AS June 2025 Paper 1 Q5
5 The locus \(L\) is defined by \(L = \{z : z \in \mathbb{C}, |z - (20 + 15\mathrm{i})| \leqslant 7\}\).
Argand diagram printed in the Printed Answer Booklet:
| Scheme | Marks | AO |
|---|---|---|
| Circle drawn in the first quadrant | M1 | 1.1 |
| Circle has radius and centre \((20 + 15\mathrm{i})\) with a solid circumference and the interior labelled \(L\). | A1 | 1.1 |
| [2] |
Notes
A1: Allow labelled on diagram or in words next to diagram.
Ignore lines clearly added by candidates when attempting (b) and/or (c).
Allow \(r = 7\) to be implied from other points marked on diagram
Allow BOD if not labelled \(L\) as long as it is clear the correct region is being identified
Allow BOD if L outside the circle as long as the circle is shaded (or statement such as “inside the circle”)
If M0 then SCB1 for centre of form \((\pm 20, \pm 15\mathrm{i})\) and radius 7 soi
| Scheme | Marks | AO |
|---|---|---|
| Distance from \(O\) to centre \(= \sqrt{20^2 + 15^2} = 25\) | B1 | 3.1a |
| \(\therefore |z|_{\min} = 25 - 7 = 18\) so | M1 | 2.2a |
| \(\therefore z = \dfrac{18}{25}(20 + 15\mathrm{i}) = \dfrac{72}{5} + \dfrac{54}{5}\mathrm{i}\) | A1 | 1.1 |
| [3] |
Notes
B1: Must be used in solution.
Can be implied by sight of 25 in working
BOD centres of the form \((\pm 20, \pm 15\mathrm{i})\)
M1: Finding the minimum value of \(|z|\) (could be embedded in calculation).
A1: \(14.4 + 10.8\mathrm{i}\)
Condone answer left as \(\dfrac{18}{25}(20 + 15\mathrm{i})\) and ISW once allowable form seen.
If M0, SCB2 in total for correct answer.
Maximum mark with no working shown (other than finding distance of O from the centre) is SCB2.
Alternative method (ALT)
| Scheme | Marks |
|---|---|
| (Let the origin be \(O\), the centre of the circle \(C\) and the closest point be \(X\)) The angle between the \(x\)-axis and line \(OC\) is \(\tan^{-1}\frac{15}{20}\) \((= 0.6435\ldots)\) \(\overrightarrow{XC} = \begin{pmatrix} 7\cos 0.6435 \\ 7\sin 0.6435 \end{pmatrix}\) | M1 |
| \(\overrightarrow{XC} = \begin{pmatrix} 7\cos 0.6435 \\ 7\sin 0.6435 \end{pmatrix}\ \left[= \begin{pmatrix} 5.6 \\ 4.2 \end{pmatrix}\right]\) | A1FT |
| \(\begin{pmatrix} 20 \\ 15 \end{pmatrix} - \begin{pmatrix} 5.6 \\ 4.2 \end{pmatrix} = \begin{pmatrix} 14.4 \\ 10.8 \end{pmatrix}\) So point is at \(14.4 + 10.8\mathrm{i}\) | A1 |
M1: Finding angle from origin to centre and attempt at finding either the vertical or horizontal distance from centre to closest point
Only one distance needs to be attempted. Must be using a radius of 7
A1FT: Both correct, FT on centres of the form \((\pm 20, \pm 15\mathrm{i})\)
\(= \begin{pmatrix} \pm 5.6 \\ \pm 4.2 \end{pmatrix}\)
A1: Not FT here
Alternative method (ALT 2)
| Scheme | Marks |
|---|---|
| \(y = \dfrac{3}{4}x\) | B1 |
| \((x - 20)^2 + (y - 15)^2 = 49\) \((x - 20)^2 + \left(\dfrac{3}{4}x - 15\right)^2 = 49\) \(x^2 - 40x + 400 + \dfrac{9}{16}x^2 - \dfrac{45}{2}x + 225 = 49\) \(\dfrac{25}{16}x^2 - \dfrac{125}{2}x + 576 = 0\) \(25x^2 - 1000x + 9216 = 0\) | M1 |
| \((5x - 72)(5x - 128) = 0\) \(x = 14.4\) So point is at \(14.4 + 10.8\mathrm{i}\) | A1 |
B1: Finding the equation of the line between \(O\) and \(C\)
M1: Solving simultaneous equations which have come from valid attempts at the equation of the straight line \(OC\) and the circle equation. Must have eliminated one variable.
Might have some errors
Could use \(x^2 + y^2 = 18^2\)
A1: If both roots seen for \(x\) must both be correct.
| Scheme | Marks | AO |
|---|---|---|
| Either \(\tan^{-1}\dfrac{15}{20}\left(= \dfrac{3}{4}\right)\) or \(\sin^{-1}\dfrac{7}{25}\) (oe) seen | M1 | 3.1a |
| So required angle \(= \tan^{-1}\dfrac{3}{4} + \sin^{-1}\dfrac{7}{25}\) | M1 | 2.2a |
| \(=\) awrt 0.927 rads | A1 | 1.1 |
| [3] |
Notes
M1: Their 25. Might see \(\tan^{-1}\dfrac{7}{24}\) or \(\cos^{-1}\dfrac{24}{25}\)
\(\arctan(3/4) = 0.6435\ldots\)
\(\arcsin(7/25) = 0.28379\ldots\)
Allow BOD for sight of 0.644 or 0.284 or degree equivalents \(36.87^\circ\), \(16.263^\circ\)
M1: For reference, relevant point is \(\dfrac{72}{5} + \dfrac{96}{5}\mathrm{i}\) or \(14.4 + 19.2\mathrm{i}\)
A1: Or \(53.1^\circ\)
If either M0, SCB2 for correct answer.
Alternative method (ALT)
| Scheme | Marks |
|---|---|
| Either \(\tan^{-1}\frac{20}{15}\left(= \frac{4}{3}\right)\) or \(\sin^{-1}\dfrac{7}{25}\) (oe) seen | M1 |
| So required angle \(= \dfrac{\pi}{2} - \tan^{-1}\dfrac{4}{3} + \sin^{-1}\dfrac{7}{25}\) | M1 |
| \(=\) awrt 0.927 rads | A1 |
Alternative method (ALT 2)
| Scheme | Marks |
|---|---|
| \(y = mx\) is a tangent to \((x - 20)^2 + (y - 15)^2 = 49\) Therefore \(x^2 - 40x + 400 + m^2x^2 - 30mx + 225 = 49\) has a repeated root, hence the discriminant is 0 | M1 |
| \((40 + 30m)^2 - 2304(m^2 + 1) = 0\) \(1404m^2 - 2400m + 704 = 0\) \(m = \frac{4}{3}\) or \(m = \frac{44}{117}\) | M1 |
| \(m = \frac{4}{3}\) \(\theta = \tan^{-1}\left(\frac{4}{3}\right) = 0.927\) | A1 |
M1: Might see \(\tan\theta\) rather than \(m\)
Setting up a quadratic equation in \(x\) and \(y\) and considering the discriminant
M1: Setting up a quadratic equation in \(m\) and solving