| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Standard +0.3 This is a straightforward multi-part circle question requiring standard techniques: distance formula for diameter, circle equation from centre and radius, and discriminant method to verify tangency. All parts are routine C2 content with clear methods, making it slightly easier than average but requiring multiple steps. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03f Circle properties: angles, chords, tangents |
| Answer | Marks |
|---|---|
| \(= 2 \times \sqrt{4+1} = 2\sqrt{5}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \((x - 5)^2 + (y - 2)^2 = (\sqrt{5})^2\) | M1 |
| \((x - 5)^2 + (y - 2)^2 = 5\) | A1 |
| Answer | Marks |
|---|---|
| Sub. \(y = 2x - 3\) into eqn of C: | M1 |
| \((x - 5)^2 + [(2x - 3) - 2]^2 = 5\) | |
| \((x - 5)^2 + (2x - 5)^2 = 5\) | A1 |
| \(x^2 - 6x + 9 = 0\) | |
| \((x - 3)^2 = 0\) | M1 |
| Repeated root \(\therefore\) tangent | A1 |
| Point of contact \((3, 3)\) | A1 |
**(a)**
$= 2 \times \sqrt{4+1} = 2\sqrt{5}$ | M1 A1 |
**(b)**
$(x - 5)^2 + (y - 2)^2 = (\sqrt{5})^2$ | M1 |
$(x - 5)^2 + (y - 2)^2 = 5$ | A1 |
**(c)**
Sub. $y = 2x - 3$ into eqn of C: | M1 |
$(x - 5)^2 + [(2x - 3) - 2]^2 = 5$ | |
$(x - 5)^2 + (2x - 5)^2 = 5$ | A1 |
$x^2 - 6x + 9 = 0$ | |
$(x - 3)^2 = 0$ | M1 |
Repeated root $\therefore$ tangent | A1 |
Point of contact $(3, 3)$ | A1 |7. The circle $C$ has centre $( 5,2 )$ and passes through the point $( 7,3 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the length of the diameter of $C$.
\item Find an equation for $C$.
\item Show that the line $y = 2 x - 3$ is a tangent to $C$ and find the coordinates of the point of contact.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [9]}}