| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring standard techniques: finding a circle equation from centre and radius (using distance formula), then finding the tangent using the perpendicular gradient property. Both are routine C2 procedures with no problem-solving insight needed, making it slightly easier than average but not trivial since it requires correct execution of multiple steps. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((8-3)^2 + (3-1)^2\) or \(\sqrt{(8-3)^2+(3-1)^2}\) | M1 A1 | Condone one slip inside a bracket e.g. \((8-3)^2+(3+1)^2\), \((8-1)^2+(1-3)^2\) |
| \((x \pm 3)^2 + (y \pm 1)^2 = k\) or \((x\pm 1)^2+(y\pm 3)^2 = k\) (\(k\) a positive value) | M1 | |
| \((x-3)^2 + (y-1)^2 = 29\) (Not \((\sqrt{29})^2\) or \(5.39^2\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of radius \(= \frac{2}{5}\) (or exact equiv.) — must be seen or used in (b) | B1 | |
| Gradient of tangent \(= -\frac{5}{2}\) (using perpendicular gradient method) | M1 | |
| \(y - 3 = -\frac{5}{2}(x-8)\) (ft gradient of radius, dependent on both M marks) | M1 A1ft | 2nd M: equation of line through \((8,3)\) in any form with any gradient (except 0 or \(\infty\)). If 8 and 3 are "wrong way round", M mark only given if correct general formula \(y-y_1=m(x-x_1)\) is quoted |
| \(5x + 2y - 46 = 0\) (or equiv. equated to zero; must have integer coefficients) | A1 | e.g. \(92 - 4y - 10x = 0\) acceptable |
## Question 5:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(8-3)^2 + (3-1)^2$ or $\sqrt{(8-3)^2+(3-1)^2}$ | M1 A1 | Condone one slip inside a bracket e.g. $(8-3)^2+(3+1)^2$, $(8-1)^2+(1-3)^2$ |
| $(x \pm 3)^2 + (y \pm 1)^2 = k$ or $(x\pm 1)^2+(y\pm 3)^2 = k$ ($k$ a positive value) | M1 | |
| $(x-3)^2 + (y-1)^2 = 29$ (Not $(\sqrt{29})^2$ or $5.39^2$) | A1 | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of radius $= \frac{2}{5}$ (or exact equiv.) — must be seen or used in (b) | B1 | |
| Gradient of tangent $= -\frac{5}{2}$ (using perpendicular gradient method) | M1 | |
| $y - 3 = -\frac{5}{2}(x-8)$ (ft gradient of radius, dependent on both M marks) | M1 A1ft | 2nd M: equation of line through $(8,3)$ in any form with any gradient (except 0 or $\infty$). If 8 and 3 are "wrong way round", M mark only given if correct general formula $y-y_1=m(x-x_1)$ is quoted |
| $5x + 2y - 46 = 0$ (or equiv. equated to zero; must have integer coefficients) | A1 | e.g. $92 - 4y - 10x = 0$ acceptable |
# June 2008 Core Mathematics C2 Mark Scheme
---5. The circle $C$ has centre $( 3,1 )$ and passes through the point $P ( 8,3 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $C$.
\item Find an equation for the tangent to $C$ at $P$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q5 [9]}}