| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Moderate -0.8 This is a straightforward C2 circle question requiring standard techniques: finding perpendicular gradient (negative reciprocal of 3), writing line equation through a point, solving simultaneous equations, and using centre-radius form. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to multiple parts. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b 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 |
|---|---|---|
| (a) Gradient of \(PQ\) is \(-\frac{1}{3}\) | B1 | |
| \(y - 2 = -\frac{1}{3}(x - 2)\) | M1, A1 | \((3y + x = 8)\) (3 marks) |
| (b) \(y = 1\): \(3 + x = 8\) | B1 | \(x = 5\) (*) (1 mark) |
| (c) \(("5" - 2)^2 + (1 - 2)^2\) | M1, A1 | M: Attempt \(PQ^2\) or \(PQ\). (4 marks) |
| \((x - 5)^2 + (y - 1)^2 = 10\) | M1, A1 | M: \((x \pm a)^2 + (y \pm b)^2 = k\) |
| (8 marks) | (a) M1: eqn. of a straight line through (2, 2) with any gradient except 3, 0 or \(\infty\). Alternative: Using (2, 2) in \(y = mx + c\) to find a value of \(c\) scores M1, but an equation (general or specific) must be seen. If the given value \(x = 5\) is used to find the gradient of \(PQ\), maximum marks are (a) B0 M1 A1 (b) B0. (c) For the first M1, condone one slip, numerical or sign, inside a bracket. The first M1 can be scored if their \(x\)-coord. is used instead of 5. For the second M1, allow any equation in this form, with non-zero \(a, b\) and \(k\). |
**(a)** Gradient of $PQ$ is $-\frac{1}{3}$ | B1 | |
| --- | --- | --- |
| $y - 2 = -\frac{1}{3}(x - 2)$ | M1, A1 | $(3y + x = 8)$ (3 marks) |
| **(b)** $y = 1$: $3 + x = 8$ | B1 | $x = 5$ (*) (1 mark) |
| **(c)** $("5" - 2)^2 + (1 - 2)^2$ | M1, A1 | M: Attempt $PQ^2$ or $PQ$. (4 marks) |
| $(x - 5)^2 + (y - 1)^2 = 10$ | M1, A1 | M: $(x \pm a)^2 + (y \pm b)^2 = k$ |
| | (8 marks) | **(a)** M1: eqn. of a straight line through (2, 2) with any gradient except 3, 0 or $\infty$. **Alternative:** Using (2, 2) in $y = mx + c$ to find a value of $c$ scores M1, but an equation (general or specific) must be seen. If the given value $x = 5$ is used to find the gradient of $PQ$, maximum marks are (a) B0 M1 A1 (b) B0. **(c)** For the first M1, condone one slip, numerical or sign, inside a bracket. The first M1 can be scored if their $x$-coord. is used instead of 5. For the second M1, allow any equation in this form, with non-zero $a, b$ and $k$. |7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{29c7baa1-6929-448a-a756-319ea75dffa7-08_611_682_296_641}
\end{center}
\end{figure}
The line $y = 3 x - 4$ is a tangent to the circle $C$, touching $C$ at the point $P ( 2,2 )$, as shown in Figure 1.
The point $Q$ is the centre of $C$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the straight line through $P$ and $Q$.
Given that $Q$ lies on the line $y = 1$,
\item show that the $x$-coordinate of $Q$ is 5,
\item find an equation for $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2006 Q7 [8]}}