| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - find equation |
| Difficulty | Standard +0.3 This is a standard C2 circles question requiring: (a) substituting a point into the circle equation to find k, (b) completing the square to find centre and radius, and (c) using the tangent property (radius perpendicular to tangent) with Pythagoras. All techniques are routine for C2 level with no novel insight required, making it slightly easier than average. |
| 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 | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \((-6,5)\): \(36 + 25 - 60 - 40 + k = 0\) | M1 | |
| \(k = 39\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \((x+5)^2 - 25 + (y-4)^2 - 16 + 39 = 0\) | M1 | |
| \((x+5)^2 + (y-4)^2 = 2\) | ||
| \(\therefore\) centre \((-5, 4)\), radius \(= \sqrt{2}\) | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| dist. \((2,3)\) to centre \(= \sqrt{49+1} = \sqrt{50}\) | B1 | |
| \(\therefore AB^2 = (\sqrt{50})^2 - (\sqrt{2})^2 = 48\) | M1 A1 | |
| \(AB = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\) | M1 A1 | (10) |
## Question 7:
**(a)**
| Answer/Working | Marks | Notes |
|---|---|---|
| $(-6,5)$: $36 + 25 - 60 - 40 + k = 0$ | M1 | |
| $k = 39$ | A1 | |
**(b)**
| Answer/Working | Marks | Notes |
|---|---|---|
| $(x+5)^2 - 25 + (y-4)^2 - 16 + 39 = 0$ | M1 | |
| $(x+5)^2 + (y-4)^2 = 2$ | | |
| $\therefore$ centre $(-5, 4)$, radius $= \sqrt{2}$ | A2 | |
**(c)**
| Answer/Working | Marks | Notes |
|---|---|---|
| dist. $(2,3)$ to centre $= \sqrt{49+1} = \sqrt{50}$ | B1 | |
| $\therefore AB^2 = (\sqrt{50})^2 - (\sqrt{2})^2 = 48$ | M1 A1 | |
| $AB = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$ | M1 A1 | **(10)** |
---7. The circle $C$ has the equation
$$x ^ { 2 } + y ^ { 2 } + 10 x - 8 y + k = 0 ,$$
where $k$ is a constant.
Given that the point with coordinates $( - 6,5 )$ lies on $C$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $k$,
\item find the coordinates of the centre and the radius of $C$.
A straight line which passes through the point $A ( 2,3 )$ is a tangent to $C$ at the point $B$.
\item Find the length $A B$ in the form $k \sqrt { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q7 [10]}}