A2 June 2022 Q5
5. The locus of points \(z\) satisfies
\[|z + a\mathrm{i}| = 3|z - a|\]where \(a\) is an integer.
The locus is a circle with its centre in the third quadrant and radius \(\dfrac{3}{2}\sqrt{2}\)
Determine
| Scheme | Marks | AO |
|---|---|---|
| \((x)^2 + (y+a)^2 = 9[(x-a)^2 + y^2]\) or \(\sqrt{(x)^2 + (y+a)^2} = 3\sqrt{(x-a)^2 + y^2}\) | M1 | 2.1 |
| \(8x^2 - 18ax + 8y^2 - 2ay + 8a^2 \{= 0\}\) o.e. | A1 | 1.1b |
| \(x^2 - \dfrac{9}{4}ax + y^2 - \dfrac{1}{4}ay + a^2 = 0\) \(\Rightarrow \left(x - \dfrac{9}{8}a\right)^2 - \left(\dfrac{9}{8}a\right)^2 + \left(y - \dfrac{1}{8}a\right)^2 - \left(\dfrac{1}{8}a\right)^2 + a^2 = 0\) \(\Rightarrow r^2 = \left(\dfrac{9}{8}a\right)^2 + \left(\dfrac{1}{8}a\right)^2 - a^2 = \left(\dfrac{3}{2}\sqrt{2}\right)^2 \Rightarrow a = \ldots\) | M1 | 1.1b |
| \(a = -4\) cso | A1 | 2.2a |
| (4) |
Notes
M1: Obtains an equation in terms of \(x\) and \(y\) using the given information. Condone \((x)^2 + (y+a)^2 = 3[(x-a)^2 + y^2]\) for this mark
A1: Expands and simplifies the algebra, collects terms and obtains a correct simplified equation. Condone missing = 0
M1: Completes the square for their equation using \(x^2 + Ax = \left(x + \frac{A}{2}\right)^2 + \ldots\) Sets their radius squared \(= \left(\frac{3}{2}\sqrt{2}\right)^2 = \frac{18}{4} = \frac{9}{2}\) or radius \(= \left(\frac{3}{2}\sqrt{2}\right)\) and finds a value for \(a\).
Note the correct values are \(r^2 = \frac{9}{32}a^2,\ r = \frac{3a\sqrt{2}}{8}\)
A1: Deduces that \(a = -4\) cso
| Scheme | Marks | AO |
|---|---|---|
| \(\left(x - \frac{9}{8}(-4)\right)^2 + \left(y - \frac{1}{8}(-4)\right)^2 = \ldots\) \((x - \alpha)^2 + (y - \beta)^2 = \ldots\) implies centre \((\alpha,\ \beta)\) | M1 | 1.1b |
| centre \(\left(-\frac{9}{2},\ -\frac{1}{2}\right)\) | A1 | 2.2a |
| (2) | ||
| (6 marks) |
Notes
M1: Substitutes their value for \(a\) into their equation and finds their centre.
A1: Deduces the correct centre.