Question 5 - A-Level Maths

AQA C1 2009 June — Question 5 11 marks

Exam Board	AQA
Module	C1 (Core Mathematics 1)
Year	2009
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circles
Type	Tangent equation at a known point on circle
Difficulty	Moderate -0.8 This is a straightforward C1 circle question testing basic recall and standard procedures: reading center/radius from equation, verifying a point lies on the circle, finding gradient and perpendicular tangent. All steps are routine textbook exercises with no problem-solving or insight required, making it easier than average but not trivial due to the multi-part structure.
Spec	1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 1.03e Complete the square: find centre and radius of circle 1.07m Tangents and normals: gradient and equations

5 A circle with centre $C$ has equation $$( x - 5 ) ^ { 2 } + ( y + 12 ) ^ { 2 } = 169$$

Write down:
1. the coordinates of $C$;
2. the radius of the circle.
1. Verify that the circle passes through the origin $O$.
2. Given that the circle also passes through the points $( 10,0 )$ and $( 0 , p )$, sketch the circle and find the value of $p$.
The point $A ( - 7 , - 7 )$ lies on the circle.
1. Find the gradient of $A C$.
2. Hence find an equation of the tangent to the circle at the point $A$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.

Show mark scheme Show mark scheme source

Question 5:

Part (a)(i)

Answer	Marks	Guidance
$C(5, -12)$	B1	Centre correct

Part (a)(ii)

Answer	Marks	Guidance
Radius $= 13$ (or $\sqrt{169}$)	B1	$\pm\sqrt{169}$ or $\pm 13$ as final answer scores B0

Part (b)(i)

Answer	Marks	Guidance
$(-5)^2 + 12^2$ or $25 + 144 = 169 \Rightarrow$ circle passes through $O$	B1	Correct arithmetic plus statement e.g. "O lies on circle", "as required" etc

Part (b)(ii)

Answer	Marks	Guidance
Sketch: freehand circle through origin, cutting positive $x$-axis, centre in 4th quadrant	B1	Condone value 10 missing or incorrect
$25 + (p+12)^2 = 169$	M1	Or doubling their $y_c$-coordinate
$(p+12) = \pm 12 \Rightarrow p = -24$	A1 [3]	Condone use of $y$ instead of $p$; SC B2 for correct value of $p$ stated or marked on diagram

Part (c)(i)

Answer	Marks	Guidance
$\text{grad } AC = \dfrac{-12+7}{5+7}$	M1	Correct expression, but ft their $C$
$= -\dfrac{5}{12}$	A1 [2]	Condone $\dfrac{5}{-12}$

Part (c)(ii)

Answer	Marks	Guidance
$\text{grad tangent} = \dfrac{12}{5}$	B1$\checkmark$	$\dfrac{-1}{\text{their grad }AC}$
$y + 7 = \dfrac{12}{5}(x+7)$	M1	ft their $\dfrac{12}{5}$, must be tangent and not $AC$
$\Rightarrow 12x - 5y + 49 = 0$	A1 [3]	OE with integer coefficients, all terms on one side

## Question 5:

**Part (a)(i)**
$C(5, -12)$ | B1 | Centre correct

**Part (a)(ii)**
Radius $= 13$ (or $\sqrt{169}$) | B1 | $\pm\sqrt{169}$ or $\pm 13$ as final answer scores B0

**Part (b)(i)**
$(-5)^2 + 12^2$ or $25 + 144 = 169 \Rightarrow$ circle passes through $O$ | B1 | Correct arithmetic plus statement e.g. "O lies on circle", "as required" etc

**Part (b)(ii)**
Sketch: freehand circle through origin, cutting positive $x$-axis, centre in 4th quadrant | B1 | Condone value 10 missing or incorrect

$25 + (p+12)^2 = 169$ | M1 | Or doubling their $y_c$-coordinate

$(p+12) = \pm 12 \Rightarrow p = -24$ | A1 [3] | Condone use of $y$ instead of $p$; SC B2 for correct value of $p$ stated or marked on diagram

**Part (c)(i)**
$\text{grad } AC = \dfrac{-12+7}{5+7}$ | M1 | Correct expression, but ft their $C$

$= -\dfrac{5}{12}$ | A1 [2] | Condone $\dfrac{5}{-12}$

**Part (c)(ii)**
$\text{grad tangent} = \dfrac{12}{5}$ | B1$\checkmark$ | $\dfrac{-1}{\text{their grad }AC}$

$y + 7 = \dfrac{12}{5}(x+7)$ | M1 | ft their $\dfrac{12}{5}$, must be **tangent** and not $AC$

$\Rightarrow 12x - 5y + 49 = 0$ | A1 [3] | OE with **integer** coefficients, all terms on one side

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Show LaTeX source

5 A circle with centre $C$ has equation

$$( x - 5 ) ^ { 2 } + ( y + 12 ) ^ { 2 } = 169$$
\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 circle passes through the origin $O$.
\item Given that the circle also passes through the points $( 10,0 )$ and $( 0 , p )$, sketch the circle and find the value of $p$.
\end{enumerate}\item The point $A ( - 7 , - 7 )$ lies on the circle.
\begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A C$.
\item Hence find an equation of the tangent to the circle at the point $A$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2009 Q5 [11]}}

This paper (7 questions)

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Q1 8 Q2 7 Q3 13 Q4 17 Q5 11 Q6 10 Q7 9

Answer	Marks	Guidance
Sketch: freehand circle through origin, cutting positive \(x\)-axis, centre in 4th quadrant	B1	Condone value 10 missing or incorrect
\(25 + (p+12)^2 = 169\)	M1	Or doubling their \(y_c\)-coordinate
\((p+12) = \pm 12 \Rightarrow p = -24\)	A1 [3]	Condone use of \(y\) instead of \(p\); SC B2 for correct value of \(p\) stated or marked on diagram

Answer	Marks	Guidance
\(\text{grad } AC = \dfrac{-12+7}{5+7}\)	M1	Correct expression, but ft their \(C\)
\(= -\dfrac{5}{12}\)	A1 [2]	Condone \(\dfrac{5}{-12}\)

Answer	Marks	Guidance
\(\text{grad tangent} = \dfrac{12}{5}\)	B1\(\checkmark\)	\(\dfrac{-1}{\text{their grad }AC}\)
\(y + 7 = \dfrac{12}{5}(x+7)\)	M1	ft their \(\dfrac{12}{5}\), must be tangent and not \(AC\)
\(\Rightarrow 12x - 5y + 49 = 0\)	A1 [3]	OE with integer coefficients, all terms on one side