AQA C1 2007 June — Question 5 14 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2007
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeTangent from external point - find equation
DifficultyModerate -0.8 This is a straightforward C1 circle question requiring only basic recall and standard procedures: reading center/radius from standard form, verifying a point satisfies the equation, finding a normal (perpendicular line through center), and applying Pythagoras for tangent length. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents1.07m Tangents and normals: gradient and equations
5 A circle with centre \(C\) has equation \(( x + 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 25\).
  1. Write down:
    1. the coordinates of \(C\);
    2. the radius of the circle.
    1. Verify that the point \(N ( 0 , - 2 )\) lies on the circle.
    2. Sketch the circle.
    3. Find an equation of the normal to the circle at the point \(N\).
  3. The point \(P\) has coordinates (2, 6).
    1. Find the distance \(P C\), leaving your answer in surd form.
    2. Find the length of a tangent drawn from \(P\) to the circle.
Question 5:
Part (a)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Centre \((-3, 2)\)M1 \(\pm 3\) or \(\pm 2\)
A1correct
Part (a)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Radius \(= 5\)B1 accept \(\sqrt{25}\) but not \(\pm\sqrt{25}\)
Part (b)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(3^2 + (-4)^2 = 9 + 16 = 25 \Rightarrow N\) lies on circleB1 must have \(9 + 16 = 25\) or a statement
Part (b)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
[Sketch of circle in correct quadrant enclosing origin]M1 must draw axes; ft their centre in correct quadrant
A1correct (reasonable freehand circle enclosing origin)
Part (b)(iii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Attempt at gradient of \(CN\)M1 withhold if subsequently finds tangent
grad \(CN = -\frac{4}{3}\)A1 CSO
\(y = -\frac{4}{3}x - 2\) (or equivalent)A1\(\checkmark\) ft their grad \(CN\)
Part (c)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(2,6)\) hence \(PC^2 = 5^2 + 4^2\)M1 "their" \(PC^2\)
\(\Rightarrow PC = \sqrt{41}\)A1
Part (c)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Use of Pythagoras correctlyM1
\(PT^2 = PC^2 - r^2 = 41 - 25\) where \(T\) is point of contactA1\(\checkmark\) ft their \(PC^2\) and \(r^2\)
\(\Rightarrow PT = 4\)A1 Alternative: sketch with vertical tangent M1, showing tangent touches circle at point \((2,2)\) A1, hence \(PT = 4\) A1 
## Question 5:

### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre $(-3, 2)$ | M1 | $\pm 3$ or $\pm 2$ |
| | A1 | correct |

### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Radius $= 5$ | B1 | accept $\sqrt{25}$ but not $\pm\sqrt{25}$ |

### Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3^2 + (-4)^2 = 9 + 16 = 25 \Rightarrow N$ lies on circle | B1 | must have $9 + 16 = 25$ or a statement |

### Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| [Sketch of circle in correct quadrant enclosing origin] | M1 | must draw axes; ft their centre in correct quadrant |
| | A1 | correct (reasonable freehand circle enclosing origin) |

### Part (b)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt at gradient of $CN$ | M1 | withhold if subsequently finds tangent |
| grad $CN = -\frac{4}{3}$ | A1 | CSO |
| $y = -\frac{4}{3}x - 2$ (or equivalent) | A1$\checkmark$ | ft their grad $CN$ |

### Part (c)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(2,6)$ hence $PC^2 = 5^2 + 4^2$ | M1 | "their" $PC^2$ |
| $\Rightarrow PC = \sqrt{41}$ | A1 | |

### Part (c)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of Pythagoras correctly | M1 | |
| $PT^2 = PC^2 - r^2 = 41 - 25$ where $T$ is point of contact | A1$\checkmark$ | ft their $PC^2$ and $r^2$ |
| $\Rightarrow PT = 4$ | A1 | **Alternative:** sketch with vertical tangent M1, showing tangent touches circle at point $(2,2)$ A1, hence $PT = 4$ A1 |
5 A circle with centre $C$ has equation $( x + 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 25$.
\begin{enumerate}[label=(\alph*)]
\item Write down:
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $C$;
\item the radius of the circle.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Verify that the point $N ( 0 , - 2 )$ lies on the circle.
\item Sketch the circle.
\item Find an equation of the normal to the circle at the point $N$.
\end{enumerate}\item The point $P$ has coordinates (2, 6).
\begin{enumerate}[label=(\roman*)]
\item Find the distance $P C$, leaving your answer in surd form.
\item Find the length of a tangent drawn from $P$ to the circle.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2007 Q5 [14]}}
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