AS June 2024 Paper 1 Q5

EdexcelCurrent spec10 marksComplex Numbers

5. Given that on an Argand diagram the locus of points defined by \(|z + 5 - 12\mathrm{i}| = 10\) is a circle,

(a) write down,

(i) the coordinates of the centre of this circle,

(ii) the radius of this circle.

(2)

(b) Show, by shading on an Argand diagram, the set of points defined by\[|z + 5 - 12\mathrm{i}| \leqslant 10\] (1)

The set of points \(A\) is defined by

\[A = \left\{z : 0 \leqslant \arg(z + 5 - 20\mathrm{i}) \leqslant \pi\right\} \cap \left\{z : |z + 5 - 12\mathrm{i}| \leqslant 10\right\}\]

(d) Determine the area of the region defined by \(A\), giving your answer to 3 significant figures. (4)

Scheme	Marks	AO
(i) \((-5, 12)\) or \(-5 + 12\mathrm{i}\)	B1	1.1b
(ii) \(r = 10\)	B1	1.1b
	(2)

Argand diagram with a shaded circle, centre in the second quadrant, radius 10 drawn to a point on the imaginary axis; the circle crosses the positive imaginary axis but not the real axis — Mark scheme (b)

Scheme	Marks	AO
	B1ft	1.1b
	(1)

Mark scheme (c)

Scheme	Marks	AO
\(OC = \sqrt{5^2 + 12^2}\)	M1	1.1b
\(\|z\|_{\max} = \sqrt{5^2 + 12^2} + 10\)	M1	3.1a
= 23	A1	1.1b
	(3)

Notes

M1: Calculates the distance from \(O\) to the centre of their circle.

M1: Fully correct strategy for the maximum. E.g. Finds distance from \(O\) to centre of their circle and adds their radius.

A1: Correct answer.
Correct answer with no working and following a correct centre and radius scores M1M1A1

Alternative

Scheme	Marks	AO
\(y = -\dfrac{12}{5}x\) and \((x + 5)^2 + (y - 12)^2 = 10^2\) \((x + 5)^2 + \left(-\dfrac{12}{5}x - 12\right)^2 = 10^2 \Rightarrow x = \ldots\) Or \(\tan\theta = \dfrac{5}{12} \Rightarrow \theta = \ldots\{22.61..\}\quad x = -5 - 10\sin\theta = \ldots\)	M1	1.1b
\(x = \text{‘}-\dfrac{115}{13}\text{’} \Rightarrow y = \ldots\left\{\dfrac{276}{13}\right\}\) \(\|z\|_{\max} = \sqrt{\left(-\dfrac{115}{13}\right)^2 + \left(\dfrac{276}{13}\right)^2}\) Or \(x = \text{‘}-\dfrac{15}{13}\text{’} \Rightarrow y = \ldots\left\{\dfrac{36}{13}\right\}\) \(\|z\|_{\max} = \sqrt{\left(-\dfrac{15}{13}\right)^2 + \left(\dfrac{36}{13}\right)^2} + 2 \times 10\)	M1	3.1a
= 23	A1	1.1b
	(3)

M1: Finds the equation of the line from the origin to centre and the Cartesian equation of the circle. Solves simultaneously to find the \(x\) values (or \(y\)) where the line intersects the circle.

M1: Selects the \(x\) coordinate to give the largest distance, find the corresponding \(y\) value and then the distance from the origin.
Selects the \(x\) coordinate to give the smallest distance, find the corresponding \(y\) value and then adds on 2 times the radius.

A1: Correct answer

Handwritten sketch of the circle with the segment above the line y = 20 shaded, showing radii 10, the distance 8 from the centre to the chord and 12 — Mark scheme (d)

Handwritten sketch of the circle with the segment above the line y = 20 shaded — Mark scheme (d)

Scheme	Marks	AO
\(\left\{z : 0 \leqslant \arg(z + 5 - 20\mathrm{i}) \leqslant \pi\right\} \Rightarrow y = 20\) \(\Rightarrow (x + 5)^2 + 8^2 = 100 \Rightarrow x = \ldots\) AND finds an angle \(\cos\theta = \dfrac{10^2 + 10^2 - 12^2}{2 \times 10 \times 10} = 0.28\) Or \(a^2 = 10^2 - 8^2 \Rightarrow a = \ldots\{6\}\quad \sin\left(\dfrac{1}{2}\theta\right) = \dfrac{6}{10}\) Or \(\cos\left(\dfrac{1}{2}\theta\right) = \dfrac{8}{10}\)	M1	3.1a
\(\theta = 1.287\ldots\) or 73.7° or \(\dfrac{1}{2}\theta = 0.6435\ldots\) or 36.9°	A1	1.1b
Area \(= \dfrac{1}{2} \times 10^2 \times \theta - \dfrac{1}{2} \times 12 \times 8\) angle in radians Area \(= \pi \times 10^2 \times \dfrac{\theta}{360} - \dfrac{1}{2} \times 12 \times 8\) angle in degrees or Area \(= \dfrac{1}{2} \times 10^2 \times \theta - \dfrac{1}{2} \times 10 \times 10 \times \sin\theta\) angle in radians Area \(= \pi \times 10^2 \times \dfrac{\theta}{360} - \dfrac{1}{2} \times 10 \times 10 \times \sin\theta\) angle in degrees Or Area \(= 2\left[\dfrac{1}{2} \times 10^2 \times \theta - \dfrac{1}{2} \times 8 \times 6\right]\)	M1	3.1a
= awrt 16.4	A1	1.1b
	(4)
	(10 marks)