Standard +0.8 This requires identifying that |z - 5 - 2i| ≤ √32 represents a disk, finding where the vertical line Re(z) = 9 intersects the circle, then calculating the area of a circular segment using geometry and trigonometry. While the individual components are standard Further Maths topics, combining them into a multi-step solution with exact form requires solid geometric visualization and careful calculation, making it moderately challenging for FP1 level.
4 Find, in exact form, the area of the region on an Argand diagram which represents the locus of points for which \(| z - 5 - 2 \mathrm { i } | \leqslant \sqrt { 32 }\) and \(\operatorname { Re } ( z ) \geqslant 9\).
Understanding that the first condition leads to a circle centre (5, 2) radius \(\sqrt{32}\) (could be stated or drawn)
\((9 - 5)^2 + (y - 2)^2 = 32\)
M1
Second condition is \(x = 9\) and attempt to solve simultaneously
\(y = 6\) or \(-2\)
A1
Area of triangle \(= \frac{1}{2} \times (6 - (-2)) \times (9 - 5) = 16\)
B1
\(\frac{\theta}{2} = \tan^{-1}\frac{4}{4}\) oe so sector angle is \(\frac{\pi}{2}\)
M1
Or observation that triangle is two right-angled isosceles triangles.
So area is \(8\pi - 16\)
A1
[6]
$(x - 5)^2 + (y - 2)^2 = 32$ oe | B1 | Understanding that the first condition leads to a circle centre (5, 2) radius $\sqrt{32}$ (could be stated or drawn)
$(9 - 5)^2 + (y - 2)^2 = 32$ | M1 | Second condition is $x = 9$ and attempt to solve simultaneously
$y = 6$ or $-2$ | A1 |
Area of triangle $= \frac{1}{2} \times (6 - (-2)) \times (9 - 5) = 16$ | B1 |
$\frac{\theta}{2} = \tan^{-1}\frac{4}{4}$ oe so sector angle is $\frac{\pi}{2}$ | M1 | Or observation that triangle is two right-angled isosceles triangles.
So area is $8\pi - 16$ | A1 |
| [6] |
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4 Find, in exact form, the area of the region on an Argand diagram which represents the locus of points for which $| z - 5 - 2 \mathrm { i } | \leqslant \sqrt { 32 }$ and $\operatorname { Re } ( z ) \geqslant 9$.
\hfill \mbox{\textit{OCR FP1 AS 2018 Q4 [6]}}