A2 June 2019 Q1
1. A complex number \(z = x + \mathrm{i}y\) is represented by the point \(P\) in an Argand diagram.
Given that
\[|z - 3| = 4|z + 1|\]| Scheme | Marks | AO |
|---|---|---|
| \((x - 3)^2 + y^2 = 16\left((x + 1)^2 + y^2\right)\) | M1 | 1.1b |
| \(x^2 - 6x + 9 + y^2 = 16x^2 + 32x + 16 + 16y^2\) \(15x^2 + 15y^2 + 38x + 7 = 0\) * | A1* | 2.1 |
| (2) |
Notes
M1: Obtains an equation in terms of \(x\) and \(y\) using the given information. Allow if the 4 is not squared, but i2 must have been dealt with correctly (ie positive \(y^2\) terms). Condone invisible brackets for the M mark.
A1*: Expands and simplifies and obtains a circle equation correctly. Accept terms in different order but must include =0. No errors seen, so bracketing errors in solution are A0.
| Scheme | Marks | AO |
|---|---|---|
| \(15x^2 + 15y^2 + 38x + 7 = 15\left(x \pm \dfrac{19}{15}\right)^2 - \ldots + 15y^2 + 7 = 0\) | M1 | 2.1 |
| Centre is \(\left(-\dfrac{\text{“}19\text{”}}{15}, 0\right)\) and radius is \(\sqrt{\left(\dfrac{\text{“}19\text{”}}{15}\right)^2 - \dfrac{7}{15}}\ \left(= \dfrac{16}{15}\right)\) | M1 | 2.2a |
| \(\max|z| = \dfrac{16}{15} + \dfrac{19}{15} = \dfrac{7}{3}\) | A1 | 3.1a |
| (3) | ||
| (5 marks) |
Notes
M1: Completes the square on the \(x\) term achieving \(A\left(x \pm \dfrac{19}{15}\right)^2 - B\), or uses other appropriate method in order to attempt the radius and/or centre of the circle. Award if correct \(x\) coordinate of centre or radius is found.
M1: Deduces both centre and radius for their completed square form, either seen used in work clearly as centre and radius, stated or labelled on a diagram, not just embedded within the equation. This is implied by the correct calculation being carried out for their centre and radius.
A1cso: Realises the need to add distance of centre from origin to radius to achieve the correct answer. Must come from correct work.
Note that completing the square as \(\left(x - \dfrac{19}{15}\right)^2 - \ldots\) can score a maximum M1M1A0
(corrected from the printed mark scheme: the centre is printed as \(\left(-\dfrac{\text{“}19\text{”}}{5}, 0\right)\))