| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.8 This is a straightforward multi-part coordinate geometry question requiring standard techniques: midpoint formula, gradient calculations, perpendicular lines, and simultaneous equations. All parts follow routine procedures with no novel problem-solving required, making it easier than average but not trivial due to the multiple steps involved. |
| 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^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Mid-point of \(AB = \left[\frac{1}{2}(-3+8), \frac{1}{2}(-2+4)\right] = \left(\frac{5}{2}, 1\right)\) | M1, A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(M_{AB} = \frac{4-(-2)}{8-(-3)} = \frac{6}{11}\) | M1, A1 | |
| Equation of \(AB\): \(y - 4 = \frac{6}{11}(x-8)\) | M1 | |
| \(\Rightarrow 11y - 44 = 6x - 48 \Rightarrow 6x - 11y - 4 = 0\) (or equivalent) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of tangent \(= -\frac{11}{6}\) | B1ft | |
| Equation: \(y - 4 = -\frac{11}{6}(x-8)\) (or \(6y + 11x - 112 = 0\)) | M1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Equation of \(l\): \(y = \frac{2}{3}x\) | B1 | |
| Substitute into part (c): \(\frac{2}{3}x - 4 = -\frac{11}{6}x + \frac{88}{6}\) | M1 | |
| \(\Rightarrow x = 7\frac{7}{15}\), \(y = 4\frac{44}{45}\) | A1, A1 | (4 marks) (13 marks) |
## Question 7:
### Part (a)
Mid-point of $AB = \left[\frac{1}{2}(-3+8), \frac{1}{2}(-2+4)\right] = \left(\frac{5}{2}, 1\right)$ | M1, A1 | (2 marks)
### Part (b)
$M_{AB} = \frac{4-(-2)}{8-(-3)} = \frac{6}{11}$ | M1, A1 |
Equation of $AB$: $y - 4 = \frac{6}{11}(x-8)$ | M1 |
$\Rightarrow 11y - 44 = 6x - 48 \Rightarrow 6x - 11y - 4 = 0$ (or equivalent) | A1 | (4 marks)
### Part (c)
Gradient of tangent $= -\frac{11}{6}$ | B1ft |
Equation: $y - 4 = -\frac{11}{6}(x-8)$ (or $6y + 11x - 112 = 0$) | M1 A1 | (3 marks)
### Part (d)
Equation of $l$: $y = \frac{2}{3}x$ | B1 |
Substitute into part (c): $\frac{2}{3}x - 4 = -\frac{11}{6}x + \frac{88}{6}$ | M1 |
$\Rightarrow x = 7\frac{7}{15}$, $y = 4\frac{44}{45}$ | A1, A1 | (4 marks) **(13 marks)**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8bae58f7-c53a-43ed-9a1d-2f718bd1e539-3_563_570_785_561}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The points $A ( - 3 , - 2 )$ and $B ( 8,4 )$ are at the ends of a diameter of the circle shown in Fig. 1.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre of the circle.
\item Find an equation of the diameter $A B$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\item Find an equation of tangent to the circle at $B$.
The line $l$ passes through $A$ and the origin.
\item Find the coordinates of the point at which $l$ intersects the tangent to the circle at $B$, giving your answer as exact fractions.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q7 [13]}}