| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.8 This is a straightforward C2 circle question with standard procedures: finding circle equation from centre and radius, verifying a point lies on the circle, and finding tangent equation using perpendicular gradient. All parts are routine textbook exercises requiring only direct application of formulas with no problem-solving insight needed. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{radius} = \sqrt{25 + 1} = \sqrt{26}\) | M1 A1 | |
| \(\therefore (x+3)^2 + (y-2)^2 = (\sqrt{26})^2\) | M1 | |
| \((x+3)^2 + (y-2)^2 = 26\) | A1 | |
| (b) \((-4, 7)\), LHS \(= (-4+3)^2 + (7-2)^2 = 1 + 25 = 26\) \(\therefore\) lies on circle | B1 | |
| (c) \(\text{grad of radius} = \frac{7-2}{-4-(-3)} = -5\) | M1 | |
| \(\therefore \text{grad of tangent} = -\frac{1}{-5} = \frac{1}{5}\) | M1 A1 | |
| \(\therefore y - 7 = \frac{1}{5}(x + 4)\) | M1 | |
| \(5y - 35 = x + 4\) | A1 | |
| \(x - 5y + 39 = 0\) | A1 | (10) |
(a) $\text{radius} = \sqrt{25 + 1} = \sqrt{26}$ | M1 A1 |
$\therefore (x+3)^2 + (y-2)^2 = (\sqrt{26})^2$ | M1 |
$(x+3)^2 + (y-2)^2 = 26$ | A1 |
(b) $(-4, 7)$, LHS $= (-4+3)^2 + (7-2)^2 = 1 + 25 = 26$ $\therefore$ lies on circle | B1 |
(c) $\text{grad of radius} = \frac{7-2}{-4-(-3)} = -5$ | M1 |
$\therefore \text{grad of tangent} = -\frac{1}{-5} = \frac{1}{5}$ | M1 A1 |
$\therefore y - 7 = \frac{1}{5}(x + 4)$ | M1 |
$5y - 35 = x + 4$ | A1 |
$x - 5y + 39 = 0$ | A1 | (10)\begin{enumerate}
\item The circle $C$ has centre $( - 3,2 )$ and passes through the point $( 2,1 )$.\\
(a) Find an equation for $C$.\\
(b) Show that the point with coordinates $( - 4,7 )$ lies on $C$.\\
(c) Find an equation for the tangent to $C$ at the point ( - 4 , 7). Give your answer in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.
\item
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c7c8cf84-06ac-4059-b8f0-d68b6d1d8dcc-3_664_1016_1276_376}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the curve $y = 2 x ^ { 2 } + 6 x + 7$ and the straight line $y = 2 x + 13$.\\
\hfill \mbox{\textit{Edexcel C2 Q6 [10]}}