AQA Further Paper 2 2019 June — Question 6 6 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeOptimization of argument on loci
DifficultyChallenging +1.8 This is a Further Maths question requiring geometric insight about circles in the complex plane, specifically that the point with minimum argument lies where the line from the origin is tangent to the circle. Students must recognize this tangency condition, apply Pythagoras to find the radius, and work with exact values involving surds. While the calculation is straightforward once the geometric insight is achieved, recognizing the tangency property and setting up the proof correctly elevates this above standard complex number exercises.
Spec4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines
A circle \(C\) in the complex plane has equation \(|z - 2 - 5\mathrm{i}| = a\) The point \(z_1\) on \(C\) has the least argument of any point on \(C\), and \(\arg(z_1) = \frac{\pi}{4}\) Prove that \(a = \frac{3\sqrt{2}}{2}\) [6 marks]
Question 6:
AnswerMarks
6Sets as equal the real and
imaginary parts of
AnswerMarks Guidance
or uses the lineAO2.2a M1
Gradient of line co𝑧𝑧n 1 n=ecπ‘˜π‘˜ti+ngπ‘˜π‘˜ (i2,5) and
∴
(π‘˜π‘˜,π‘˜π‘˜)=βˆ’1
5βˆ’π‘˜π‘˜
=βˆ’1
2βˆ’π‘˜π‘˜
5βˆ’π‘˜π‘˜ =π‘˜π‘˜βˆ’2
π‘˜π‘˜ =3.5
2 2 2
π‘Žπ‘Ž =(5βˆ’3.5) +(2βˆ’ 3.5) =4.5
3√2
π‘Žπ‘Ž =
2
Uses the fact that th𝑧𝑧e 1 half-
line is a tangent 𝑦𝑦to= thπ‘Žπ‘Že
AnswerMarks Guidance
circleAO2.2a M1
Forms an equation for the
gradient of the line joining
(2, 5) and (k, k)
or the gradient of a
tangent at any point on C
or
uses a right angled
triangle containing the line
joining (2, 5) and (k, k)
or
correctly substitutes
into the equation of the
AnswerMarks Guidance
circle 𝑦𝑦 = π‘Žπ‘ŽAO3.1a M1
Forms a correct equation
based on their method.
AnswerMarks Guidance
PI byAO2.2a M1
Obtains the value of
AnswerMarks Guidance
from tπ‘˜π‘˜he=ir e3q.5uationAO1.1b A1
Produces a completeπ‘Žπ‘Žly
correct, rigorous proof
leading to exact value of .
Must show all steps clearly
AnswerMarks Guidance
π‘Žπ‘ŽAO2.1 R1
Total6
QMarking Instructions AO 
Question 6:
6 | Sets as equal the real and
imaginary parts of
or uses the line | AO2.2a | M1 | Radius is perpendicular to tangent
Gradient of line co𝑧𝑧n 1 n=ecπ‘˜π‘˜ti+ngπ‘˜π‘˜ (i2,5) and
∴
(π‘˜π‘˜,π‘˜π‘˜)=βˆ’1
5βˆ’π‘˜π‘˜
=βˆ’1
2βˆ’π‘˜π‘˜
5βˆ’π‘˜π‘˜ =π‘˜π‘˜βˆ’2
π‘˜π‘˜ =3.5
2 2 2
π‘Žπ‘Ž =(5βˆ’3.5) +(2βˆ’ 3.5) =4.5
3√2
π‘Žπ‘Ž =
2
Uses the fact that th𝑧𝑧e 1 half-
line is a tangent 𝑦𝑦to= thπ‘Žπ‘Že
circle | AO2.2a | M1
Forms an equation for the
gradient of the line joining
(2, 5) and (k, k)
or the gradient of a
tangent at any point on C
or
uses a right angled
triangle containing the line
joining (2, 5) and (k, k)
or
correctly substitutes
into the equation of the
circle 𝑦𝑦 = π‘Žπ‘Ž | AO3.1a | M1
Forms a correct equation
based on their method.
PI by | AO2.2a | M1
Obtains the value of
from tπ‘˜π‘˜he=ir e3q.5uation | AO1.1b | A1
Produces a completeπ‘Žπ‘Žly
correct, rigorous proof
leading to exact value of .
Must show all steps clearly
π‘Žπ‘Ž | AO2.1 | R1
Total | 6
Q | Marking Instructions | AO | Marks | Typical solution
A circle $C$ in the complex plane has equation $|z - 2 - 5\mathrm{i}| = a$

The point $z_1$ on $C$ has the least argument of any point on $C$, and $\arg(z_1) = \frac{\pi}{4}$

Prove that $a = \frac{3\sqrt{2}}{2}$
[6 marks]

\hfill \mbox{\textit{AQA Further Paper 2 2019 Q6 [6]}}
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