| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Circle of Apollonius locus |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring conversion of modulus equation to Cartesian form (standard technique), sketching (routine), and finding tangent conditions using complex number line representation. Part (c) requires some geometric insight connecting the line equation to tangency, but follows established FP2 methods. More challenging than typical A-level pure maths but standard for FP2 loci work. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((x+6)^2 + y^2 = 4[(x-6) + y^2]\) | M1 | Obtains an equation in terms of \(x\) and \(y\) using the given information |
| \(x^2 + y^2 - 20x + 36 = 0\) (equation of a circle) | A1* | Expands and simplifies, collecting terms to obtain circle equation, deducing this is a circle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Circle drawn with centre \((10, 0)\) | M1 | Draws a circle with centre at \((10, 0)\) |
| Circle does not cross the \(y\)-axis | A1 | Radius is \(8\) so circle does not cross the \(y\)-axis |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Let \(a = c + id\) and \(a^* = c - id\), then \((c + id)(x - iy) + (c - id)(x + iy) = 0\) | M1 | Attempts to convert line equation into Cartesian form |
| \(y = -\dfrac{c}{d}x\) | A1 | Obtains a simplified line equation |
| Line through origin drawn intersecting circle in diagram | B1 | Uses geometry to deduce the gradients of the tangents |
| Gradients of tangents (from geometry) are \(\pm\dfrac{4}{3}\), so \(-\dfrac{c}{d} = \pm\dfrac{4}{3}\) and \(\dfrac{d}{c} = \mp\dfrac{3}{4}\) | M1 | Understands connection between \(\arg a\) and gradient of tangents and uses this connection |
| \(\tan\theta = \pm\dfrac{3}{4}\) | A1 | Correct answers |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(x+6)^2 + y^2 = 4[(x-6) + y^2]$ | M1 | Obtains an equation in terms of $x$ and $y$ using the given information |
| $x^2 + y^2 - 20x + 36 = 0$ (equation of a circle) | A1* | Expands and simplifies, collecting terms to obtain circle equation, deducing this is a circle |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Circle drawn with centre $(10, 0)$ | M1 | Draws a circle with centre at $(10, 0)$ |
| Circle does not cross the $y$-axis | A1 | Radius is $8$ so circle does not cross the $y$-axis |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Let $a = c + id$ and $a^* = c - id$, then $(c + id)(x - iy) + (c - id)(x + iy) = 0$ | M1 | Attempts to convert line equation into Cartesian form |
| $y = -\dfrac{c}{d}x$ | A1 | Obtains a simplified line equation |
| Line through origin drawn intersecting circle in diagram | B1 | Uses geometry to deduce the gradients of the tangents |
| Gradients of tangents (from geometry) are $\pm\dfrac{4}{3}$, so $-\dfrac{c}{d} = \pm\dfrac{4}{3}$ and $\dfrac{d}{c} = \mp\dfrac{3}{4}$ | M1 | Understands connection between $\arg a$ and gradient of tangents and uses this connection |
| $\tan\theta = \pm\dfrac{3}{4}$ | A1 | Correct answers |
---\begin{enumerate}
\item A curve has equation
\end{enumerate}
$$| z + 6 | = 2 | z - 6 | \quad z \in \mathbb { C }$$
(a) Show that the curve is a circle with equation $x ^ { 2 } + y ^ { 2 } - 20 x + 36 = 0$\\
(b) Sketch the curve on an Argand diagram.
The line $l$ has equation $a z ^ { * } + a ^ { * } z = 0$, where $a \in \mathbb { C }$ and $z \in \mathbb { C }$\\
Given that the line $l$ is a tangent to the curve and that $\arg a = \theta$\\
(c) find the possible values of $\tan \theta$
\hfill \mbox{\textit{Edexcel FP2 Q6 [9]}}