| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Circle of Apollonius locus |
| Difficulty | Standard +0.8 This FP2 locus question requires converting a modulus equality to Cartesian form, completing the square to identify circle parameters, then working backwards from given geometric constraints (quadrant and radius) to find the unknown constant. It combines algebraic manipulation with geometric interpretation, going beyond routine locus problems by requiring reverse-engineering from the answer properties. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x)^2 + (y+a)^2 = 9[(x-a)^2 + y^2]\) or \(\sqrt{(x)^2+(y+a)^2} = 3\sqrt{(x-a)^2+y^2}\) | M1 | Obtains equation in \(x\) and \(y\) using given information; condone \((x)^2+(y+a)^2=3[(x-a)^2+y^2]\) |
| \(8x^2 - 18ax + 8y^2 - 2ay + 8a^2 = 0\) | A1 | Expands, simplifies, collects terms; condone missing \(=0\) |
| \(x^2 - \frac{9}{4}ax + y^2 - \frac{1}{4}ay + a^2 = 0\); \(\Rightarrow \left(x - \frac{9}{8}a\right)^2 - \left(\frac{9}{8}a\right)^2 + \left(y - \frac{1}{8}a\right)^2 - \left(\frac{1}{8}a\right)^2 + a^2 = 0\); \(\Rightarrow r^2 = \left(\frac{9}{8}a\right)^2 + \left(\frac{1}{8}a\right)^2 - a^2 = \left(\frac{3}{2}\sqrt{2}\right)^2\) | M1 | Completes the square; sets radius squared \(= \left(\frac{3}{2}\sqrt{2}\right)^2 = \frac{18}{4} = \frac{9}{2}\); finds value for \(a\). Note: \(r^2 = \frac{9}{32}a^2\), \(r = \frac{3a\sqrt{2}}{8}\) |
| \(a = -4\) cso | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(x - \frac{9}{8}(-4)\right)^2 + \left(y - \frac{1}{8}(-4)\right)^2 = \ldots\); \((x-\alpha)^2 + (y-\beta)^2 = \ldots\) implies centre \((\alpha, \beta)\) | M1 | Substitutes their value for \(a\) into equation and finds centre |
| centre \(\left(-\frac{9}{2}, -\frac{1}{2}\right)\) | A1 |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x)^2 + (y+a)^2 = 9[(x-a)^2 + y^2]$ or $\sqrt{(x)^2+(y+a)^2} = 3\sqrt{(x-a)^2+y^2}$ | M1 | Obtains equation in $x$ and $y$ using given information; condone $(x)^2+(y+a)^2=3[(x-a)^2+y^2]$ |
| $8x^2 - 18ax + 8y^2 - 2ay + 8a^2 = 0$ | A1 | Expands, simplifies, collects terms; condone missing $=0$ |
| $x^2 - \frac{9}{4}ax + y^2 - \frac{1}{4}ay + a^2 = 0$; $\Rightarrow \left(x - \frac{9}{8}a\right)^2 - \left(\frac{9}{8}a\right)^2 + \left(y - \frac{1}{8}a\right)^2 - \left(\frac{1}{8}a\right)^2 + a^2 = 0$; $\Rightarrow r^2 = \left(\frac{9}{8}a\right)^2 + \left(\frac{1}{8}a\right)^2 - a^2 = \left(\frac{3}{2}\sqrt{2}\right)^2$ | M1 | Completes the square; sets radius squared $= \left(\frac{3}{2}\sqrt{2}\right)^2 = \frac{18}{4} = \frac{9}{2}$; finds value for $a$. Note: $r^2 = \frac{9}{32}a^2$, $r = \frac{3a\sqrt{2}}{8}$ |
| $a = -4$ cso | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(x - \frac{9}{8}(-4)\right)^2 + \left(y - \frac{1}{8}(-4)\right)^2 = \ldots$; $(x-\alpha)^2 + (y-\beta)^2 = \ldots$ implies centre $(\alpha, \beta)$ | M1 | Substitutes their value for $a$ into equation and finds centre |
| centre $\left(-\frac{9}{2}, -\frac{1}{2}\right)$ | A1 | |
---\begin{enumerate}
\item The locus of points $z$ satisfies
\end{enumerate}
$$| z + a \mathrm { i } | = 3 | z - a |$$
where $a$ is an integer.\\
The locus is a circle with its centre in the third quadrant and radius $\frac { 3 } { 2 } \sqrt { 2 }$\\
Determine\\
(a) the value of $a$,\\
(b) the coordinates of the centre of the circle.
\hfill \mbox{\textit{Edexcel FP2 2022 Q5 [6]}}