Question 8 - A-Level Maths

AQA Paper 1 2022 June — Question 8 11 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2022
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circles
Type	Distance from centre to line
Difficulty	Standard +0.3 This is a straightforward coordinate geometry question requiring standard techniques: finding intersection points, perpendicular distance between parallel lines, and using the property that tangent lines are equidistant from a circle's centre. All steps are routine applications of formulas with no novel problem-solving required, making it slightly easier than average.
Spec	1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 1.03b Straight lines: parallel and perpendicular relationships 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2

8 The lines $L _ { 1 }$ and $L _ { 2 }$ are parallel. $L _ { 1 }$ has equation $$5 x + 3 y = 15$$ and $L _ { 2 }$ has equation $$5 x + 3 y = 83$$ $L _ { 1 }$ intersects the $y$-axis at the point $P$.
The point $Q$ is the point on $L _ { 2 }$ closest to $P$, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-10_849_917_945_561} 8

1. Find the coordinates of $Q$.
  8
  1. (ii) Hence show that $P Q = k \sqrt { 34 }$, where $k$ is an integer to be found. 8
2. A circle, $C$, has centre ( $a , - 17$ ). $L _ { 1 }$ and $L _ { 2 }$ are both tangents to $C$.
  8
3. 1. Find $a$.
    8
4. (ii) Find the equation of $C$. \includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-13_2493_1732_214_139}

Show mark scheme Show mark scheme source

Question 8(a)(i):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
At $P$: $x = 0 \Rightarrow y = 5$, so $P = (0,5)$	B1	Correct $y$-intercept at $P$; $(0,5)$ or $y=5$; PI by correct equation for line $PQ$
Line $PQ$: $3x - 5y = -25$ (gradient $\frac{3}{5}$)	M1	Equation of $PQ$ with correct gradient; e.g. $y = \frac{3}{5}x + c$ or $5y - 3x = k$; or forms distance/distance squared equation from $(0,5)$ to point on $L_2$
$3x - 5y = -25$	A1	Correct equation; ACF
Solves $PQ$ and $L_2$: $5x + 3y = 83$ simultaneously	M1	Solves simultaneous equations; $PQ$ must not be horizontal or vertical; condone rearrangement errors; or minimises distance squared to find one coordinate
$Q = (10, 11)$	A1	$x = 10$, $y = 11$

Question 8(a)(ii):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$PQ^2 = (10-0)^2 + (11-5)^2$	M1	Uses distance formula for $PQ$ or $PQ^2$; or Pythagoras with $10^2 + 6^2$; if coordinates of $P$ and $Q$ are incorrect, differences in $x$ and $y$ must be clearly shown
$PQ = \sqrt{136} = 2\sqrt{34}$	R1	Completes demonstration that $k=2$; must show clear use of distance formula; condone not seeing $x=0$ substituted; answer of $2\sqrt{34}$ with no working scores M1 R0

Question 8(b)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Valid method to find $a$: e.g. $5x + 3y = 49$, then $5a + 3(-17) = 49$	M1	AO 3.1a. Evidence could be: forming equation of line mid-way between $L_1$ and $L_2$; using $(a, -17)$ as midpoint of segment from $L_1$ to $L_2$; finding midpoint of $PQ$ using gradient $-\frac{5}{3}$; substituting $y=17$ and $x=a$ into $y=\frac{3}{5}x+5$
$a = 20$	R1	AO 2.2a. Deduces $a=20$

Question 8(b)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Forms $(x \pm a)^2 + (y \pm 17)^2$ using $a$ or their value of $a$	M1	AO 1.1a
$(x-20)^2 + (y+17)^2 = 34$	A1F	AO 1.1b. Correct equation for their $a$; radius$^2 = \frac{17k^2}{2}$ from part (a)(ii) for integer $k$; condone $(\sqrt{34})^2$

## Question 8(a)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| At $P$: $x = 0 \Rightarrow y = 5$, so $P = (0,5)$ | B1 | Correct $y$-intercept at $P$; $(0,5)$ or $y=5$; PI by correct equation for line $PQ$ |
| Line $PQ$: $3x - 5y = -25$ (gradient $\frac{3}{5}$) | M1 | Equation of $PQ$ with correct gradient; e.g. $y = \frac{3}{5}x + c$ or $5y - 3x = k$; or forms distance/distance squared equation from $(0,5)$ to point on $L_2$ |
| $3x - 5y = -25$ | A1 | Correct equation; ACF |
| Solves $PQ$ and $L_2$: $5x + 3y = 83$ simultaneously | M1 | Solves simultaneous equations; $PQ$ must not be horizontal or vertical; condone rearrangement errors; or minimises distance squared to find one coordinate |
| $Q = (10, 11)$ | A1 | $x = 10$, $y = 11$ |

---

## Question 8(a)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $PQ^2 = (10-0)^2 + (11-5)^2$ | M1 | Uses distance formula for $PQ$ or $PQ^2$; or Pythagoras with $10^2 + 6^2$; if coordinates of $P$ and $Q$ are incorrect, differences in $x$ and $y$ must be clearly shown |
| $PQ = \sqrt{136} = 2\sqrt{34}$ | R1 | Completes demonstration that $k=2$; must show clear use of distance formula; condone not seeing $x=0$ substituted; answer of $2\sqrt{34}$ with no working scores M1 R0 |

## Question 8(b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Valid method to find $a$: e.g. $5x + 3y = 49$, then $5a + 3(-17) = 49$ | M1 | AO 3.1a. Evidence could be: forming equation of line mid-way between $L_1$ and $L_2$; using $(a, -17)$ as midpoint of segment from $L_1$ to $L_2$; finding midpoint of $PQ$ using gradient $-\frac{5}{3}$; substituting $y=17$ and $x=a$ into $y=\frac{3}{5}x+5$ |
| $a = 20$ | R1 | AO 2.2a. Deduces $a=20$ |

---

## Question 8(b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Forms $(x \pm a)^2 + (y \pm 17)^2$ using $a$ or their value of $a$ | M1 | AO 1.1a |
| $(x-20)^2 + (y+17)^2 = 34$ | A1F | AO 1.1b. Correct equation for their $a$; radius$^2 = \frac{17k^2}{2}$ from part (a)(ii) for integer $k$; condone $(\sqrt{34})^2$ |

---

Show LaTeX source

8 The lines $L _ { 1 }$ and $L _ { 2 }$ are parallel.\\
$L _ { 1 }$ has equation

$$5 x + 3 y = 15$$

and $L _ { 2 }$ has equation

$$5 x + 3 y = 83$$

$L _ { 1 }$ intersects the $y$-axis at the point $P$.\\
The point $Q$ is the point on $L _ { 2 }$ closest to $P$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-10_849_917_945_561}

8
\begin{enumerate}[label=(\alph*)]
\item (i) Find the coordinates of $Q$.\\

8 (a) (ii) Hence show that $P Q = k \sqrt { 34 }$, where $k$ is an integer to be found.

8
\item A circle, $C$, has centre ( $a , - 17$ ).\\
$L _ { 1 }$ and $L _ { 2 }$ are both tangents to $C$.\\
8 (b) (i) Find $a$.\\

8 (b) (ii) Find the equation of $C$.\\

\includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-13_2493_1732_214_139}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2022 Q8 [11]}}

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