Edexcel C12 2017 January — Question 2 7 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2017
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeFind centre and radius from equation
DifficultyModerate -0.8 This is a straightforward completing-the-square exercise to find centre and radius, followed by a simple substitution (x=0) to find y-intercepts. All steps are routine procedures covered early in C1/C2 with no problem-solving or insight required, making it easier than average but not trivial.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle

A circle, with centre \(C\) and radius \(r\), has equation $$x ^ { 2 } + y ^ { 2 } - 8 x + 4 y - 12 = 0$$ Find
  1. the coordinates of \(C\),
  2. the exact value of \(r\). The circle cuts the \(y\)-axis at the points \(A\) and \(B\).
  3. Find the coordinates of the points \(A\) and \(B\).

Question 2:
Parts (a) and (b) marked together
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((x \pm 4)\ldots(y \pm 2)\)M1 Attempts to complete the square on \(x\) and \(y\), or sight of \((x \pm 4)\) and \((y \pm 2)\). May be implied by centre \((\pm 4, \pm 2)\). Or if considering \(x^2 + y^2 + 2gx + 2fy + c = 0\), centre is \((\pm g, \pm f)\)
Centre \(C = (4, -2)\)A1 Correct centre (allow \(x=4\), \(y=-2\)). But not \(g = \ldots\), \(f = \ldots\) or \(p = \ldots\), \(q = \ldots\) etc.
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(r^2 = 12 + (\pm 4)^2 + (\pm 2)^2\)M1 Must reach \(r^2 = 12 +\) their \((\pm 4)^2 +\) their \((\pm 2)^2\), or \(r = \sqrt{12 + \text{their}(\pm4)^2 + \text{their}(\pm2)^2}\). Or if using \(x^2+y^2+2gx+2fy+c=0\): \(r^2 = g^2 + f^2 - c\). Must clearly identify radius or \(r^2\). May be implied by correct exact value or awrt 5.66
\(r = \sqrt{32}\)A1 Accept exact equivalents such as \(4\sqrt{2}\). Do not allow \(\pm\sqrt{32}\) unless minus is rejected
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x = 0 \Rightarrow y^2 + 4y - 12 = 0\)B1 Correct quadratic. Allow \(16 + (y+2)^2 = 32\)
\((y+6)(y-2) = 0 \Rightarrow y = \ldots\)M1 Attempts to solve a 3TQ from substituting \(x=0\) or \(y=0\) into given or changed equation. May be implied by correct answers
\(y = 2, -6\) or \((0, 2)\) and \((0, -6)\)A1 Correct \(y\) values or correct coordinates. Accept sight of these for all 3 marks if no incorrect working; must clearly be \(y\) values or correct coordinates. May be implied by correct roots of quadratic in \(y\)
## Question 2:

### Parts (a) and (b) marked together

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x \pm 4)\ldots(y \pm 2)$ | M1 | Attempts to complete the square on $x$ and $y$, or sight of $(x \pm 4)$ **and** $(y \pm 2)$. May be implied by centre $(\pm 4, \pm 2)$. Or if considering $x^2 + y^2 + 2gx + 2fy + c = 0$, centre is $(\pm g, \pm f)$ |
| Centre $C = (4, -2)$ | A1 | Correct centre (allow $x=4$, $y=-2$). But not $g = \ldots$, $f = \ldots$ or $p = \ldots$, $q = \ldots$ etc. |

---

### Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r^2 = 12 + (\pm 4)^2 + (\pm 2)^2$ | M1 | Must reach $r^2 = 12 +$ their $(\pm 4)^2 +$ their $(\pm 2)^2$, or $r = \sqrt{12 + \text{their}(\pm4)^2 + \text{their}(\pm2)^2}$. Or if using $x^2+y^2+2gx+2fy+c=0$: $r^2 = g^2 + f^2 - c$. Must clearly identify radius or $r^2$. May be implied by correct exact value or awrt 5.66 |
| $r = \sqrt{32}$ | A1 | Accept exact equivalents such as $4\sqrt{2}$. Do not allow $\pm\sqrt{32}$ unless minus is rejected |

---

### Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 0 \Rightarrow y^2 + 4y - 12 = 0$ | B1 | Correct quadratic. Allow $16 + (y+2)^2 = 32$ |
| $(y+6)(y-2) = 0 \Rightarrow y = \ldots$ | M1 | Attempts to solve a 3TQ from substituting $x=0$ or $y=0$ into given or changed equation. May be implied by correct answers |
| $y = 2, -6$ or $(0, 2)$ and $(0, -6)$ | A1 | Correct $y$ values or correct coordinates. Accept sight of these for all 3 marks if no incorrect working; must clearly be $y$ values or correct coordinates. May be implied by correct roots of quadratic in $y$ |

---
A circle, with centre $C$ and radius $r$, has equation

$$x ^ { 2 } + y ^ { 2 } - 8 x + 4 y - 12 = 0$$

Find
\begin{enumerate}[label=(\alph*)]
\item the coordinates of $C$,
\item the exact value of $r$.

The circle cuts the $y$-axis at the points $A$ and $B$.
\item Find the coordinates of the points $A$ and $B$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C12 2017 Q2 [7]}}