| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic circle equations and line-circle intersection. Part (a) uses Pythagoras, (b) writes the standard circle equation, (c) substitutes point A into the line equation, and (d) solves simultaneous equations. All steps are routine procedures with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation in part (d). |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d 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 | Marks | Guidance |
| \((9)^2 + (\pm13)^2 = r^2\) | M1 | Allow for correct expression for \(r^2\) or \(r\); distance from \((9,-13)\) to origin |
| \(r = \sqrt{250} = 5\sqrt{10}\) | A1 | Either \(\sqrt{250}\) or \(5\sqrt{10}\) implies M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 + y^2 = 250\) | B1 | Accept any multiple e.g. \((x\pm0)^2+(y\pm0)^2=250\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \(x=9\), \(y=-13\) to give \(k=1\) | B1 | \(k=1\) stated or implied by \(2y+3x=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts to combine \(2y+3x=k\) and \(x^2+y^2=r^2\) | M1 | Must eliminate \(x\) or \(y\); allow numerical or algebraic \(k\) |
| \(13x^2 - 6x - 999 = 0\) or \(13y^2 - 4y - 2249 = 0\) | A1 | Correct 3TQ in \(x\) or \(y\) |
| Solve to give \(x=\) (or \(y=\)) | M1 | Solve 3TQ by correct method |
| Substitute to give \(y=\) (or \(x=\)) | M1 | Substitute to find other coordinate |
| \(\left(-\dfrac{111}{13},\ \dfrac{173}{13}\right)\) | A1 A1 | A1: one correct coordinate. A1: both correct answers |
# Question 13:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(9)^2 + (\pm13)^2 = r^2$ | M1 | Allow for correct expression for $r^2$ or $r$; distance from $(9,-13)$ to origin |
| $r = \sqrt{250} = 5\sqrt{10}$ | A1 | Either $\sqrt{250}$ or $5\sqrt{10}$ implies M1 |
**[2 marks]**
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 + y^2 = 250$ | B1 | Accept any multiple e.g. $(x\pm0)^2+(y\pm0)^2=250$ |
**[1 mark]**
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $x=9$, $y=-13$ to give $k=1$ | B1 | $k=1$ stated or implied by $2y+3x=1$ |
**[1 mark]**
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts to combine $2y+3x=k$ and $x^2+y^2=r^2$ | M1 | Must eliminate $x$ or $y$; allow numerical or algebraic $k$ |
| $13x^2 - 6x - 999 = 0$ **or** $13y^2 - 4y - 2249 = 0$ | A1 | Correct 3TQ in $x$ or $y$ |
| Solve to give $x=$ (or $y=$) | M1 | Solve 3TQ by correct method |
| Substitute to give $y=$ (or $x=$) | M1 | Substitute to find other coordinate |
| $\left(-\dfrac{111}{13},\ \dfrac{173}{13}\right)$ | A1 A1 | A1: one correct coordinate. A1: both correct answers |
**[6 marks]**
---13. The point $A ( 9 , - 13 )$ lies on a circle $C$ with centre the origin and radius $r$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $r$.
\item Find an equation of the circle $C$.
A straight line through point $A$ has equation $2 y + 3 x = k$, where $k$ is a constant.
\item Find the value of $k$.
This straight line cuts the circle again at the point $B$.
\item Find the exact coordinates of point $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q13 [10]}}