Question 6 - A-Level Maths

AQA C1 2011 January — Question 6 13 marks

Exam Board	AQA
Module	C1 (Core Mathematics 1)
Year	2011
Session	January
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circles
Type	Tangent equation at a known point on circle
Difficulty	Easy -1.2 This is a routine multi-part question testing standard circle techniques: writing equations in different forms, finding intercepts, verifying points lie on circles, and using perpendicular gradients for tangents. All parts follow textbook procedures with no problem-solving or novel insight required. Slightly easier than average due to straightforward calculations.
Spec	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

6 A circle has centre $C ( - 3,1 )$ and radius $\sqrt { 13 }$.

1. Express the equation of the circle in the form $$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$
2. Hence find the equation of the circle in the form $$x ^ { 2 } + y ^ { 2 } + m x + n y + p = 0$$ where $m , n$ and $p$ are integers.
The circle cuts the $y$-axis at the points $A$ and $B$. Find the distance $A B$.
1. Verify that the point $D ( - 5 , - 2 )$ lies on the circle.
2. Find the gradient of $C D$.
3. Hence find an equation of the tangent to the circle at the point $D$.

Show mark scheme Show mark scheme source

6(a)(i)

Answer	Marks	Guidance
$(x + 3)^2 + (y - 1)^2$	B1	condone $(x - (-3))^2$
$= 13$	B1	condone $(\sqrt{13})^2$

6(a)(ii)

Answer	Marks	Guidance
$x^2 + 6x + 9 + y^2 - 2y + 1$	M1	attempt to multiply out both of their brackets; must have x and y terms
$x^2 + y^2 + 6x - 2y$	A1	both $m = 6$ and $n = -2$
$- 3 = 0$	A1	All correct, $p = -3$ and $\ldots = 0$

6(b)

Answer	Marks	Guidance
$x = 0 \Rightarrow y^2 - 2y - 3 = 0$	M1	putting $x = 0$ PI and attempt to solve or factorise
$\Rightarrow (y - 3)(y + 1) = 0$
$y = 3, y = -1$	A1
$\Rightarrow$ Distance $AB = 3 + 1 = 4$	A1cso	OR Pythagoras $d^2 = 13 - 3^2$ M1, $d = 2$ A1, distance = 2 × 2 = 4 A1

6(c)(i)

Answer	Marks	Guidance
$(-5 + 3)^2 + (-2 - 1)^2 = 4 + 9$		Substitution $x = -5, y = -2$ into any correct circle equation
$= 13$
$\Rightarrow D$ lies on circle	B1	convincing verification plus statement

6(c)(ii)

Answer	Marks	Guidance
grad $CD = \frac{1 + 2}{-3 + 5}$	M1	condone one sign slip
$= \frac{3}{2}$ (or 1.5)	A1	not $\frac{-3}{-2}$

6(c)(iii)

Answer	Marks	Guidance
Perpendicular gradient $= -\frac{2}{3}$	M1	ft their grad CD or $m_1m_2 = -1$ stated
Tangent has equation $y + 2 = -\frac{2}{3}(x + 5)$	A1	any form of correct equation; eg $2x + 3y + 16 = 0$; $y = -\frac{2}{3}x + c, c = -\frac{16}{3}$
Total	13

**6(a)(i)**
$(x + 3)^2 + (y - 1)^2$ | B1 | condone $(x - (-3))^2$

$= 13$ | B1 | condone $(\sqrt{13})^2$ | 2

**6(a)(ii)**
$x^2 + 6x + 9 + y^2 - 2y + 1$ | M1 | attempt to multiply out both of their brackets; must have x and y terms

$x^2 + y^2 + 6x - 2y$ | A1 | both $m = 6$ and $n = -2$

$- 3 = 0$ | A1 | All correct, $p = -3$ and $\ldots = 0$ | 3

**6(b)**
$x = 0 \Rightarrow y^2 - 2y - 3 = 0$ | M1 | putting $x = 0$ PI and attempt to solve or factorise

$\Rightarrow (y - 3)(y + 1) = 0$ | | 

$y = 3, y = -1$ | A1 | 

$\Rightarrow$ Distance $AB = 3 + 1 = 4$ | A1cso | OR Pythagoras $d^2 = 13 - 3^2$ M1, $d = 2$ A1, distance = 2 × 2 = 4 A1 | 3

