Edexcel P2 2018 Specimen — Question 7 10 marks

Exam BoardEdexcel
ModuleP2 (Pure Mathematics 2)
Year2018
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeTangent from external point - find equation
DifficultyModerate -0.5 This is a straightforward multi-part circle question testing standard techniques: completing the square to find centre/radius, substituting a line equation to find intersection points, and using Pythagoras with a tangent. All parts are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-19_739_871_260_532} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} The circle with equation $$x ^ { 2 } + y ^ { 2 } - 20 x - 16 y + 139 = 0$$ had centre \(C\) and radius \(r\).
  1. Find the coordinates of \(C\).
  2. Show that \(r = 5\) The line with equation \(x = 13\) crosses the circle at the points \(P\) and \(Q\) as shown in Figure 1 .
  3. Find the \(y\) coordinate of \(P\) and the \(y\) coordinate of \(Q\). A tangent to the circle from \(O\) touches the circle at point \(X\).
  4. Find, in surd form, the length \(O X\). \includegraphics[max width=\textwidth, alt={}, center]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-22_2673_1948_107_118}
Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Obtain \((x \pm 10)^2\) and \((y \pm 8)^2\)M1 May be implied by one correct coordinate
Centre = \((10, 8)\)A1 Answer only scores both marks
Alternative Method 2: From \(x^2 + y^2 + 2gx + 2fy + c = 0\), centre is \((\pm g, \pm f)\)
AnswerMarks
Obtains \((\pm 10, \pm 8)\)M1
Centre is \((-g, -f)\), so centre is \((10, 8)\)A1
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
See \((x \pm 10)^2 + (y \pm 8)^2 = 25\) or \((r^2=)\) "\(100\)"\(+\)"\(64\)"\(- 139\)M1 Allow attempt using \((x\pm10)^2+(y\pm8)^2=r^2\) to identify \(r\)
\(r = 5\)A1* Printed answer, correct method must be seen
Alternative:
AnswerMarks
Attempts \(\sqrt{g^2 + f^2 - c}\) or \((r^2=)\)"\(100\)"\(+\)"\(64\)"\(-139\)M1
\(r = 5\) following correct methodA1*
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Substitute \(x = 13\) into circle equation, solve quadratic for \(y\)M1 Substitutes \(x=13\) into either form
\(x=13 \Rightarrow (13-10)^2 + (y-8)^2 = 25 \Rightarrow (y-8)^2 = 16\)A1 Either \(y=4\) or \(12\)
\(y = 4\) or \(12\)A1 Both \(y=4\) and \(12\)
*N.B. Can be attempted via 3,4,5 triangle: spotting this and achieving one value for \(y\) is M1 A1. Both values scores M1 A1 A1*
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(OC = \sqrt{10^2 + 8^2} = \sqrt{164}\)M1 Uses Pythagoras to find length \(OC\) using \((10,8)\)
Length of tangent \(= \sqrt{164 - 5^2} = \sqrt{139}\)M1 A1 Uses Pythagoras to find \(OX\); look for \(\sqrt{OC^2 - r^2}\); \(\sqrt{139}\) only 
## Question 7:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain $(x \pm 10)^2$ **and** $(y \pm 8)^2$ | M1 | May be implied by one correct coordinate |
| Centre = $(10, 8)$ | A1 | Answer only scores both marks |

**Alternative Method 2:** From $x^2 + y^2 + 2gx + 2fy + c = 0$, centre is $(\pm g, \pm f)$
| Obtains $(\pm 10, \pm 8)$ | M1 | |
| Centre is $(-g, -f)$, **so centre is** $(10, 8)$ | A1 | |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| See $(x \pm 10)^2 + (y \pm 8)^2 = 25$ **or** $(r^2=)$ "$100$"$+$"$64$"$- 139$ | M1 | Allow attempt using $(x\pm10)^2+(y\pm8)^2=r^2$ to identify $r$ |
| $r = 5$ | A1* | Printed answer, correct method must be seen |

**Alternative:**
| Attempts $\sqrt{g^2 + f^2 - c}$ or $(r^2=)$"$100$"$+$"$64$"$-139$ | M1 | |
| $r = 5$ following correct method | A1* | |

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x = 13$ into circle equation, solve quadratic for $y$ | M1 | Substitutes $x=13$ into either form |
| $x=13 \Rightarrow (13-10)^2 + (y-8)^2 = 25 \Rightarrow (y-8)^2 = 16$ | A1 | Either $y=4$ **or** $12$ |
| $y = 4$ **or** $12$ | A1 | Both $y=4$ **and** $12$ |

*N.B. Can be attempted via 3,4,5 triangle: spotting this and achieving one value for $y$ is M1 A1. Both values scores M1 A1 A1*

### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $OC = \sqrt{10^2 + 8^2} = \sqrt{164}$ | M1 | Uses Pythagoras to find length $OC$ using $(10,8)$ |
| Length of tangent $= \sqrt{164 - 5^2} = \sqrt{139}$ | M1 A1 | Uses Pythagoras to find $OX$; look for $\sqrt{OC^2 - r^2}$; $\sqrt{139}$ only |

---
7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-19_739_871_260_532}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

The circle with equation

$$x ^ { 2 } + y ^ { 2 } - 20 x - 16 y + 139 = 0$$

had centre $C$ and radius $r$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $C$.
\item Show that $r = 5$

The line with equation $x = 13$ crosses the circle at the points $P$ and $Q$ as shown in Figure 1 .
\item Find the $y$ coordinate of $P$ and the $y$ coordinate of $Q$.

A tangent to the circle from $O$ touches the circle at point $X$.
\item Find, in surd form, the length $O X$.\\

\includegraphics[max width=\textwidth, alt={}, center]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-22_2673_1948_107_118}
\end{enumerate}

\hfill \mbox{\textit{Edexcel P2 2018 Q7 [10]}}
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