| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Distance from centre to line |
| Difficulty | Moderate -0.3 This is a straightforward application of completing the square to find centre and radius, followed by using the perpendicular distance formula from a point to a line. All techniques are standard AS-level procedures with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-step nature. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x\pm5)^2+(y\pm4)^2\) | M1 | Attempts to complete the square for both \(x\) and \(y\) terms; may be implied by centre \((\pm5,\pm4)\) |
| Centre is \((5,4)\) | A1 | |
| Radius is \(3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2y+x+6=0 \Rightarrow y=-\frac{1}{2}x+... \Rightarrow -\frac{1}{2}\rightarrow 2\) | B1 | Deduces gradient of perpendicular to \(l\) is \(2\); may be seen in equation for perpendicular line |
| \(m_N=2 \Rightarrow y-4=2(x-5)\) and \(2y+x+6=0 \Rightarrow x=..., y=...\) | M1 | Fully correct strategy for finding intersection; must use perpendicular gradient (not gradient of \(l\)); look for \(y-"4"="2"(x-"5")\) solved simultaneously with equation of \(l\); do not penalise rearrangement mechanics; note: finding \(y\)-intercept of \(l\) at \((0,-3)\) is M0 |
| Intersection is at \(\left(\frac{6}{5},-\frac{18}{5}\right)\) | A1 | |
| Distance from centre to intersection is \(\sqrt{\left(5-\frac{6}{5}\right)^2+\left(4+\frac{18}{5}\right)^2}\) so distance required is \(\sqrt{\left("5"-"\frac{6}{5}"\right)^2+\left("4"+"\frac{18}{5}"\right)^2}-"3"\) | dM1 | Dependent on previous M1 |
| \(=\frac{19\sqrt{5}}{5}-3\) (or awrt \(5.50\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Centre \(\left(\frac{6}{5}, -\frac{18}{5}\right)\) or equivalent e.g. \((1.2, -3.6)\) | A1 | Coordinates need not be explicit; may appear in working or on diagram |
| Correct strategy: use Pythagoras to find distance between centre and intersection, then subtract radius | dM1 | Dependent on previous method mark. Condone sign slip subtracting \(-\frac{18}{5}\). Alt: solve \(y=2x-6\) simultaneously with circle equation, find intersection, choose smaller positive root |
| \((x-5)^2+(2x-6-4)^2=9 \Rightarrow 5x^2-50x+125=9\), \(x=\frac{25-3\sqrt{5}}{5}\), \(y=\frac{20-6\sqrt{5}}{5}\) | — | Example working for alternative method |
| \(\sqrt{\frac{361}{5}}-3\) or \(\frac{19\sqrt{5}-15}{5}\), also allow awrt 5.50 | A1 | Isw after correct answer seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempt perpendicular distance from \((5,4)\) to \(x+2y+6=0\) using \(\frac{\ | ax+by+c\ | }{\sqrt{a^2+b^2}}\) |
| \(\frac{\ | 5\times1+4\times2+6\ | }{\sqrt{1^2+2^2}} = \frac{19}{\sqrt{5}}\) |
| Distance \(= \frac{19\sqrt{5}}{5} - 3\) | dM1 | |
| \(\frac{19\sqrt{5}-15}{5}\) | A1 |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x\pm5)^2+(y\pm4)^2$ | M1 | Attempts to complete the square for both $x$ and $y$ terms; may be implied by centre $(\pm5,\pm4)$ |
| Centre is $(5,4)$ | A1 | |
| Radius is $3$ | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2y+x+6=0 \Rightarrow y=-\frac{1}{2}x+... \Rightarrow -\frac{1}{2}\rightarrow 2$ | B1 | Deduces gradient of perpendicular to $l$ is $2$; may be seen in equation for perpendicular line |
| $m_N=2 \Rightarrow y-4=2(x-5)$ and $2y+x+6=0 \Rightarrow x=..., y=...$ | M1 | Fully correct strategy for finding intersection; must use perpendicular gradient (not gradient of $l$); look for $y-"4"="2"(x-"5")$ solved simultaneously with equation of $l$; do not penalise rearrangement mechanics; note: finding $y$-intercept of $l$ at $(0,-3)$ is M0 |
| Intersection is at $\left(\frac{6}{5},-\frac{18}{5}\right)$ | A1 | |
| Distance from centre to intersection is $\sqrt{\left(5-\frac{6}{5}\right)^2+\left(4+\frac{18}{5}\right)^2}$ so distance required is $\sqrt{\left("5"-"\frac{6}{5}"\right)^2+\left("4"+"\frac{18}{5}"\right)^2}-"3"$ | dM1 | Dependent on previous M1 |
| $=\frac{19\sqrt{5}}{5}-3$ (or awrt $5.50$) | A1 | |
# Question 11 (Circle/Distance):
**Part (b):**
| Working | Mark | Guidance |
|---------|------|----------|
| Centre $\left(\frac{6}{5}, -\frac{18}{5}\right)$ or equivalent e.g. $(1.2, -3.6)$ | A1 | Coordinates need not be explicit; may appear in working or on diagram |
| Correct strategy: use Pythagoras to find distance between centre and intersection, then subtract radius | dM1 | Dependent on previous method mark. Condone sign slip subtracting $-\frac{18}{5}$. Alt: solve $y=2x-6$ simultaneously with circle equation, find intersection, choose smaller positive root |
| $(x-5)^2+(2x-6-4)^2=9 \Rightarrow 5x^2-50x+125=9$, $x=\frac{25-3\sqrt{5}}{5}$, $y=\frac{20-6\sqrt{5}}{5}$ | — | Example working for alternative method |
| $\sqrt{\frac{361}{5}}-3$ or $\frac{19\sqrt{5}-15}{5}$, also allow awrt 5.50 | A1 | Isw after correct answer seen |
**Alt (b) Vector Method:**
| Working | Mark | Guidance |
|---------|------|----------|
| Attempt perpendicular distance from $(5,4)$ to $x+2y+6=0$ using $\frac{\|ax+by+c\|}{\sqrt{a^2+b^2}}$ | B1M1 | Substituting values into point-to-line distance formula |
| $\frac{\|5\times1+4\times2+6\|}{\sqrt{1^2+2^2}} = \frac{19}{\sqrt{5}}$ | A1 | |
| Distance $= \frac{19\sqrt{5}}{5} - 3$ | dM1 | |
| $\frac{19\sqrt{5}-15}{5}$ | A1 | |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d31369fa-9532-4a09-b67d-a3a3cbf7d586-34_833_1033_248_516}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the circle $C$ with equation
$$x ^ { 2 } + y ^ { 2 } - 10 x - 8 y + 32 = 0$$
and the line $l$ with equation
$$2 y + x + 6 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the coordinates of the centre of $C$,
\item the radius of $C$.
\end{enumerate}\item Find the shortest distance between $C$ and $l$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2022 Q11 [8]}}