| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Point position relative to circle |
| Difficulty | Moderate -0.8 This is a straightforward circle question requiring basic recall of circle equation form (parts a(i) and a(ii)), substitution of a point into the inequality for points inside a circle (part b), and solving a quadratic inequality (part c). All steps are routine Core 1/2 techniques with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3,-4)\) | B1 | Accept as \(x=\) , \(y=\) or even without brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sqrt{30}\) | B1 [2] | Do not accept decimals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempts \((6-3)^2+(k+4)^2 < 30\) | M1, M1 | M1: attempts length or length² from \(P(6,k)\) to centre \(C(3,-4)\); M1: forms inequality using length from \(P\) to centre \(< \) radius |
| \(k^2+8k-5<0\) | A1* [3] | Given answer; must see intermediate line with \(<30\) before \(<0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solves \(k^2+8k-5=0\) by formula or completing the square | M1 | Factorisation to integer roots not suitable; answers could just appear from graphical calculator; accept decimals for M marks only |
| \(k=-4\pm\sqrt{21}\) | A1 | Accept \(k=-4\pm\sqrt{21}\) or exact equivalent \(k=\frac{-8\pm\sqrt{84}}{2}\); do not accept decimal equivalents \(k=-8.58, (+)0.58\) for this mark |
| Chooses region between two values | M1 | Chooses inside region from their two roots |
| \(-4-\sqrt{21} < k < -4+\sqrt{21}\) | A1cao [4] | Accept equivalents such as \((-4-\sqrt{21}, -4+\sqrt{21})\); do not accept \(-4-\sqrt{21} |
# Question 13:
## Part (a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3,-4)$ | B1 | Accept as $x=$ , $y=$ or even without brackets |
## Part (a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sqrt{30}$ | B1 [2] | Do not accept decimals |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempts $(6-3)^2+(k+4)^2 < 30$ | M1, M1 | M1: attempts length or length² from $P(6,k)$ to centre $C(3,-4)$; M1: forms inequality using length from $P$ to centre $< $ radius |
| $k^2+8k-5<0$ | A1* [3] | Given answer; must see intermediate line with $<30$ before $<0$ |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solves $k^2+8k-5=0$ by formula or completing the square | M1 | Factorisation to integer roots not suitable; answers could just appear from graphical calculator; accept decimals for M marks only |
| $k=-4\pm\sqrt{21}$ | A1 | Accept $k=-4\pm\sqrt{21}$ or exact equivalent $k=\frac{-8\pm\sqrt{84}}{2}$; do not accept decimal equivalents $k=-8.58, (+)0.58$ for this mark |
| Chooses region between two values | M1 | Chooses inside region from their two roots |
| $-4-\sqrt{21} < k < -4+\sqrt{21}$ | A1cao [4] | Accept equivalents such as $(-4-\sqrt{21}, -4+\sqrt{21})$; do not accept $-4-\sqrt{21}<x<-4+\sqrt{21}$ for final mark |
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\begin{enumerate}
\item The circle $C$ has equation
\end{enumerate}
$$( x - 3 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 30$$
Write down\\
(a) (i) the coordinates of the centre of $C$,\\
(ii) the exact value of the radius of $C$.
Given that the point $P$ with coordinates $( 6 , k )$, where $k$ is a constant, lies inside circle $C$, (b) show that
$$k ^ { 2 } + 8 k - 5 < 0$$
(c) Hence find the exact set of values of $k$ for which $P$ lies inside $C$.\\
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\hfill \mbox{\textit{Edexcel C12 2017 Q13 [9]}}