| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Perpendicular bisector of chord |
| Difficulty | Moderate -0.3 This is a structured multi-part question testing standard circle geometry concepts (perpendicular bisector of chord passes through center). Each part guides students through the solution methodically: finding gradient, midpoint, using perpendicularity to find center, then verifying the equation. While it requires multiple techniques (coordinate geometry, perpendicular gradients, circle equations), the scaffolding makes it slightly easier than average for A-level. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(= \frac{-8 - 4}{8 - 2} = -2\) | M1 A1 | |
| (b) \(= (\frac{2+8}{2}, \frac{4+8}{2}) = (5, -2)\) | M1 A1 | |
| (c) perp. grad \(= -\frac{-1}{2} = \frac{1}{2}\) | M1 | |
| perp. bisector: \(y + 2 = \frac{1}{2}(x - 5)\) | M1 A1 | |
| centre where \(y = 0 \therefore x = 9 \Rightarrow (9, 0)\) | M1 A1 | |
| (d) radius \(= \text{dist. }(2, 4) \text{ to } (9, 0) = \sqrt{49 + 16} = \sqrt{65}\) | B1 | |
| \(\therefore (x - 9)^2 + (y - 0)^2 = (\sqrt{65})^2\) | M1 | |
| \(x^2 - 18x + 81 + y^2 = 65\) | ||
| \(x^2 + y^2 - 18x + 16 = 0\) | A1 | (12 marks) |
**(a)** $= \frac{-8 - 4}{8 - 2} = -2$ | M1 A1 |
**(b)** $= (\frac{2+8}{2}, \frac{4+8}{2}) = (5, -2)$ | M1 A1 |
**(c)** perp. grad $= -\frac{-1}{2} = \frac{1}{2}$ | M1 |
perp. bisector: $y + 2 = \frac{1}{2}(x - 5)$ | M1 A1 |
centre where $y = 0 \therefore x = 9 \Rightarrow (9, 0)$ | M1 A1 |
**(d)** radius $= \text{dist. }(2, 4) \text{ to } (9, 0) = \sqrt{49 + 16} = \sqrt{65}$ | B1 |
$\therefore (x - 9)^2 + (y - 0)^2 = (\sqrt{65})^2$ | M1 |
$x^2 - 18x + 81 + y^2 = 65$ | |
$x^2 + y^2 - 18x + 16 = 0$ | A1 | (12 marks)8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{857bf144-b03e-4b46-b043-1119b30f9e78-4_533_685_242_497}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the circle $C$ and the straight line $l$. The centre of $C$ lies on the $x$-axis and $l$ intersects $C$ at the points $A ( 2,4 )$ and $B ( 8 , - 8 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $l$.
\item Find the coordinates of the mid-point of $A B$.
\item Find the coordinates of the centre of $C$.
\item Show that $C$ has the equation $x ^ { 2 } + y ^ { 2 } - 18 x + 16 = 0$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q8 [12]}}