| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Convert Cartesian to polar equation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question involving standard conversions between Cartesian and polar forms. Part (a) requires finding the other end of a diameter (routine circle geometry), part (b)(i) involves direct conversion to polar coordinates, and part (b)(ii) uses the standard substitutions x=r cos θ, y=r sin θ to convert a circle equation—a textbook exercise with clear steps and no novel insight required. Slightly above average difficulty only because it's Further Maths content. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Centre of circle is \(M(3, 4)\) | B1 | PI |
| \(A(6, 8)\) | B1 | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(k = OA = 10\) | B1 | |
| \(\tan\alpha = \frac{y_A}{x_A} = \frac{4}{3}\) | B1 | Total: 2 marks. SC "\(r = 10\) and \(\tan\theta = \frac{8}{6}\)" = B1 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 + y^2 - 6x - 8y + 25 = 25\) | B1 | If polar form before expansion award B1 for correct expansions of both \((r\cos\theta - m)^2\) and \((r\sin\theta - n)^2\) where \((m,n)=(3,4)\) or \((m,n)=(4,3)\) |
| \(r^2 - 6r\cos\theta - 8r\sin\theta = 0\) | M1M1 | 1st M1 for use of any one of \(x^2+y^2=r^2\), \(x=r\cos\theta\), \(y=r\sin\theta\). 2nd M1 for use of these to convert \(x^2+y^2+ax+by=0\) to \(r^2+ar\cos\theta+br\sin\theta=0\) |
| \(\{r=0, \text{origin}\}\) Circle: \(r = 6\cos\theta + 8\sin\theta\) | A1 | Total: 4 marks. NMS Mark as 4 or 0 |
| ALTn: Circle has eqn \(r = OA\cos(\alpha - \theta)\) | (M2) | |
| \(r = OA\cos\alpha\cos\theta + OA\sin\alpha\sin\theta\) | (m1) | OE |
| Circle: \(r = 6\cos\theta + 8\sin\theta\) | (A1) |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre of circle is $M(3, 4)$ | B1 | PI |
| $A(6, 8)$ | B1 | Total: 2 marks |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k = OA = 10$ | B1 | |
| $\tan\alpha = \frac{y_A}{x_A} = \frac{4}{3}$ | B1 | Total: 2 marks. SC "$r = 10$ and $\tan\theta = \frac{8}{6}$" = B1 only |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 + y^2 - 6x - 8y + 25 = 25$ | B1 | If polar form before expansion award B1 for correct expansions of both $(r\cos\theta - m)^2$ and $(r\sin\theta - n)^2$ where $(m,n)=(3,4)$ or $(m,n)=(4,3)$ |
| $r^2 - 6r\cos\theta - 8r\sin\theta = 0$ | M1M1 | 1st M1 for use of any one of $x^2+y^2=r^2$, $x=r\cos\theta$, $y=r\sin\theta$. 2nd M1 for use of these to convert $x^2+y^2+ax+by=0$ to $r^2+ar\cos\theta+br\sin\theta=0$ |
| $\{r=0, \text{origin}\}$ Circle: $r = 6\cos\theta + 8\sin\theta$ | A1 | Total: 4 marks. NMS Mark as 4 or 0 |
| **ALTn:** Circle has eqn $r = OA\cos(\alpha - \theta)$ | (M2) | |
| $r = OA\cos\alpha\cos\theta + OA\sin\alpha\sin\theta$ | (m1) | OE |
| Circle: $r = 6\cos\theta + 8\sin\theta$ | (A1) | |
---3 The diagram shows a sketch of a circle which passes through the origin $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{13cfb9ca-9495-4b69-80c5-9fb7e8e0f957-3_423_451_356_794}
The equation of the circle is $( x - 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 25$ and $O A$ is a diameter.
\begin{enumerate}[label=(\alph*)]
\item Find the cartesian coordinates of the point $A$.
\item Using $O$ as the pole and the positive $x$-axis as the initial line, the polar coordinates of $A$ are $( k , \alpha )$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $k$ and the value of $\tan \alpha$.
\item Find the polar equation of the circle $( x - 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 25$, giving your answer in the form $r = p \cos \theta + q \sin \theta$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2009 Q3 [8]}}