| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Intersection of two loci |
| Difficulty | Standard +0.3 This is a standard two-part question on circle loci in the Argand diagram. Part (a) requires finding the center (midpoint) and radius from diameter endpoints—straightforward application of formulas. Part (b) involves solving simultaneous circle equations, which is routine algebraic manipulation. While it requires multiple steps and careful calculation, it follows standard techniques taught in Core Pure 2 with no novel insight required, making it slightly easier than average. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02k Argand diagrams: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Centre of circle \(C\) is \((1, -1)\) | B1 | Correct coordinates of centre |
| \(r = \sqrt{(5-1)^2 + (-4+1)^2} = 5\) or \(r = \sqrt{(-3-1)^2 + (2+1)^2} = 5\) or \(r = \frac{1}{2}\sqrt{(-3-5)^2 + (2+4)^2} = 5\) | M1 | Fully correct strategy for identifying radius. If diameter calculated must be halved |
| \(\ | z - 1 + \text{i}\ | = 5\) or \(\ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-1)^2 + (y+1)^2 = 25\), \((x-2)^2 + (y-3)^2 = 4\), expanding to give \(\Rightarrow 2x + 8y = 32\) | M1 | Begins finding \(z_1\) and \(z_2\) using Cartesian equations to obtain line of intersection |
| \((16-4y)^2 - 4(16-4y) + 4 + y^2 - 6y + 9 = 4\) or \(x^2 - 4x + 4 + \left(\frac{16-x}{4}\right)^2 - 6\left(\frac{16-x}{4}\right) + 9 = 4\) | M1 | Substitutes back into equation of one circle to obtain equation in one variable |
| \(17y^2 - 118y + 201 = 0\) or \(17x^2 - 72x + 16 = 0\) | A1 | Correct 3-term quadratic |
| \(17y^2 - 118y + 201 = 0 \Rightarrow (17y-67)(y-3) = 0 \Rightarrow y = \frac{67}{17}, 3\) or \(17x^2 - 72x + 16 = 0 \Rightarrow (17x-4)(x-4) = 0 \Rightarrow x = \frac{4}{17}, 4\) | M1 | Solves their 3TQ |
| \(y = \frac{67}{17}, 3 \Rightarrow x = \frac{4}{17}, 4\) or \(x = \frac{4}{17}, 4 \Rightarrow y = \frac{67}{17}, 3\) | M1 | Substitutes to find values of other variable |
| \(4 + 3\text{i}\), \(\frac{4}{17} + \frac{67}{17}\text{i}\) | A1 | Correct complex numbers |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Centre of circle $C$ is $(1, -1)$ | B1 | Correct coordinates of centre |
| $r = \sqrt{(5-1)^2 + (-4+1)^2} = 5$ or $r = \sqrt{(-3-1)^2 + (2+1)^2} = 5$ or $r = \frac{1}{2}\sqrt{(-3-5)^2 + (2+4)^2} = 5$ | M1 | Fully correct strategy for identifying radius. If diameter calculated must be halved |
| $\|z - 1 + \text{i}\| = 5$ or $\|z - (1-\text{i})\| = 5$ | A1 | Correct equation using required notation |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-1)^2 + (y+1)^2 = 25$, $(x-2)^2 + (y-3)^2 = 4$, expanding to give $\Rightarrow 2x + 8y = 32$ | M1 | Begins finding $z_1$ and $z_2$ using Cartesian equations to obtain line of intersection |
| $(16-4y)^2 - 4(16-4y) + 4 + y^2 - 6y + 9 = 4$ or $x^2 - 4x + 4 + \left(\frac{16-x}{4}\right)^2 - 6\left(\frac{16-x}{4}\right) + 9 = 4$ | M1 | Substitutes back into equation of one circle to obtain equation in one variable |
| $17y^2 - 118y + 201 = 0$ or $17x^2 - 72x + 16 = 0$ | A1 | Correct 3-term quadratic |
| $17y^2 - 118y + 201 = 0 \Rightarrow (17y-67)(y-3) = 0 \Rightarrow y = \frac{67}{17}, 3$ or $17x^2 - 72x + 16 = 0 \Rightarrow (17x-4)(x-4) = 0 \Rightarrow x = \frac{4}{17}, 4$ | M1 | Solves their 3TQ |
| $y = \frac{67}{17}, 3 \Rightarrow x = \frac{4}{17}, 4$ or $x = \frac{4}{17}, 4 \Rightarrow y = \frac{67}{17}, 3$ | M1 | Substitutes to find values of other variable |
| $4 + 3\text{i}$, $\frac{4}{17} + \frac{67}{17}\text{i}$ | A1 | Correct complex numbers |
---\begin{enumerate}
\item In an Argand diagram, the points $A$ and $B$ are represented by the complex numbers $- 3 + 2 \mathrm { i }$ and $5 - 4 \mathrm { i }$ respectively. The points $A$ and $B$ are the end points of a diameter of a circle $C$.\\
(a) Find the equation of $C$, giving your answer in the form
\end{enumerate}
$$| z - a | = b \quad a \in \mathbb { C } , \quad b \in \mathbb { R }$$
The circle $D$, with equation $| z - 2 - 3 i | = 2$, intersects $C$ at the points representing the complex numbers $z _ { 1 }$ and $z _ { 2 }$\\
(b) Find the complex numbers $z _ { 1 }$ and $z _ { 2 }$
\hfill \mbox{\textit{Edexcel CP2 2020 Q2 [9]}}