AS June 2024 Q4
4. A circle \(C\) in the complex plane has equation
\[|z - (-3 + 3\mathrm{i})| = \alpha|z - (1 + 3\mathrm{i})|\]where \(\alpha\) is a real constant with \(\alpha \gt 1\)
Given that the imaginary axis is a tangent to \(C\)
The circle \(C\) is contained in the region
\[R = \left\{z \in \mathbb{C} : \beta \leqslant \arg z \leqslant \frac{\pi}{2}\right\}\]Give your answer in radians to 3 significant figures. (6)
| Scheme | Marks | AO |
|---|---|---|
 | M1 | 1.1b |
| Correct circle, centre on line \(y = 3\) | A1 | 1.1b |
| (2) |
Notes
M1: For any circle drawn tangentially to imaginary axis. May be any quadrant.
A1: Correct circle, in quadrant 1, with centre on line \(y = 3\) and tangent to imaginary axis.
| Scheme | Marks | AO |
|---|---|---|
| Circle must touch at 3i \(\Rightarrow |3\mathrm{i} - (-3 + 3\mathrm{i})| = \alpha|3\mathrm{i} - (1 + 3\mathrm{i})| \Rightarrow |3| = \alpha\lvert -1 \rvert \Rightarrow \alpha = 3\) | B1 | 2.4 |
| (1) |
Notes
B1: Correct explanation given.
| Scheme | Marks | AO |
|---|---|---|
| E.g. Diametrically opposite point to 3i is \(x + 3\mathrm{i}\) where \(|x + 3| = 3|x - 1| \Rightarrow x + 3 = 3x - 3 \Rightarrow x = \ldots\) Or \(|x + \mathrm{i}y - (-3 + 3\mathrm{i})| = 3|x + \mathrm{i}y - (1 + 3\mathrm{i})|\) \(\Rightarrow (x + 3)^2 + (y - 3)^2 = \text{“}9\text{”}(x - 1)^2 + \text{“}9\text{”}(y - 3)^2\) \(\Rightarrow x^2 - 3x + y^2 - 6y + 9 = 0 \Rightarrow \left(x - \dfrac{3}{2}\right)^2 + (y - 3)^2 = \dfrac{9}{4}\) | M1 | 3.1a |
| Centre is \(\left(\dfrac{3}{2}, 3\right)\) | A1 | 1.1b |
| Radius is \(\dfrac{3}{2}\) | A1 | 1.1b |
| \(\beta = \theta - \phi\) where \(\theta = \arctan\left(\dfrac{3}{3/2}\right) = \ldots\) or \(\phi = \arcsin\left(\dfrac{3/2}{\sqrt{\left(3/2\right)^2 + 3^2}}\right)\) | M1 | 3.1a |
| Attempts both \(\theta = \arctan\left(\dfrac{3}{3/2}\right) = \ldots\) and \(\phi = \arcsin\left(\dfrac{3/2}{\sqrt{\left(3/2\right)^2 + 3^2}}\right)\) \(\Rightarrow \beta = \arctan 2 - \arcsin\dfrac{\sqrt{5}}{5}\) oe method. | M1 | 2.1 |
| = awrt 0.644 | A1 | 1.1b |
| (6) | ||
| (9 marks) |
Notes
M1: Correct method to deduce the centre or radius of the circle. May use geometry, or substitute \(z = x + \mathrm{i}y\) and find Cartesian equation first etc.
A1: Correct centre stated or implied.
A1: Correct radius stated or implied.
M1: Realises the right angle triangle and use it to find one relevant angle.
M1: Full method to find the maximum value for \(\beta\). E.g. Finds both relevant angles and subtracts as shown in scheme, or may use \(\dfrac{\pi}{2} - 2\arcsin\dfrac{1}{\sqrt{5}}\).
A1: Correct answer, awrt.