 20 - k$ | |
$\therefore 20 - k = 2^2$ | M1 |
$k = 16$ | A1 | (5)A circle has the equation
$$x^2 + y^2 + 8x - 4y + k = 0,$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre of the circle. [2]
\end{enumerate}
Given that the $x$-axis is a tangent to the circle,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $k$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q2 [5]}}