| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Intersection of two loci |
| Difficulty | Challenging +1.2 This is a standard Further Maths loci question requiring the circle of Apollonius formula and intersection with a half-line. Part (a) involves algebraic manipulation to find circle parameters (routine for F2), while part (b) requires solving simultaneous equations with the argument condition. The techniques are well-practiced in Further Maths courses, though the multi-step nature and coordinate geometry involved make it moderately above average difficulty. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ | z+1-13i\ | = 3\ |
| \(\Rightarrow x^2 + y^2 - 16x - 8y + 62 = 0\) | A1 | Correct equation in any form with terms collected |
| Centre \((8, 4)\) | A1 | Correct centre. i included scores A0 |
| \(r^2 = 64 + 16 - 62 = \ldots\) | M1 | Correct method for \(r\) or \(r^2\) |
| \(r = \sqrt{18}\) or \(3\sqrt{2}\) | A1 | Correct radius. Must be exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\arg(z-8-6i) = -\frac{3\pi}{4} \Rightarrow y-6 = x-8\) | B1 | Converts the given locus to the correct Cartesian form |
| \(\Rightarrow x^2 + y^2 - 16x - 8y + 62 = 0\) \(\Rightarrow x^2 + (x-2)^2 - 16x - 8(x-2) + 62 = 0 \Rightarrow x = \ldots\) or \(\Rightarrow (y+2)^2 + y^2 - 16x - 8(y+2) + 62 = 0 \Rightarrow y = \ldots\) | M1 | Uses both Cartesian equations to obtain an equation in one variable and attempts to solve |
| \(x = 7 - 2\sqrt{2}\) or \(y = 5 - 2\sqrt{2}\) | A1 | One correct "coordinate" |
| \(R\) is \(7-2\sqrt{2} + (5-2\sqrt{2})i\) or \(x = 7-2\sqrt{2}\) and \(y = 5-2\sqrt{2}\) | A1 | Correct complex number or coordinates and no others. Must be exact |
## Question 6:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|z+1-13i\| = 3\|z-7-5i\| \Rightarrow (x+1)^2+(y-13)^2 = 9\{(x-7)^2+(y-5)^2\}$ | M1 | Correct application of Pythagoras. Accept 3 or 9 on RHS |
| $\Rightarrow x^2 + y^2 - 16x - 8y + 62 = 0$ | A1 | Correct equation in any form with terms collected |
| Centre $(8, 4)$ | A1 | Correct centre. i included scores A0 |
| $r^2 = 64 + 16 - 62 = \ldots$ | M1 | Correct method for $r$ or $r^2$ |
| $r = \sqrt{18}$ or $3\sqrt{2}$ | A1 | Correct radius. Must be exact |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\arg(z-8-6i) = -\frac{3\pi}{4} \Rightarrow y-6 = x-8$ | B1 | Converts the given locus to the correct Cartesian form |
| $\Rightarrow x^2 + y^2 - 16x - 8y + 62 = 0$ $\Rightarrow x^2 + (x-2)^2 - 16x - 8(x-2) + 62 = 0 \Rightarrow x = \ldots$ or $\Rightarrow (y+2)^2 + y^2 - 16x - 8(y+2) + 62 = 0 \Rightarrow y = \ldots$ | M1 | Uses both Cartesian equations to obtain an equation in one variable and attempts to solve |
| $x = 7 - 2\sqrt{2}$ or $y = 5 - 2\sqrt{2}$ | A1 | One correct "coordinate" |
| $R$ is $7-2\sqrt{2} + (5-2\sqrt{2})i$ or $x = 7-2\sqrt{2}$ and $y = 5-2\sqrt{2}$ | A1 | Correct complex number or coordinates and no others. Must be exact |
---6. The complex number $z$ on an Argand diagram is represented by the point $P$ where
$$| z + 1 - 13 i | = 3 | z - 7 - 5 i |$$
Given that the locus of $P$ is a circle,
\begin{enumerate}[label=(\alph*)]
\item determine the centre and radius of this circle.
The complex number $w$, on the same Argand diagram, is represented by the point $Q$, where
$$\arg ( w - 8 - 6 \mathrm { i } ) = - \frac { 3 \pi } { 4 }$$
Given that the locus of $P$ intersects the locus of $Q$ at the point $R$,
\item determine the complex number representing $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2021 Q6 [9]}}