| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find centre and radius from equation |
| Difficulty | Moderate -0.8 This is a straightforward question testing completing the square for circles (parts a,b) and substitution to find intersection points (part c). All techniques are standard Core 2 material with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation needed. |
| 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 | Marks | Guidance |
| Obtain \((x\pm5)^2\) and \((y\pm3)^2\) | M1 | Can be implied by \((\pm5, \pm3)\) |
| Centre is \((-5, 3)\) | A1 | Correct answer with no working implies M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x+5)^2 + (y-3)^2 = 16\) \((=r^2)\) or \((r^2=)\) "25"+"9"\(-18\) | M1 | Complete and correct method leading to \(r^2 =\) "25"+"9"\(-18\) |
| \(r = 4\) | A1 | Only \(r=4\) (not \(r=-4\)); correct answer with no working implies M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use \(x = -3\) in either form of equation of circle to obtain simplified quadratic in \(y\) | M1 | Substituting \(x=-3\) and attempt to simplify to 3 term quadratic or \((y-a)^2 = b\) |
| e.g. \((-3+5)^2 + (y-3)^2 = 16 \Rightarrow (y-3)^2 = 12\) | ||
| Solve resulting quadratic | M1 | Attempting to solve their quadratic |
| \(y = 3 \pm 2\sqrt{3}\) | A1, A1 | Answers must be given as surds; A1 for each correct answer; both must be simplified |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Obtain $(x\pm5)^2$ and $(y\pm3)^2$ | M1 | Can be implied by $(\pm5, \pm3)$ |
| Centre is $(-5, 3)$ | A1 | Correct answer with no working implies M1A1 |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x+5)^2 + (y-3)^2 = 16$ $(=r^2)$ or $(r^2=)$ "25"+"9"$-18$ | M1 | Complete and correct method leading to $r^2 =$ "25"+"9"$-18$ |
| $r = 4$ | A1 | Only $r=4$ (not $r=-4$); correct answer with no working implies M1A1 |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $x = -3$ in either form of equation of circle to obtain simplified quadratic in $y$ | M1 | Substituting $x=-3$ and attempt to simplify to 3 term quadratic or $(y-a)^2 = b$ |
| e.g. $(-3+5)^2 + (y-3)^2 = 16 \Rightarrow (y-3)^2 = 12$ | | |
| Solve resulting quadratic | M1 | Attempting to solve their quadratic |
| $y = 3 \pm 2\sqrt{3}$ | A1, A1 | Answers must be given as surds; A1 for each correct answer; both must be simplified |
---\begin{enumerate}
\item The circle $C$ has equation
\end{enumerate}
$$x ^ { 2 } + y ^ { 2 } + 10 x - 6 y + 18 = 0$$
Find\\
(a) the coordinates of the centre of $C$,\\
(b) the radius of $C$.
The circle $C$ meets the line with equation $x = - 3$ at two points.\\
(c) Find the exact values for the $y$ coordinates of these two points, giving your answers as fully simplified surds.\\
\hfill \mbox{\textit{Edexcel C12 2015 Q7 [8]}}