| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - find equation |
| Difficulty | Standard +0.8 This is a multi-part question requiring completion of the square to find centre/radius (routine), distance calculation (standard), then applying Pythagoras to a tangent-radius configuration to find triangle area (requires geometric insight and careful algebraic manipulation with surds). The final part elevates this above typical C1/C2 questions. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x \pm 4)^2\) and \((y \pm 5)^2\) | M1 | Attempts completing the square on \(x\) and \(y\), or sight of \((x\pm4)^2\) and \((y\pm5)^2\). May be implied by centre \((\pm4, \pm5)\). If using \(x^2+y^2+2gx+2fy+c=0\), centre is \((\pm g, \pm f)\) |
| Centre is \((4, 5)\) | A1 | Correct centre |
| \(r^2 = (\pm\text{"4"})^2 + (\pm\text{"5"})^2 - 16\) (Must be \(-16\)) | M1 | Must reach \(r^2 = \text{their}(\pm4)^2 + \text{their}(\pm5)^2 - 16\) or \(r = \sqrt{\text{their}(\pm4)^2 + \text{their}(\pm5)^2 - 16}\); must clearly identify radius or \(r^2\) |
| \(r = 5\) | A1 | Correct answer scores both marks |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(MT^2 = (20 - \text{"4"})^2 + (12 - \text{"5"})^2 (= 305)\) | M1 | Fully correct method using Pythagoras for \(MT\) or \(MT^2\) |
| \(MT = \sqrt{305}\) | A1 | Must be exact |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((MP^2) = MT^2 - \text{"5"}^2\) | M1 | Correct method for \(MP\) or \(MP^2\) where \(MT > \text{"5"}\) |
| Area \(MTP = \frac{1}{2} \times \text{"5"} \times \text{"5"} \times \sqrt{280}\)" | M1 | Correct triangle area method |
| \(5\sqrt{70}\) | A1 | cao |
| Alternative: \(\cos PTM = \frac{\text{"5"}}{\sqrt{\text{"305"}}}\), \(\sin PMT = \frac{\text{"5"}}{\sqrt{\text{"305"}}}\) | M1 | Correct method for angle \(PTM\) or \(PMT\) |
| Area \(MTP = \frac{1}{2} \times \text{"5"} \times \sqrt{305} \times \sqrt{\frac{56}{61}}\) | M1 | Correct triangle area method |
| \(5\sqrt{70}\) | A1 | cao. Note \(5\sqrt{70} = 41.83..\) |
| Total: 3 marks |
## Question 11:
### Part (a)(i) and (ii) together:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x \pm 4)^2$ and $(y \pm 5)^2$ | M1 | Attempts completing the square on $x$ and $y$, or sight of $(x\pm4)^2$ and $(y\pm5)^2$. May be implied by centre $(\pm4, \pm5)$. If using $x^2+y^2+2gx+2fy+c=0$, centre is $(\pm g, \pm f)$ |
| Centre is $(4, 5)$ | A1 | Correct centre |
| $r^2 = (\pm\text{"4"})^2 + (\pm\text{"5"})^2 - 16$ (Must be $-16$) | M1 | Must reach $r^2 = \text{their}(\pm4)^2 + \text{their}(\pm5)^2 - 16$ or $r = \sqrt{\text{their}(\pm4)^2 + \text{their}(\pm5)^2 - 16}$; must clearly identify radius or $r^2$ |
| $r = 5$ | A1 | Correct answer scores both marks |
| **Total: 4 marks** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $MT^2 = (20 - \text{"4"})^2 + (12 - \text{"5"})^2 (= 305)$ | M1 | Fully correct method using Pythagoras for $MT$ or $MT^2$ |
| $MT = \sqrt{305}$ | A1 | Must be exact |
| **Total: 2 marks** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(MP^2) = MT^2 - \text{"5"}^2$ | M1 | Correct method for $MP$ or $MP^2$ where $MT > \text{"5"}$ |
| Area $MTP = \frac{1}{2} \times \text{"5"} \times \text{"5"} \times \sqrt{280}$" | M1 | Correct triangle area method |
| $5\sqrt{70}$ | A1 | cao |
| **Alternative:** $\cos PTM = \frac{\text{"5"}}{\sqrt{\text{"305"}}}$, $\sin PMT = \frac{\text{"5"}}{\sqrt{\text{"305"}}}$ | M1 | Correct method for angle $PTM$ or $PMT$ |
| Area $MTP = \frac{1}{2} \times \text{"5"} \times \sqrt{305} \times \sqrt{\frac{56}{61}}$ | M1 | Correct triangle area method |
| $5\sqrt{70}$ | A1 | cao. Note $5\sqrt{70} = 41.83..$ |
| **Total: 3 marks** | | |
---11. The circle $C$ has equation
$$x ^ { 2 } + y ^ { 2 } - 8 x - 10 y + 16 = 0$$
The centre of $C$ is at the point $T$.
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the coordinates of the point $T$,
\item the radius of the circle $C$.
The point $M$ has coordinates $( 20,12 )$.
\end{enumerate}\item Find the exact length of the line $M T$.
Point $P$ lies on the circle $C$ such that the tangent at $P$ passes through the point $M$.
\item Find the exact area of triangle $M T P$, giving your answer as a simplified surd.
\begin{center}
\end{center}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q11 [9]}}