**6(c)(i)**
$(-5 + 3)^2 + (-2 - 1)^2 = 4 + 9$ | | Substitution $x = -5, y = -2$ into any correct circle equation

$= 13$ | | 

$\Rightarrow D$ lies on circle | B1 | convincing verification plus statement | 1

**6(c)(ii)**
grad $CD = \frac{1 + 2}{-3 + 5}$ | M1 | condone one sign slip

$= \frac{3}{2}$ (or 1.5) | A1 | not $\frac{-3}{-2}$ | 2

**6(c)(iii)**
Perpendicular gradient $= -\frac{2}{3}$ | M1 | ft their grad CD or $m_1m_2 = -1$ stated

Tangent has equation $y + 2 = -\frac{2}{3}(x + 5)$ | A1 | any form of correct equation; eg $2x + 3y + 16 = 0$; $y = -\frac{2}{3}x + c, c = -\frac{16}{3}$ | 2

| | Total | 13

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Show LaTeX source

6 A circle has centre $C ( - 3,1 )$ and radius $\sqrt { 13 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express the equation of the circle in the form

$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$
\item Hence find the equation of the circle in the form

$$x ^ { 2 } + y ^ { 2 } + m x + n y + p = 0$$

where $m , n$ and $p$ are integers.
\end{enumerate}\item The circle cuts the $y$-axis at the points $A$ and $B$. Find the distance $A B$.
\item \begin{enumerate}[label=(\roman*)]
\item Verify that the point $D ( - 5 , - 2 )$ lies on the circle.
\item Find the gradient of $C D$.
\item Hence find an equation of the tangent to the circle at the point $D$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2011 Q6 [13]}}

This paper (7 questions)

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Q1 10 Q2 5 Q3 11 Q4 12 Q5 13 Q6 13 Q7 11

Answer	Marks	Guidance
\((x + 3)^2 + (y - 1)^2\)	B1	condone \((x - (-3))^2\)
\(= 13\)	B1	condone \((\sqrt{13})^2\)

Answer	Marks	Guidance
\(x^2 + 6x + 9 + y^2 - 2y + 1\)	M1	attempt to multiply out both of their brackets; must have x and y terms
\(x^2 + y^2 + 6x - 2y\)	A1	both \(m = 6\) and \(n = -2\)
\(- 3 = 0\)	A1	All correct, \(p = -3\) and \(\ldots = 0\)

Answer	Marks	Guidance
\(x = 0 \Rightarrow y^2 - 2y - 3 = 0\)	M1	putting \(x = 0\) PI and attempt to solve or factorise
\(\Rightarrow (y - 3)(y + 1) = 0\)
\(y = 3, y = -1\)	A1
\(\Rightarrow\) Distance \(AB = 3 + 1 = 4\)	A1cso	OR Pythagoras \(d^2 = 13 - 3^2\) M1, \(d = 2\) A1, distance = 2 × 2 = 4 A1

Answer	Marks	Guidance
\((-5 + 3)^2 + (-2 - 1)^2 = 4 + 9\)		Substitution \(x = -5, y = -2\) into any correct circle equation
\(= 13\)
\(\Rightarrow D\) lies on circle	B1	convincing verification plus statement

Answer	Marks	Guidance
grad \(CD = \frac{1 + 2}{-3 + 5}\)	M1	condone one sign slip
\(= \frac{3}{2}\) (or 1.5)	A1	not \(\frac{-3}{-2}\)

Answer	Marks	Guidance
Perpendicular gradient \(= -\frac{2}{3}\)	M1	ft their grad CD or \(m_1m_2 = -1\) stated
Tangent has equation \(y + 2 = -\frac{2}{3}(x + 5)\)	A1	any form of correct equation; eg \(2x + 3y + 16 = 0\); \(y = -\frac{2}{3}x + c, c = -\frac{16}{3}\)
Total	13