A2 October 2020 Paper 2 Q2
2. 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 \(C\).
The circle \(D\), with equation \(|z - 2 - 3\mathrm{i}| = 2\), intersects \(C\) at the points representing the complex numbers \(z_1\) and \(z_2\)
| Scheme | Marks | AO |
|---|---|---|
| Centre of circle \(C\) is \((1, -1)\) | B1 | 1.1b |
| \(r = \sqrt{(5 - 1)^2 + (-4 + 1)^2} = 5\) or \(r = \sqrt{(-3 - 1)^2 + (2 + 1)^2} = 5\) or \(r = \dfrac{1}{2}\sqrt{(-3 - 5)^2 + (2 + 4)^2} = 5\) | M1 | 3.1a |
| \(|z - 1 + \mathrm{i}| = 5\) or \(|z - (1 - \mathrm{i})| = 5\) | A1 | 2.5 |
| (3) |
Notes
(a)
B1: Correct coordinates of centre
M1: Fully correct strategy for identifying the radius. If the diameter is calculated this must be halved to achieve this mark.
A1: Correct equation using the required notation
| Scheme | Marks | AO |
|---|---|---|
| \((x - 1)^2 + (y + 1)^2 = 25,\quad (x - 2)^2 + (y - 3)^2 = 4\) \(x^2 - 2x + 1 + y^2 + 2y + 1 = 25\) \(x^2 - 4x + 4 + y^2 - 6y + 9 = 4\) \(\Rightarrow 2x + 8y = 32\) | M1 | 3.1a |
| \((16 - 4y)^2 - 4(16 - 4y) + 4 + y^2 - 6y + 9 = 4\) or \(x^2 - 4x + 4 + \left(\dfrac{16 - x}{4}\right)^2 - 6\left(\dfrac{16 - x}{4}\right) + 9 = 4\) | M1 | 1.1b |
| \(17y^2 - 118y + 201 = 0\) or \(17x^2 - 72x + 16 = 0\) | A1 | 1.1b |
| \(17y^2 - 118y + 201 = 0 \Rightarrow (17y - 67)(y - 3) = 0 \Rightarrow y = \dfrac{67}{17}, 3\) or \(17x^2 - 72x + 16 = 0 \Rightarrow (17x - 4)(x - 4) = 0 \Rightarrow x = \dfrac{4}{17}, 4\) | M1 | 1.1b |
| \(y = \dfrac{67}{17}, 3 \Rightarrow x = \dfrac{4}{17}, 4\) or \(x = \dfrac{4}{17}, 4 \Rightarrow y = \dfrac{67}{17}, 3\) | M1 | 2.1 |
| \(4 + 3\mathrm{i},\ \dfrac{4}{17} + \dfrac{67}{17}\mathrm{i}\) | A1 | 2.2a |
| (6) | ||
| (9 marks) |
Notes
(b)
M1: Begins the process of finding \(z_1\) and \(z_2\) by using the Cartesian equations to obtain the equation of the line of intersection
M1: Substitutes back into the equation of one of the circles to obtain an equation in one variable
A1: Correct 3 term quadratic
M1: Solves their 3TQ
M1: Substitutes to find values of the other variable to complete the process of finding \(z_1\) and \(z_2\)
A1: Correct complex numbers