| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Standard +0.3 This is a straightforward application of the tangent-radius perpendicularity property and Pythagoras' theorem. Students need to recognize that OT ⊥ QT, apply Pythagoras to triangle OQT to find OQ, then use distance formula to find k. Part (b) is routine circle equation writing. Slightly above average due to the coordinate geometry setup and exact value requirement, but still a standard C2 exercise with clear structure. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(OQ^2 = (6\sqrt{5})^2 + 4^2\) or \(OQ = \sqrt{(6\sqrt{5})^2 + 4^2}\ \{=14\}\) | M1 | Uses addition form of Pythagoras on \(6\sqrt{5}\) and 4; condone missing brackets on \((6\sqrt{5})^2\); working or 14 may be on diagram |
| \(y_Q = \sqrt{14^2 - 11^2}\) | dM1 | \(y_Q = \sqrt{(\text{their }OQ)^2 - 11^2}\); must include \(\sqrt{}\); dependent on first M1; requires \(OQ > 11\) |
| \(= \sqrt{75}\) or \(5\sqrt{3}\) | A1cso | \(\sqrt{75}\) or \(5\sqrt{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-11)^2 + (y - 5\sqrt{3})^2 = 16\) | M1 | \((x \pm 11)^2 + (y \pm \text{their }k)^2 = 4^2\); equation must use \(x\) and \(y\); \(k\) could be their answer to (a); allow \(k \neq 0\) or just the letter \(k\) |
| \((x-11)^2 + (y-5\sqrt{3})^2 = 16\) | A1 | \(5\sqrt{3}\) must come from correct work in (a); allow awrt 8.66; allow expanded form |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $OQ^2 = (6\sqrt{5})^2 + 4^2$ or $OQ = \sqrt{(6\sqrt{5})^2 + 4^2}\ \{=14\}$ | M1 | Uses addition form of Pythagoras on $6\sqrt{5}$ and 4; condone missing brackets on $(6\sqrt{5})^2$; working or 14 may be on diagram |
| $y_Q = \sqrt{14^2 - 11^2}$ | dM1 | $y_Q = \sqrt{(\text{their }OQ)^2 - 11^2}$; must include $\sqrt{}$; dependent on first M1; requires $OQ > 11$ |
| $= \sqrt{75}$ or $5\sqrt{3}$ | A1cso | $\sqrt{75}$ or $5\sqrt{3}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-11)^2 + (y - 5\sqrt{3})^2 = 16$ | M1 | $(x \pm 11)^2 + (y \pm \text{their }k)^2 = 4^2$; equation must use $x$ and $y$; $k$ could be their answer to (a); allow $k \neq 0$ or just the letter $k$ |
| $(x-11)^2 + (y-5\sqrt{3})^2 = 16$ | A1 | $5\sqrt{3}$ must come from correct work in (a); allow awrt 8.66; allow expanded form |9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e6b490c0-80c4-4e15-b587-ac052ee27db7-15_761_1082_210_424}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a circle $C$ with centre $Q$ and radius 4 and the point $T$ which lies on $C$. The tangent to $C$ at the point $T$ passes through the origin $O$ and $O T = 6 \sqrt { } 5$ Given that the coordinates of $Q$ are $( 11 , k )$, where $k$ is a positive constant, (a) find the exact value of $k$,\\
(b) find an equation for $C$.\\
\hfill \mbox{\textit{Edexcel C2 2014 Q9 [5]}}