Question 7 - A-Level Maths

AQA C1 2014 June — Question 7 14 marks

Exam Board	AQA
Module	C1 (Core Mathematics 1)
Year	2014
Session	June
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circles
Type	Tangent equation at a known point on circle
Difficulty	Moderate -0.5 This is a structured multi-part C1 question covering standard circle techniques: completing the square, finding center/radius, and tangent equation using perpendicular gradients. While it requires multiple steps and some geometric reasoning in part (d) using Pythagoras, each individual technique is routine for C1 level. The scaffolding through parts (a)-(c) makes it slightly easier than average.
Spec	1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 1.03e Complete the square: find centre and radius of circle 1.03f Circle properties: angles, chords, tangents

Express the equation of the circle in the form $$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$
1. Write down the coordinates of $C$.
2. Show that the circle has radius $n \sqrt { 5 }$, where $n$ is an integer.
Find the equation of the tangent to the circle at the point $A$, giving your answer in the form $x + p y = q$, where $p$ and $q$ are integers.
The point $B$ lies on the tangent to the circle at $A$ and the length of $B C$ is 6. Find the length of $A B$.
[0pt] [3 marks]
\includegraphics[max width=\textwidth, alt={}]{f2124c89-79de-4758-b7b8-ff273345b9dd-8_1421_1709_1286_153}

Show mark scheme Show mark scheme source

Question 7:

Part (a):

Answer	Marks	Guidance
$x^2 - 10x + y^2 + 12y = -41$	M1	Attempting to complete the square
$(x-5)^2 - 25 + (y+6)^2 - 36 = -41$	A1	Both brackets correct
$(x-5)^2 + (y+6)^2 = 20$	A1	Correct final form

Part (b)(i):

Answer	Marks	Guidance
$C = (5, -6)$	B1	Correct coordinates

Part (b)(ii):

Answer	Marks	Guidance
$k = 20$, $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$	M1	Simplifying $\sqrt{20}$
$n = 2$ (integer confirmed)	A1	Conclusion stated

Part (c):

Answer	Marks	Guidance
Gradient of $CA$: $\frac{-2-(-6)}{3-5} = \frac{4}{-2} = -2$	M1	Finding gradient of radius
Gradient of tangent $= \frac{1}{2}$	M1	Using perpendicular gradient
$y - (-2) = \frac{1}{2}(x - 3)$	M1	Using point $A(3,-2)$
$2y + 4 = x - 3 \Rightarrow x - 2y = 7$	A1	Correct rearrangement
$p = -2$, $q = 7$	A1	Final answer in required form

Part (d):

Answer	Marks	Guidance
$BC^2 = AB^2 + AC^2$ (tangent-radius perpendicular, so right angle at $A$)	M1	Using Pythagoras / right angle at tangent point
$AC = \sqrt{20}$, $BC = 6$	M1	Correct values identified
$AB = \sqrt{36 - 20} = \sqrt{16} = 4$	A1	Correct answer

# Question 7:

## Part (a):
| $x^2 - 10x + y^2 + 12y = -41$ | M1 | Attempting to complete the square |
|---|---|---|
| $(x-5)^2 - 25 + (y+6)^2 - 36 = -41$ | A1 | Both brackets correct |
| $(x-5)^2 + (y+6)^2 = 20$ | A1 | Correct final form |

## Part (b)(i):
| $C = (5, -6)$ | B1 | Correct coordinates |
|---|---|---|

## Part (b)(ii):
| $k = 20$, $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$ | M1 | Simplifying $\sqrt{20}$ |
|---|---|---|
| $n = 2$ (integer confirmed) | A1 | Conclusion stated |

## Part (c):
| Gradient of $CA$: $\frac{-2-(-6)}{3-5} = \frac{4}{-2} = -2$ | M1 | Finding gradient of radius |
|---|---|---|
| Gradient of tangent $= \frac{1}{2}$ | M1 | Using perpendicular gradient |
| $y - (-2) = \frac{1}{2}(x - 3)$ | M1 | Using point $A(3,-2)$ |
| $2y + 4 = x - 3 \Rightarrow x - 2y = 7$ | A1 | Correct rearrangement |
| $p = -2$, $q = 7$ | A1 | Final answer in required form |

## Part (d):
| $BC^2 = AB^2 + AC^2$ (tangent-radius perpendicular, so right angle at $A$) | M1 | Using Pythagoras / right angle at tangent point |
|---|---|---|
| $AC = \sqrt{20}$, $BC = 6$ | M1 | Correct values identified |
| $AB = \sqrt{36 - 20} = \sqrt{16} = 4$ | A1 | Correct answer |

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Show LaTeX source

\begin{enumerate}[label=(\alph*)]
\item Express the equation of the circle in the form

$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$
\item \begin{enumerate}[label=(\roman*)]
\item Write down the coordinates of $C$.
\item Show that the circle has radius $n \sqrt { 5 }$, where $n$ is an integer.
\end{enumerate}\item Find the equation of the tangent to the circle at the point $A$, giving your answer in the form $x + p y = q$, where $p$ and $q$ are integers.
\item The point $B$ lies on the tangent to the circle at $A$ and the length of $B C$ is 6. Find the length of $A B$.\\[0pt]
[3 marks]

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{f2124c89-79de-4758-b7b8-ff273345b9dd-8_1421_1709_1286_153}
\end{center}
\end{enumerate}

\hfill \mbox{\textit{AQA C1 2014 Q7 [14]}}

This paper (8 questions)

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Q1 13 Q2 4 Q3 12 Q4 7 Q5 7 Q6 12 Q7 14 Q8 6

Answer	Marks	Guidance
\(x^2 - 10x + y^2 + 12y = -41\)	M1	Attempting to complete the square
\((x-5)^2 - 25 + (y+6)^2 - 36 = -41\)	A1	Both brackets correct
\((x-5)^2 + (y+6)^2 = 20\)	A1	Correct final form

Answer	Marks	Guidance
\(k = 20\), \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\)	M1	Simplifying \(\sqrt{20}\)
\(n = 2\) (integer confirmed)	A1	Conclusion stated

Answer	Marks	Guidance
Gradient of \(CA\): \(\frac{-2-(-6)}{3-5} = \frac{4}{-2} = -2\)	M1	Finding gradient of radius
Gradient of tangent \(= \frac{1}{2}\)	M1	Using perpendicular gradient
\(y - (-2) = \frac{1}{2}(x - 3)\)	M1	Using point \(A(3,-2)\)
\(2y + 4 = x - 3 \Rightarrow x - 2y = 7\)	A1	Correct rearrangement
\(p = -2\), \(q = 7\)	A1	Final answer in required form

Answer	Marks	Guidance
\(BC^2 = AB^2 + AC^2\) (tangent-radius perpendicular, so right angle at \(A\))	M1	Using Pythagoras / right angle at tangent point
\(AC = \sqrt{20}\), \(BC = 6\)	M1	Correct values identified
\(AB = \sqrt{36 - 20} = \sqrt{16} = 4\)	A1	Correct answer