| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Intersection of two loci |
| Difficulty | Challenging +1.2 This question requires understanding that the first locus is a circle centered at (3,5) with radius 2r, and the second is a half-line from (2,0) at angle 3π/4. Finding when they intersect in exactly two points involves calculating the perpendicular distance from the center to the line (which equals 4√2) and applying the condition r > distance for two intersections. While it requires geometric visualization and careful calculation, it's a standard loci intersection problem with straightforward application of distance formulas—typical of AS Core Pure material but slightly above average due to the optimization aspect. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-3)^2+(y-5)^2=(2r)^2\) and \(y=-x+2\) | B1 | Correct equations for each locus of points |
| \((x-3)^2+(-x+2-5)^2=(2r)^2\) or \((-y+2-3)^2+(y-5)^2=(2r)^2\) | M1 | Complete method to find 3TQ in one variable using equations of the form \((x\pm3)^2+(y\pm5)^2=(2r)^2\) or \(2r^2\) or \(r^2\) and \(y=\pm x\pm 2\) |
| \(2x^2+18-4r^2=0\) or \(2y^2-8y+26-4r^2=0\) | A1 | Correct quadratic equation |
| \(b^2-4ac>0 \Rightarrow 0^2-4(2)(18-4r^2)>0 \Rightarrow r>\ldots\) or \(x^2=9-2r^2 \Rightarrow 9-2r^2>0 \Rightarrow r>\ldots\) or \(b^2-4ac>0 \Rightarrow (-8)^2-4(2)(26-4r^2)>0 \Rightarrow r>\ldots\) | dM1 | Uses discriminant \(>0\) or rearranges to find minimum \(r\); dependent on M1 |
| Finds maximum value for \(r\): \((2r)^2=5^2+(3-2)^2 \Rightarrow r=\ldots\) | M1 | Finds maximum value for \(r\) |
| \(\frac{3\sqrt{2}}{2} < r < \frac{\sqrt{26}}{2}\) | A1, A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Circle with centre \((3,5)\) and radius \(2r\); \(y=-x+2\) | B1 | Correct equations for each locus |
| \(y-5=1(x-3) \Rightarrow y=x+2\); \(x+2=-x+2 \Rightarrow x=\ldots\); \((0,2)\) | M1, A1 | Finds perpendicular from centre to line; correct intersection point |
| \(2r > \sqrt{(3-0)^2+(5-2)^2} \Rightarrow r>\ldots\) | dM1 | Uses distance condition; dependent on M1 |
| \((2r)^2=5^2+(3-2)^2 \Rightarrow r=\ldots\) | M1 | Finds maximum \(r\) |
| \(\frac{3\sqrt{2}}{2} < r < \frac{\sqrt{26}}{2}\) | A1, A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(b^2 - 4ac > 0\) or rearranges to find \(x^2 = f(r)\) and uses \(f(r) > 0\) to find minimum value of \(r\) | dM1 | Dependent on previous method mark. Complete method required. |
| Realises there will be an upper limit for \(r\) and uses Pythagoras theorem: \((2r)^2 = (y \text{ coord of centre})^2 + (x \text{ coord of centre} - 2)^2\) | M1 | Condone \((r)^2 = (y \text{ coord of centre})^2 + (x \text{ coord of centre} - 2)^2\) |
| One correct limit, either \(\frac{3\sqrt{2}}{2} < r\) or \(r < \frac{\sqrt{26}}{2}\) | A1 | o.e. |
| Fully correct inequality: \(\frac{3\sqrt{2}}{2} < r < \frac{\sqrt{26}}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Circle with centre \((3, 5)\) and radius \(2r\), and line \(y = -x + 2\) | B1 | |
| Complete method to find point of intersection of line \(y = \pm x \pm 2\) and circle where line is tangent to circle | M1 | |
| Correct point of intersection | A1 | |
| Finds distance between point of intersection and centre, uses this to find minimum value of \(r\). Condone radius of \(r\) | dM1 | |
| \((2r)^2 = (y \text{ coord of centre})^2 + (x \text{ coord of centre} - 2)^2\) | M1 | Realises upper limit exists; uses Pythagoras |
