AS June 2019 Q3
3. A curve \(C\) in the complex plane is described by the equation
\[|z - 1 - 8\mathrm{i}| = 3|z - 1|\]| Scheme | Marks | AO |
|---|---|---|
| \((x-1)^2 + (y-8)^2 = 9\left[(x-1)^2 + y^2\right]\) Or \(\sqrt{(x-1)^2 + (y-8)^2} = 3\sqrt{(x-1)^2 + y^2}\) | M1 | 2.1 |
| \(8x^2 - 16x + 8y^2 + 16y - 56 = 0\) | A1 | 1.1b |
| \(x^2 - 2x + y^2 + 2y - 7 = 0\) so \((x-1)^2 + (y+1)^2 = 9\) and finds the centre and radius | M1 | 1.1b |
| Therefore, a circle with centre \((1, -1)\) and radius = 3 | A1 | 2.2a |
| (4) |
Notes
M1: Obtains an equation in terms of \(x\) and \(y\) using the given information. Condone \((x-1)^2 + (y-8)^2 = 3\left[(x-1)^2 + y^2\right]\) for this mark.
A1: Expands and simplifies the algebra, collecting terms and obtains a correct equation.
M1: Completes the square for their equation to find the centre and radius.
A1: Deduces that it is a circle (may be seen anywhere in their solution) with centre \((1, -1)\) and radius = 3
| Scheme | Marks | AO |
|---|---|---|
| Distance \(= \sqrt{(3-1)^2 + (-3--1)^2} = \ldots\) or finds \((d^2 =)\,(3-1)^2 + (-3--1)^2 = \ldots\) | M1 | 1.1b |
| Distance \(= \sqrt{8} = 2.828 \lt 3 \;\therefore\; z = 3 - 3\mathrm{i}\) satisfies the inequality Or \(8 \lt 9 \;\therefore\; z = 3 - 3\mathrm{i}\) satisfies the inequality | A1 | 2.2a |
| (2) |
Notes
M1: Finds the distance between \((3, -3)\) and their centre or \(d^2\) (note: correct centre is \((1, -1)\))
A1: Compares distance with 3 or compares \(d^2\) with 9 and deduces that the inequality is satisfied – must be using correct centre and radius.
| Scheme | Marks | AO |
|---|---|---|
 | M1 | 1.1b |
| Circle with centre in the fourth quadrant | A1 | 1.1b |
| Half line drawn from \((0, -1)\) and passes through the \(x\)-axis within the circle | M1 | 1.1b |
| Correct region shaded | A1 | 2.2a |
| (4) | ||
| (10 marks) |
Notes
M1: Circle for their centre and radius.
A1: Correct circle with centre in the fourth quadrant and passing through all four quadrants. Condone dotted circle.
M1: Half line drawn from \((0, -1)\) and passing the \(x\)-axis within the circle. Condone dotted line.
A1: Correct region shaded with both half-line and circle correct and not dotted.
Special case: M1A1M1A0 if no coordinates stated throughout and it is clear that the half-line intersects the coordinate axes level with the correct centre of the circle.