Edexcel C12 2014 January — Question 15 14 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2014
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypePerpendicular bisector of chord
DifficultyModerate -0.8 This is a straightforward multi-part question testing standard circle geometry concepts: finding gradients, perpendicular bisectors, and circle equations. All parts follow routine procedures with no problem-solving insight required—students simply apply learned formulas (gradient formula, midpoint, perpendicular gradient = -1/m, distance formula). The scaffolding through parts (a)-(d) makes it easier than average.
Spec1.03a 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
15. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e878227b-d625-4ef2-ac49-a9dc05c5321a-40_883_824_212_568} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Diagram NOT drawn to scale The points \(X\) and \(Y\) have coordinates \(( 0,3 )\) and \(( 6,11 )\) respectively. \(X Y\) is a chord of a circle \(C\) with centre \(Z\), as shown in Figure 3.
  1. Find the gradient of \(X Y\). The point \(M\) is the midpoint of \(X Y\).
  2. Find an equation for the line which passes through \(Z\) and \(M\). Given that the \(y\) coordinate of \(Z\) is 10 ,
  3. find the \(x\) coordinate of \(Z\),
  4. find the equation of the circle \(C\), giving your answer in the form $$x ^ { 2 } + y ^ { 2 } + a x + b y + c = 0$$ where \(a\), \(b\) and \(c\) are constants.
Question 15:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
gradient \(= \frac{11-3}{6-0} = \frac{4}{3}\)M1 A1
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Mid-point of \(XY = (3, 7)\)M1 A1
\(ZM\) has gradient \(-\frac{1}{m}\left(= -\frac{3}{4}\right)\)B1ft
Either: \(y-\text{"}7\text{"} = \text{"}-\frac{3}{4}\text{"}(x-\text{"}3\text{"})\) or \(y=\text{"}-\frac{3}{4}\text{"}x+c\) and \(\text{"7"} = \text{"}-\frac{3}{4}\text{"}(\text{"3"})+c \Rightarrow c = \text{"}9\frac{1}{4}\text{"}\)M1
\(4y + 3x - 37 = 0\) or \(y-7=-\frac{3}{4}(x-3)\) or \(y=-\frac{3}{4}x+9\frac{1}{4}\)A1
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Substitute \(y=10\) into their line equation to give \(x=\)M1
\(x = -1\)A1
Part (d):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((r^2) = (-1-0)^2 + (10-3)^2\) or \((r^2)=(-1-6)^2+(10-11)^2\)M1
\(r^2 = 50\)A1
\(\text{"}50\text{"} = (x\pm\text{"(-1)"})^2 + (y\pm\text{"10"})^2\)M1
\(\text{"}50\text{"} = (x-\text{"(-1)"})^2 + (y-\text{"10"})^2\)A1ft
\(x^2 + y^2 + 2x - 20y + 51 = 0\)A1
Question 15:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
States gradient equation or uses correctlyM1
\(\frac{4}{3}\) or \(\frac{8}{6}\) or decimal equivalentA1
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Uses midpoint formula, or implied by \(y\) coordinate of 7M1
\((3, 7)\)A1 cao
Uses negative reciprocal of their gradientB1 Follow through their gradient
Line equation with their midpoint and perpendicular gradientM1
Correct equation at any stageA1 May be unsimplified, isw. Should be linear
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Substitute \(y = 10\) into line equation to give \(x =\)M1
Correct value of \(x\)A1 cao — answer only with no working may have M1A1
Part (d)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Finds radius or diameter or \(r^2\) using any valid method — probably distance from centre to one of the points. Need not state \(r =\)M1
\(r^2 = 50\) or \(r = \sqrt{50}\) etcA1 Numeric answer must be identified. If they halve or double it, this is M1 A0
Attempt to use a true circle equation with their centre and their radius or letter \(r\) — allow sign slips in brackets. Do not allow use of \(r\) instead of \(r^2\) in the equationM1
Correct work following their centre and genuine attempt at radiusA1ft
Correct equation in final formA1
Alternative: \(a = 2\), \(b = -20\), \(c = 51\) is sufficient Do not need to write out full equation at the end 
## Question 15:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| gradient $= \frac{11-3}{6-0} = \frac{4}{3}$ | M1 A1 | |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mid-point of $XY = (3, 7)$ | M1 A1 | |
| $ZM$ has gradient $-\frac{1}{m}\left(= -\frac{3}{4}\right)$ | B1ft | |
| Either: $y-\text{"}7\text{"} = \text{"}-\frac{3}{4}\text{"}(x-\text{"}3\text{"})$ or $y=\text{"}-\frac{3}{4}\text{"}x+c$ and $\text{"7"} = \text{"}-\frac{3}{4}\text{"}(\text{"3"})+c \Rightarrow c = \text{"}9\frac{1}{4}\text{"}$ | M1 | |
| $4y + 3x - 37 = 0$ or $y-7=-\frac{3}{4}(x-3)$ or $y=-\frac{3}{4}x+9\frac{1}{4}$ | A1 | |

### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $y=10$ into their line equation to give $x=$ | M1 | |
| $x = -1$ | A1 | |

### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(r^2) = (-1-0)^2 + (10-3)^2$ or $(r^2)=(-1-6)^2+(10-11)^2$ | M1 | |
| $r^2 = 50$ | A1 | |
| $\text{"}50\text{"} = (x\pm\text{"(-1)"})^2 + (y\pm\text{"10"})^2$ | M1 | |
| $\text{"}50\text{"} = (x-\text{"(-1)"})^2 + (y-\text{"10"})^2$ | A1ft | |
| $x^2 + y^2 + 2x - 20y + 51 = 0$ | A1 | |

# Question 15:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| States gradient equation or uses correctly | M1 | |
| $\frac{4}{3}$ or $\frac{8}{6}$ or decimal equivalent | A1 | |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses midpoint formula, or implied by $y$ coordinate of 7 | M1 | |
| $(3, 7)$ | A1 | cao |
| Uses negative reciprocal of their gradient | B1 | Follow through their gradient |
| Line equation with their midpoint and perpendicular gradient | M1 | |
| Correct equation at any stage | A1 | **May be unsimplified**, isw. Should be linear |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $y = 10$ into line equation to give $x =$ | M1 | |
| Correct value of $x$ | A1 | cao — answer only with no working may have M1A1 |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds radius or diameter or $r^2$ using any valid method — probably distance from centre to one of the points. Need not state $r =$ | M1 | |
| $r^2 = 50$ or $r = \sqrt{50}$ etc | A1 | Numeric answer must be identified. If they halve or double it, this is M1 A0 |
| Attempt to use a true circle equation with their centre and their radius or letter $r$ — allow sign slips in brackets. Do not allow use of $r$ instead of $r^2$ in the equation | M1 | |
| Correct work following their centre and genuine attempt at radius | A1ft | |
| Correct equation in final form | A1 | |
| **Alternative:** $a = 2$, $b = -20$, $c = 51$ is sufficient | | Do not need to write out full equation at the end |
15.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{e878227b-d625-4ef2-ac49-a9dc05c5321a-40_883_824_212_568}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

Diagram NOT drawn to scale

The points $X$ and $Y$ have coordinates $( 0,3 )$ and $( 6,11 )$ respectively. $X Y$ is a chord of a circle $C$ with centre $Z$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $X Y$.

The point $M$ is the midpoint of $X Y$.
\item Find an equation for the line which passes through $Z$ and $M$.

Given that the $y$ coordinate of $Z$ is 10 ,
\item find the $x$ coordinate of $Z$,
\item find the equation of the circle $C$, giving your answer in the form

$$x ^ { 2 } + y ^ { 2 } + a x + b y + c = 0$$

where $a$, $b$ and $c$ are constants.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C12 2014 Q15 [14]}}
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