| One correct limit, either \(\frac{3\sqrt{2}}{2} < r\) or \(r < \frac{\sqrt{26}}{2}\) | A1 | o.e. |
| Fully correct inequality | A1 |
## Question 10:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-3)^2+(y-5)^2=(2r)^2$ and $y=-x+2$ | B1 | Correct equations for each locus of points |
| $(x-3)^2+(-x+2-5)^2=(2r)^2$ or $(-y+2-3)^2+(y-5)^2=(2r)^2$ | M1 | Complete method to find 3TQ in one variable using equations of the form $(x\pm3)^2+(y\pm5)^2=(2r)^2$ or $2r^2$ or $r^2$ and $y=\pm x\pm 2$ |
| $2x^2+18-4r^2=0$ or $2y^2-8y+26-4r^2=0$ | A1 | Correct quadratic equation |
| $b^2-4ac>0 \Rightarrow 0^2-4(2)(18-4r^2)>0 \Rightarrow r>\ldots$ or $x^2=9-2r^2 \Rightarrow 9-2r^2>0 \Rightarrow r>\ldots$ or $b^2-4ac>0 \Rightarrow (-8)^2-4(2)(26-4r^2)>0 \Rightarrow r>\ldots$ | dM1 | Uses discriminant $>0$ or rearranges to find minimum $r$; dependent on M1 |
| Finds maximum value for $r$: $(2r)^2=5^2+(3-2)^2 \Rightarrow r=\ldots$ | M1 | Finds maximum value for $r$ |
| $\frac{3\sqrt{2}}{2} < r < \frac{\sqrt{26}}{2}$ | A1, A1 | Correct final answer |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Circle with centre $(3,5)$ and radius $2r$; $y=-x+2$ | B1 | Correct equations for each locus |
| $y-5=1(x-3) \Rightarrow y=x+2$; $x+2=-x+2 \Rightarrow x=\ldots$; $(0,2)$ | M1, A1 | Finds perpendicular from centre to line; correct intersection point |
| $2r > \sqrt{(3-0)^2+(5-2)^2} \Rightarrow r>\ldots$ | dM1 | Uses distance condition; dependent on M1 |
| $(2r)^2=5^2+(3-2)^2 \Rightarrow r=\ldots$ | M1 | Finds maximum $r$ |
| $\frac{3\sqrt{2}}{2} < r < \frac{\sqrt{26}}{2}$ | A1, A1 | Correct final answer |
## Question (Circle/Line Intersection with radius bounds):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $b^2 - 4ac > 0$ or rearranges to find $x^2 = f(r)$ and uses $f(r) > 0$ to find minimum value of $r$ | dM1 | Dependent on previous method mark. Complete method required. |
| Realises there will be an upper limit for $r$ and uses Pythagoras theorem: $(2r)^2 = (y \text{ coord of centre})^2 + (x \text{ coord of centre} - 2)^2$ | M1 | Condone $(r)^2 = (y \text{ coord of centre})^2 + (x \text{ coord of centre} - 2)^2$ |
| One correct limit, either $\frac{3\sqrt{2}}{2} < r$ or $r < \frac{\sqrt{26}}{2}$ | A1 | o.e. |
| Fully correct inequality: $\frac{3\sqrt{2}}{2} < r < \frac{\sqrt{26}}{2}$ | A1 | |
**Alternative Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Circle with centre $(3, 5)$ and radius $2r$, and line $y = -x + 2$ | B1 | |
| Complete method to find point of intersection of line $y = \pm x \pm 2$ and circle where line is tangent to circle | M1 | |
| Correct point of intersection | A1 | |
| Finds distance between point of intersection and centre, uses this to find minimum value of $r$. Condone radius of $r$ | dM1 | |
| $(2r)^2 = (y \text{ coord of centre})^2 + (x \text{ coord of centre} - 2)^2$ | M1 | Realises upper limit exists; uses Pythagoras |
| One correct limit, either $\frac{3\sqrt{2}}{2} < r$ or $r < \frac{\sqrt{26}}{2}$ | A1 | o.e. |
| Fully correct inequality | A1 | |\begin{enumerate}
\item Given that there are two distinct complex numbers $z$ that satisfy
\end{enumerate}
$$\{ z : | z - 3 - 5 i | = 2 r \} \cap \quad z : \arg ( z - 2 ) = \frac { 3 \pi } { 4 }$$
determine the exact range of values for the real constant $r$.
\hfill \mbox{\textit{Edexcel CP AS 2020 Q10 [7]}}