| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Range of parameter for intersection |
| Difficulty | Moderate -0.3 Part (a) is routine completion of the square to find centre and radius. Part (b) requires recognizing that a vertical line x=k is tangent when the horizontal distance from the centre equals the radius, leading to a simple equation. This is a standard AS-level circle question with straightforward application of techniques, slightly easier than average due to the geometric simplicity of vertical tangent lines. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \((x-2)^2+(y+4)^2-4-16-8=0\) | M1 | Completing the square; look for \((x\pm2)^2+(y\pm4)^2\ldots\) |
| Centre \((2,-4)\) | A1 | May be written as \(x=2, y=-4\); but \(a=2, b=-4\) is A0 |
| Radius \(\sqrt{28}\) or \(2\sqrt{7}\) | A1 | isw after correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to add/subtract \(r\) from \(2\); \(k=2\pm\sqrt{28}\) | M1 | Alternatively substitutes \(y=-4\) into circle equation and finds \(x/k\) by solving quadratic |
| \(k=2+\sqrt{28}\) and \(k=2-\sqrt{28}\) | A1ft | Follow through on their 2 and their \(\sqrt{28}\); accept \(2\pm2\sqrt{7}\); condone \(x=2\pm\sqrt{28}\) but not \(y=2\pm\sqrt{28}\) |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $(x-2)^2+(y+4)^2-4-16-8=0$ | M1 | Completing the square; look for $(x\pm2)^2+(y\pm4)^2\ldots$ |
| Centre $(2,-4)$ | A1 | May be written as $x=2, y=-4$; but $a=2, b=-4$ is A0 |
| Radius $\sqrt{28}$ or $2\sqrt{7}$ | A1 | isw after correct answer |
## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to add/subtract $r$ from $2$; $k=2\pm\sqrt{28}$ | M1 | Alternatively substitutes $y=-4$ into circle equation and finds $x/k$ by solving quadratic |
| $k=2+\sqrt{28}$ and $k=2-\sqrt{28}$ | A1ft | Follow through on their 2 and their $\sqrt{28}$; accept $2\pm2\sqrt{7}$; condone $x=2\pm\sqrt{28}$ but not $y=2\pm\sqrt{28}$ |
---\begin{enumerate}
\item A circle $C$ has equation
\end{enumerate}
$$x ^ { 2 } + y ^ { 2 } - 4 x + 8 y - 8 = 0$$
(a) Find\\
(i) the coordinates of the centre of $C$,\\
(ii) the exact radius of $C$.
The straight line with equation $x = k$, where $k$ is a constant, is a tangent to $C$.\\
(b) Find the possible values for $k$.
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q10 [5]}}