| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Moderate -0.3 This is a structured multi-part question testing standard C2 circle geometry. Part (a) requires calculating two gradients and showing their product is -1 (routine perpendicular lines). Parts (b) and (c) use the angle-in-semicircle theorem to identify the diameter, then find the centre as midpoint and verify the circle equation—all standard techniques with clear scaffolding and no novel problem-solving required. |
| 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 circle1.10g Problem solving with vectors: in geometry |
| Answer | Marks |
|---|---|
| (a) grad \(PQ = \frac{-8-2}{-3-(-5)} = 3\), grad \(QR = \frac{4-8}{9-(-3)} = -\frac{1}{3}\) | M1 A1 |
| grad \(PQ \times\) grad \(QR = 3 \times (-\frac{1}{3}) = -1\) | M1 |
| \(\therefore PQ\) perp. to \(QR\). \(\therefore \angle PQR = 90°\) | A1 |
| (b) \(\angle PQR = 90° \therefore PR\) is a diameter | M1 |
| \(\therefore\) centre \(=\) mid-point of \(PR = (\frac{-1+9}{2}, \frac{2+4}{2}) = (2, 3)\) | M1 A1 |
| (c) radius \(= \) dist. \((-5, 2)\) to \((2, 3) = \sqrt{49 + 1} = \sqrt{50}\) | B1 |
| \(\therefore (x-2)^2 + (y-3)^2 = (\sqrt{50})^2\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 + y^2 - 4x - 6y = 37\) | A1 | [\(k = 37\)] |
**(a)** grad $PQ = \frac{-8-2}{-3-(-5)} = 3$, grad $QR = \frac{4-8}{9-(-3)} = -\frac{1}{3}$ | M1 A1 |
grad $PQ \times$ grad $QR = 3 \times (-\frac{1}{3}) = -1$ | M1 |
$\therefore PQ$ perp. to $QR$. $\therefore \angle PQR = 90°$ | A1 |
**(b)** $\angle PQR = 90° \therefore PR$ is a diameter | M1 |
$\therefore$ centre $=$ mid-point of $PR = (\frac{-1+9}{2}, \frac{2+4}{2}) = (2, 3)$ | M1 A1 |
**(c)** radius $= $ dist. $(-5, 2)$ to $(2, 3) = \sqrt{49 + 1} = \sqrt{50}$ | B1 |
$\therefore (x-2)^2 + (y-3)^2 = (\sqrt{50})^2$ | M1 |
$x^2 - 4x + 4 + y^2 - 6y + 9 = 50$
$x^2 + y^2 - 4x - 6y = 37$ | A1 | [$k = 37$] | (10 marks)The points $P$, $Q$ and $R$ have coordinates $(-5, 2)$, $(-3, 8)$ and $(9, 4)$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that $\angle PQR = 90°$. [4]
\end{enumerate}
Given that $P$, $Q$ and $R$ all lie on circle $C$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the coordinates of the centre of $C$, [3]
\item show that the equation of $C$ can be written in the form
$$x^2 + y^2 - 4x - 6y = k,$$
where $k$ is an integer to be found. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [10]}}