| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.8 This is a straightforward C2 circle question requiring distance formula, circle equation, and tangent perpendicularity. All steps are routine: (a) apply distance formula, (b) write (x-7)²+(y-8)²=r², (c) find radius gradient, use perpendicular gradient -1/m, apply point-slope form. Standard textbook exercise with no problem-solving insight needed, making it easier than average. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.03d 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 | Marks | Guidance |
| (a) \(\{PQ = \} \sqrt{(7-10)^2 + (8-13)^2}\) or \(\sqrt{(10-7)^2 + (13-8)^2}\) | M1 | Applies distance formula. Can be implied. |
| \(\{PQ\} = \sqrt{34}\) | A1 | \(\sqrt{34}\) or \(\sqrt{17}.\sqrt{2}\) |
| [2] | ||
| (b) Way 1: \((x-7)^2 + (y-8)^2 = 34\) (or \((\sqrt{34})^2\)) | M1 | \((x-7)^2 + (y-8)^2 = k\), where \(k\) is a positive value |
| A1 oe | \((x-7)^2 + (y-8)^2 = 34\) | |
| [2] | ||
| Way 2: \(x^2 + y^2 - 14x - 16y + 79 = 0\) | M1 | \(x^2 + y^2 - 14x - 16y + c = 0\), where \(c\) is any value \(< 113\) |
| A1 oe | \(x^2 + y^2 - 14x - 16y + 79 = 0\) | |
| [2] | ||
| (c) Way 1: Gradient of radius \(= \frac{13-8}{10-7}\) or \(\frac{5}{3}\) | B1 | This must be seen or implied in part (c) |
| Gradient of tangent \(= -\frac{1}{m}\left(= -\frac{3}{5}\right)\) | M1 | Using perpendicular gradient method on their gradient |
| \(y - 13 = -\frac{3}{5}(x-10)\) | M1 | \(y - 13 = (\text{their changed gradient})(x-10)\) |
| \(3x + 5y - 95 = 0\) o.e. | A1 | |
| [4] | ||
| Way 2: Correct differentiation (or equivalent). \(2(x-7) + 2(y-8)\frac{dy}{dx} = 0\) | B1 | |
| Substituting both \(x = 10\) and \(y = 13\) into valid differentiation to find value for \(\frac{dy}{dx}\) | M1 | |
| \(2(10-7) + 2(13-8)\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{3}{5}\) | ||
| \(y - 13 = -\frac{3}{5}(x-10)\) | M1 | \(y - 13 = (\text{their gradient})(x-10)\) |
| \(3x + 5y - 95 = 0\) o.e. | A1 | |
| [4] | ||
| Way 3: \(10x + 13y - 7(x+10) - 8(y+13) + 79 = 0\) | B1 | |
| \(10x + 13y - 7(x+10) - 8(y+13) + c = 0\) where \(c\) is any value \(< 113\) | ||
| \(3x + 5y - 95 = 0\) | M2 | |
| A1 | ||
| [4] |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $\{PQ = \} \sqrt{(7-10)^2 + (8-13)^2}$ or $\sqrt{(10-7)^2 + (13-8)^2}$ | M1 | Applies distance formula. Can be implied. |
| $\{PQ\} = \sqrt{34}$ | A1 | $\sqrt{34}$ or $\sqrt{17}.\sqrt{2}$ |
| | [2] | |
| **(b) Way 1:** $(x-7)^2 + (y-8)^2 = 34$ (or $(\sqrt{34})^2$) | M1 | $(x-7)^2 + (y-8)^2 = k$, where $k$ is a positive value |
| | A1 oe | $(x-7)^2 + (y-8)^2 = 34$ |
| | [2] | |
| **Way 2:** $x^2 + y^2 - 14x - 16y + 79 = 0$ | M1 | $x^2 + y^2 - 14x - 16y + c = 0$, where $c$ is any value $< 113$ |
| | A1 oe | $x^2 + y^2 - 14x - 16y + 79 = 0$ |
| | [2] | |
| **(c) Way 1:** Gradient of radius $= \frac{13-8}{10-7}$ or $\frac{5}{3}$ | B1 | This must be seen or implied in part (c) |
| Gradient of tangent $= -\frac{1}{m}\left(= -\frac{3}{5}\right)$ | M1 | Using perpendicular gradient method on their gradient |
| $y - 13 = -\frac{3}{5}(x-10)$ | M1 | $y - 13 = (\text{their changed gradient})(x-10)$ |
| $3x + 5y - 95 = 0$ o.e. | A1 | |
| | [4] | |
| **Way 2:** Correct differentiation (or equivalent). $2(x-7) + 2(y-8)\frac{dy}{dx} = 0$ | B1 | |
| Substituting both $x = 10$ and $y = 13$ into valid differentiation to find value for $\frac{dy}{dx}$ | M1 | |
| $2(10-7) + 2(13-8)\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{3}{5}$ | | |
| $y - 13 = -\frac{3}{5}(x-10)$ | M1 | $y - 13 = (\text{their gradient})(x-10)$ |
| $3x + 5y - 95 = 0$ o.e. | A1 | |
| | [4] | |
| **Way 3:** $10x + 13y - 7(x+10) - 8(y+13) + 79 = 0$ | B1 | |
| $10x + 13y - 7(x+10) - 8(y+13) + c = 0$ where $c$ is any value $< 113$ | | |
| $3x + 5y - 95 = 0$ | M2 | |
| | A1 | |
| | [4] | |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{582cda45-80fc-43a8-90e6-1cae08cb1534-05_791_917_121_484}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Diagram not drawn to scale
The circle $C$ has centre $P ( 7,8 )$ and passes through the point $Q ( 10,13 )$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the length $P Q$, giving your answer as an exact value.
\item Hence write down an equation for $C$.
The line $l$ is a tangent to $C$ at the point $Q$, as shown in Figure 2.
\item Find an equation for $l$, 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 2016 Q3 [8]}}