A2 June 2019 Paper 2 Q6
6 A circle \(C\) in the complex plane has equation \(|z - 2 - 5\mathrm{i}| = a\)
The point \(z_1\) on \(C\) has the least argument of any point on \(C\), and \(\arg(z_1) = \dfrac{\pi}{4}\)
Prove that \(a = \dfrac{3\sqrt{2}}{2}\) [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Sets as equal the real and imaginary parts of \(z_1\) or uses the line \(y = x\) | M1 | 2.2a |
| Uses the fact that the half-line is a tangent to the circle | M1 | 2.2a |
| Forms an equation for the gradient of the line joining (2, 5) and \((k, k)\) or the gradient of a tangent at any point on \(C\) or uses a right angled triangle containing the line joining (2, 5) and \((k, k)\) or correctly substitutes \(y = x\) into the equation of the circle | M1 | 3.1a |
| Forms a correct equation based on their method. PI by \(k = 3.5\) | M1 | 2.2a |
| Obtains the value of \(a\) from their equation | A1 | 1.1b |
| Produces a completely correct, rigorous proof leading to exact value of \(a\). Must show all steps clearly | R1 | 2.1 |
| (6 marks) |
Typical solution

Radius is perpendicular to tangent \(\therefore\)
Gradient of line connecting (2,5) and \((k, k) = -1\)