AS October 2020 Paper 1 Q10
10. Given that there are two distinct complex numbers \(z\) that satisfy
\[\left\{z : |z - 3 - 5\mathrm{i}| = 2r\right\} \cap \left\{z : \arg(z - 2) = \frac{3\pi}{4}\right\}\]determine the exact range of values for the real constant \(r\). (7)
| Scheme | Marks | AO |
|---|---|---|
| \((x - 3)^2 + (y - 5)^2 = (2r)^2\) and \(y = -x + 2\) | B1 | 1.1b |
| \[(x - 3)^2 + (-x + 2 - 5)^2 = (2r)^2\]or\[(-y + 2 - 3)^2 + (y - 5)^2 = (2r)^2\] | M1 | 3.1a |
| \[2x^2 + 18 - 4r^2 = 0\]or\[2y^2 - 8y + 26 - 4r^2 = 0\] | A1 | 1.1b |
| \[b^2 - 4ac \gt 0 \Rightarrow 0^2 - 4(2)\left(18 - 4r^2\right) \gt 0 \Rightarrow r \gt \ldots\]or\[x^2 = 2r^2 - 9 \Rightarrow 2r^2 - 9 \gt 0 \Rightarrow r \gt \ldots\]or\[b^2 - 4ac \gt 0 \Rightarrow (-8)^2 - 4(2)\left(26 - 4r^2\right) \gt 0 \Rightarrow r \gt \ldots\] | dM1 | 3.1a |
| Finds a maximum value for \(r\) \((2r)^2 = 5^2 + (3 - 2)^2 \Rightarrow r = \ldots\) | M1 | 3.1a |
| \(\dfrac{3\sqrt{2}}{2} \lt r \lt \dfrac{\sqrt{26}}{2}\) o.e. | A1 A1 | 1.1b 1.1b |
| (7) | ||
| (7 marks) |
Notes
(Corrected from the printed mark scheme: the second method prints \(x^2 = 9 - 2r^2 \Rightarrow 9 - 2r^2 \gt 0\); from \(2x^2 + 18 - 4r^2 = 0\), \(x^2 = 2r^2 - 9\), so the condition is \(2r^2 - 9 \gt 0\).)
B1: Correct equations for each loci of points
M1: A complete method to find a 3TQ involving one variable using equations of the form \((x \pm 3)^2 + (y \pm 5)^2 = (2r)^2\) or \(2r^2\) or \(r^2\) and \(y = \pm x \pm 2\)
A1: Correct quadratic equation
dM1: Dependent on previous method mark. A complete method uses \(b^2 - 4ac \gt 0\) or rearranges to find \(x^2 = \mathrm{f}(r)\) and uses \(\mathrm{f}(r) \gt 0\) to the minimum value of \(r\).
M1: Realises there will be an upper limit for \(r\) and uses Pythagoras theorem
\((2r)^2 = (y\text{ coord of centre})^2 + (x\text{ coord of centre} - 2)^2\)
condone \((r)^2 = (y\text{ coord of centre})^2 + (x\text{ coord of centre} - 2)^2\)
A1: One correct limit, either \(\dfrac{3\sqrt{2}}{2} \lt r\) or \(r \lt \dfrac{\sqrt{26}}{2}\) o.e.
A1: Fully correct inequality
Alternative
| Scheme | Marks | AO |
|---|---|---|
| Using a circle with centre (3, 5) and radius \(2r\) and \(y = -x + 2\) | B1 | 1.1b |
| \(y - 5 = 1(x - 3) \Rightarrow y = x + 2\) \(x + 2 = -x + 2 \Rightarrow x = \ldots\) | M1 | 3.1a |
| \((0, 2)\) | A1 | 1.1b |
| \(2r \gt \sqrt{(3 - 0)^2 + (5 - 2)^2} \Rightarrow r \gt \ldots\) | dM1 | 3.1a |
| Finds a maximum value for \(r\) \((2r)^2 = 5^2 + (3 - 2)^2 \Rightarrow r = \ldots\) | M1 | 3.1a |
| \(\dfrac{3\sqrt{2}}{2} \lt r \lt \dfrac{\sqrt{26}}{2}\) o.e. | A1 A1 | 1.1b 1.1b |
| (7) |
B1: Using a circle with centre (3, 5) and radius \(2r\) and \(y = -x + 2\)
M1: A complete method to find the point of intersection of the line \(y = \pm x \pm 2\) and circle where the line is a tangent to the circle.
A1: Correct point of intersection
dM1: Finds the distance between the point of intersection and the centre and uses this to find the minimum value of \(r\). Condone radius of r.
M1: Realises there will be an upper limit for \(r\) and uses Pythagoras theorem
\((2r)^2 = (y\text{ coord of centre})^2 + (x\text{ coord of centre} - 2)^2\)
A1: One correct limit, either \(\dfrac{3\sqrt{2}}{2} \lt r\) or \(r \lt \dfrac{\sqrt{26}}{2}\) o.e.
A1: Fully correct inequality