| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Standard +0.3 This is a straightforward multi-part circle question requiring standard techniques: reading center/radius from equation, substituting a point, solving a quadratic, finding a tangent line, and calculating triangle area. While it has multiple steps (7 marks typical), each individual step uses routine P2 methods with no novel insight required, making it slightly easier than average. |
| 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 |
| Centre \(= (k, 2k)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Radius \(= \sqrt{k+7}\) | B1 | isw after correct answer seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2-k)^2 + (3-2k)^2 = k+7 \Rightarrow 4-4k+k^2+9-12k+4k^2 = k+7\) | M1 | Substitutes \((2,3)\) into equation of circle and attempts to multiply out |
| \(5k^2 - 17k + 6 = 0\ *\) | A1* | Proceeds to \(5k^2 - 17k + 6 = 0\) with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = \frac{2}{5},\ 3\) | B1 | Only these two values. Do not isw if they go on to give a range of values for \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Centre is \(\left(\frac{2}{5}, \frac{4}{5}\right)\) | B1ft | Uses correct value of \(k\) and finds centre. Follow through \((k, 2k)\) with \(k=\frac{2}{5}\). Must be clearly identified as centre. NB: method marks may be scored with any \(k\) |
| Gradient of tangent \(= \pm\dfrac{2 - \frac{2}{5}}{3 - \frac{4}{5}} = \left(-\frac{8}{11}\right)\) | M1 | Attempts gradient of tangent. Look for \(\pm\)(change in \(x\)/change in \(y\)) using their centre and \(P\). May find gradient from centre to \(P\) first then apply negative reciprocal |
| \(y - 3 = -\frac{8}{11}(x-2) \Rightarrow\) sets \(y=0 \Rightarrow x = \ldots\) | M1 | Attempts to find point \(T\) via equation of tangent at \(P\), setting \(y=0\). Gradient need not be correct but must come from attempt with their centre and \(P\) |
| Area \(OPT = \frac{1}{2} \times \frac{49}{8} \times 3 = \ldots\) | dM1 | Correct method to find area of triangle. Dependent on previous M mark. Must be complete method with valid attempts to find appropriate lengths |
| \(= \dfrac{147}{16}\) oe | A1 | Exact answer required (decimal 9.1875) |
## Question 9:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Centre $= (k, 2k)$ | B1 | |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Radius $= \sqrt{k+7}$ | B1 | isw after correct answer seen |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2-k)^2 + (3-2k)^2 = k+7 \Rightarrow 4-4k+k^2+9-12k+4k^2 = k+7$ | M1 | Substitutes $(2,3)$ into equation of circle and attempts to multiply out |
| $5k^2 - 17k + 6 = 0\ *$ | A1* | Proceeds to $5k^2 - 17k + 6 = 0$ with no errors |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = \frac{2}{5},\ 3$ | B1 | Only these two values. Do not isw if they go on to give a range of values for $k$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Centre is $\left(\frac{2}{5}, \frac{4}{5}\right)$ | B1ft | Uses correct value of $k$ and finds centre. Follow through $(k, 2k)$ with $k=\frac{2}{5}$. Must be clearly identified as centre. NB: method marks may be scored with any $k$ |
| Gradient of tangent $= \pm\dfrac{2 - \frac{2}{5}}{3 - \frac{4}{5}} = \left(-\frac{8}{11}\right)$ | M1 | Attempts gradient of tangent. Look for $\pm$(change in $x$/change in $y$) using their centre and $P$. May find gradient from centre to $P$ first then apply negative reciprocal |
| $y - 3 = -\frac{8}{11}(x-2) \Rightarrow$ sets $y=0 \Rightarrow x = \ldots$ | M1 | Attempts to find point $T$ via equation of tangent at $P$, setting $y=0$. Gradient need not be correct but must come from attempt with their centre and $P$ |
| Area $OPT = \frac{1}{2} \times \frac{49}{8} \times 3 = \ldots$ | dM1 | Correct method to find area of triangle. Dependent on previous M mark. Must be complete method with valid attempts to find appropriate lengths |
| $= \dfrac{147}{16}$ oe | A1 | Exact answer required (decimal 9.1875) |9. A circle $C$ has equation
$$( x - k ) ^ { 2 } + ( y - 2 k ) ^ { 2 } = k + 7$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Write down, in terms of $k$,
\begin{enumerate}[label=(\roman*)]
\item the coordinates of the centre of $C$,
\item the radius of $C$.
Given that the point $P ( 2,3 )$ lies on $C$
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item show that $5 k ^ { 2 } - 17 k + 6 = 0$
\item hence find the possible values of $k$.
The tangent to the circle at $P$ intersects the $x$-axis at point $T$.\\
Given that $k < 2$
\end{enumerate}\item calculate the exact area of triangle $O P T$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2021 Q9 [10]}}