Question 6 - A-Level Maths

WJEC Further Unit 1 2023 June — Question 6 6 marks

Exam Board	WJEC
Module	Further Unit 1 (Further Unit 1)
Year	2023
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex Numbers Argand & Loci
Type	Circle of Apollonius locus
Difficulty	Standard +0.3 This is a standard locus problem requiring substitution of z = x + iy, expanding modulus expressions, and algebraic manipulation to reach circle form. The technique is routine for Further Maths students, though the algebra with the factor of 2 requires careful execution. Slightly easier than average due to being a well-practiced question type with clear methodology.
Spec	4.02k Argand diagrams: geometric interpretation 4.02o Loci in Argand diagram: circles, half-lines

6. The complex number $z$ is represented by the point $P ( x , y )$ in an Argand diagram. Given that $$| z - 3 + \mathrm { i } | = 2 | z - 5 - 2 \mathrm { i } |$$ show that the locus of $P$ is a circle and write down the coordinates of its centre.

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Answer	Marks	Guidance
Putting $z = x + iy$: \(	x + iy - 3 + i	= 2
$\sqrt{(x - 3)^2 + (y + 1)^2} = 2\sqrt{(x - 5)^2 + (y - 2)^2}$	m1
$(x - 3)^2 + (y + 1)^2 = 4[(x - 5)^2 + (y - 2)^2]$	A1
$x^2 - 6x + 9 + y^2 + 2y + 1 = 4x^2 - 40x + 100 + 4y^2 - 16y + 16$	A1	oe
$3x^2 + 3y^2 - 34x - 18y + 106 = 0$, which is the standard form of a circle.	A1	or equivalent form of a circle: $\left(x - \frac{17}{3}\right)^2 + (y - 3)^2 = \frac{52}{9}$
Centre $= \left(\frac{17}{3}, 3\right)$	A1	FT provided coefficients of $x^2$ and $y^2$ are equal

Total: [6]

Putting $z = x + iy$: $|x + iy - 3 + i| = 2|x + iy - 5 - 2i|$, $|(x - 3) + i(y + 1)| = 2|(x - 5) + i(y - 2)|$ | M1 |

$\sqrt{(x - 3)^2 + (y + 1)^2} = 2\sqrt{(x - 5)^2 + (y - 2)^2}$ | m1 |

$(x - 3)^2 + (y + 1)^2 = 4[(x - 5)^2 + (y - 2)^2]$ | A1 |

$x^2 - 6x + 9 + y^2 + 2y + 1 = 4x^2 - 40x + 100 + 4y^2 - 16y + 16$ | A1 | oe

$3x^2 + 3y^2 - 34x - 18y + 106 = 0$, which is the standard form of a circle. | A1 | or equivalent form of a circle: $\left(x - \frac{17}{3}\right)^2 + (y - 3)^2 = \frac{52}{9}$

Centre $= \left(\frac{17}{3}, 3\right)$ | A1 | FT provided coefficients of $x^2$ and $y^2$ are equal

**Total: [6]**

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6. The complex number $z$ is represented by the point $P ( x , y )$ in an Argand diagram. Given that

$$| z - 3 + \mathrm { i } | = 2 | z - 5 - 2 \mathrm { i } |$$

show that the locus of $P$ is a circle and write down the coordinates of its centre.\\

\hfill \mbox{\textit{WJEC Further Unit 1 2023 Q6 [6]}}

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Q1 4 Q2 5 Q3 6 Q4 7 Q5 6 Q6 6 Q7 7 Q8 9 Q9 12 Q10 8

Answer	Marks	Guidance
Putting \(z = x + iy\): \(	x + iy - 3 + i	= 2
\(\sqrt{(x - 3)^2 + (y + 1)^2} = 2\sqrt{(x - 5)^2 + (y - 2)^2}\)	m1
\((x - 3)^2 + (y + 1)^2 = 4[(x - 5)^2 + (y - 2)^2]\)	A1
\(x^2 - 6x + 9 + y^2 + 2y + 1 = 4x^2 - 40x + 100 + 4y^2 - 16y + 16\)	A1	oe
\(3x^2 + 3y^2 - 34x - 18y + 106 = 0\), which is the standard form of a circle.	A1	or equivalent form of a circle: \(\left(x - \frac{17}{3}\right)^2 + (y - 3)^2 = \frac{52}{9}\)
Centre \(= \left(\frac{17}{3}, 3\right)\)	A1	FT provided coefficients of \(x^2\) and \(y^2\) are equal