A2 June 2022 Q8
8. The locus of points \(z = x + \mathrm{i}y\) that satisfy
\[\arg\left(\frac{z - 8 - 5\mathrm{i}}{z - 2 - 5\mathrm{i}}\right) = \frac{\pi}{3}\]is an arc of a circle \(C\).
| Scheme | Marks | AO |
|---|---|---|
 | B1 | 1.1b |
| The end points of their arc (2, 5) and (8, 5) and the arc drawn above the coordinates | B1 | 1.1b |
| (2) |
Notes
B1: Major arc drawn anywhere
B1: Correct end points for their arc drawn above the end points, condone written as complex numbers
| Scheme | Marks | AO |
|---|---|---|
| The centre lies on the perpendicular bisector/midpoint/equidistant of 2 and 8 | B1 | 2.4 |
| (1) |
Notes
B1: States perpendicular bisector or midpoint of 2 and 8. Condone “in between 2 and 8” if they write \(\frac{2+8}{2} = 5\)
Note: \(\frac{2+8}{2} = 5\) on its own is B0
In between 2 and 8 on its own is B0
| Scheme | Marks | AO |
|---|---|---|
| A complete method to find the radius of the circle\[\sin\left(\frac{\pi}{3}\right) = \frac{3}{r} \Rightarrow r = \ldots\]Or\[6^2 = r^2 + r^2 - 2 \times r \times r \times \cos\left(\frac{2\pi}{3}\right) \Rightarrow r = \ldots\]Or\[h = \frac{3}{\tan\left(\frac{\pi}{3}\right)} = \sqrt{3} \Rightarrow r = \sqrt{\left(\sqrt{3}\right)^2 + 3^2} = \ldots\]or\[\tan[(\arg(x + yi - (8 + 5i)) - \arg(x + yi - (2 + 5i))] = \tan\left(\frac{\pi}{3}\right)\]\[\frac{\tan[\arg(x + yi - (8 + 5i))] - \tan[\arg(x + yi - (2 + 5i))]}{1 + \tan[\arg(x + yi - (8 + 5i))]\tan[\arg(x + yi - (2 + 5i))]} = \sqrt{3}\]\[\frac{\frac{y-5}{x-8} - \frac{y-5}{x-2}}{1 + \frac{y-5}{x-8} \times \frac{y-5}{x-2}} = \sqrt{3}\]Leading to an equation of a circle by completing the square \((x - a)^2 + (y - b)^2 = r^2\) leading to \(r = \ldots\) | M1 | 3.1a |
| \(r = \dfrac{6}{\sqrt{3}}\) or \(2\sqrt{3}\) o.e. | A1 | 1.1b |
| (2) |
Notes
M1: A complete method to find the radius of the circle
A1: Correct radius
(corrected from the printed mark scheme: the last fraction is printed with \(\frac{y-8}{x-2}\) in two places; since \(\tan[\arg(x + yi - (2 + 5i))] = \frac{y-5}{x-2}\), it should read \(\frac{y-5}{x-2}\))
| Scheme | Marks | AO |
|---|---|---|
| \(y = 5 + h\) where \(h = \dfrac{3}{\tan\left(\frac{\pi}{3}\right)}\) or \(h = \text{‘}2\sqrt{3}\text{’}\cos\left(\frac{\pi}{3}\right)\) or \(h = \sqrt{\left(\text{‘}2\sqrt{3}\text{’}\right)^2 - 3^2}\) | M1 | 3.1a |
| \(y = 5 + \sqrt{3}\) | A1 | 2.2a |
| (2) | ||
| (7 Marks) |
Notes
M1: Any correct complete strategy. If they attempt to find the height (even if incorrect method) in part (c) then \(y = 5\) + their height
A1: Correct answer