| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Standard +0.3 This is a standard AS-level circle question requiring perpendicular gradients for tangent/radius (part a), distance formula for radius (part b), and perpendicular distance from center to parallel tangent (part c). All techniques are routine applications of core coordinate geometry, making it slightly easier than average but still requiring multiple connected steps. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces gradient \(= -3\), point \((7,4)\); e.g. \(y - 4 = -3(x-7)\) | M1 | Uses perpendicular gradients; sight of \(y-4=-3(x-7)\) scores M1; if \(y=mx+c\) used, must proceed to \(c=\ldots\) |
| \(y = -3x + 25\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solves \(y = -3x+25\) and \(y = \frac{1}{3}x\) simultaneously | M1 | Valid attempt to find both coordinates of \(P\) |
| \(P = \left(\frac{15}{2}, \frac{5}{2}\right)\) | A1 | — |
| Length \(PN = \sqrt{\left(\frac{15}{2}-7\right)^2 + \left(4 - \frac{5}{2}\right)^2} = \sqrt{\frac{5}{2}}\) | M1 | Uses Pythagoras with their \(P\) and \((7,4)\); must attempt difference of coordinates |
| Equation of \(C\): \((x-7)^2 + (y-4)^2 = \frac{5}{2}\) | A1 | Full correct equation; do not accept \(= \left(\sqrt{\frac{5}{2}}\right)^2\) form without simplification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to find where \(y = \frac{1}{3}x + k\) meets \(C\) using vectors: e.g. \(\binom{7.5}{2.5} + 2\times\binom{-0.5}{1.5}\) | M1 | Vector approach to find second intersection |
| Substitutes their \(\left(\frac{13}{2}, \frac{11}{2}\right)\) into \(y = \frac{1}{3}x + k\) to find \(k\) | M1 | Full method leading to \(k\) |
| \(k = \frac{10}{3}\) | A1 | Only \(k = \frac{10}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solves \(y = \frac{1}{3}x+k\) with \((x-7)^2+(y-4)^2=\frac{5}{2}\), forming quadratic \(\frac{10}{9}x^2 + \left(\frac{2}{3}k - \frac{50}{3}\right)x + k^2 - 8k + \frac{125}{2} = 0\) | M1 | Both \(b\) and \(c\) dependent on \(k\); terms in \(x^2\) and \(x\) collected |
| Uses \(b^2 - 4ac = 0\) to find \(k\) | M1 | Not dependent on previous M; correct discriminant formula implied by \(\pm\) correct roots |
| \(k = \frac{10}{3}\) | A1 | Only \(k=\frac{10}{3}\) |
## Question 15:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces gradient $= -3$, point $(7,4)$; e.g. $y - 4 = -3(x-7)$ | M1 | Uses perpendicular gradients; sight of $y-4=-3(x-7)$ scores M1; if $y=mx+c$ used, must proceed to $c=\ldots$ |
| $y = -3x + 25$ | A1 | — |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $y = -3x+25$ and $y = \frac{1}{3}x$ simultaneously | M1 | Valid attempt to find both coordinates of $P$ |
| $P = \left(\frac{15}{2}, \frac{5}{2}\right)$ | A1 | — |
| Length $PN = \sqrt{\left(\frac{15}{2}-7\right)^2 + \left(4 - \frac{5}{2}\right)^2} = \sqrt{\frac{5}{2}}$ | M1 | Uses Pythagoras with their $P$ and $(7,4)$; must attempt difference of coordinates |
| Equation of $C$: $(x-7)^2 + (y-4)^2 = \frac{5}{2}$ | A1 | Full correct equation; do not accept $= \left(\sqrt{\frac{5}{2}}\right)^2$ form without simplification |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find where $y = \frac{1}{3}x + k$ meets $C$ using vectors: e.g. $\binom{7.5}{2.5} + 2\times\binom{-0.5}{1.5}$ | M1 | Vector approach to find second intersection |
| Substitutes their $\left(\frac{13}{2}, \frac{11}{2}\right)$ into $y = \frac{1}{3}x + k$ to find $k$ | M1 | Full method leading to $k$ |
| $k = \frac{10}{3}$ | A1 | Only $k = \frac{10}{3}$ |
**Alternative I (discriminant):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $y = \frac{1}{3}x+k$ with $(x-7)^2+(y-4)^2=\frac{5}{2}$, forming quadratic $\frac{10}{9}x^2 + \left(\frac{2}{3}k - \frac{50}{3}\right)x + k^2 - 8k + \frac{125}{2} = 0$ | M1 | Both $b$ and $c$ dependent on $k$; terms in $x^2$ and $x$ collected |
| Uses $b^2 - 4ac = 0$ to find $k$ | M1 | Not dependent on previous M; correct discriminant formula implied by $\pm$ correct roots |
| $k = \frac{10}{3}$ | A1 | Only $k=\frac{10}{3}$ |15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{235cd1dc-a3ab-473a-bf77-3e41b274dfd8-38_655_929_248_568}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of a circle $C$ with centre $N ( 7,4 )$\\
The line $l$ with equation $y = \frac { 1 } { 3 } x$ is a tangent to $C$ at the point $P$.\\
Find
\begin{enumerate}[label=(\alph*)]
\item the equation of line $P N$ in the form $y = m x + c$, where $m$ and $c$ are constants,
\item an equation for $C$.
The line with equation $y = \frac { 1 } { 3 } x + k$, where $k$ is a non-zero constant, is also a tangent to $C$.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2021 Q15 [9]}}