| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - find equation |
| Difficulty | Standard +0.3 This is a straightforward multi-part circle question testing standard techniques: finding intercepts by substitution, identifying radius from equation, distance formula, and Pythagoras with tangent-radius perpendicularity. All parts are routine C1 exercises requiring no problem-solving insight, though the tangent length calculation in (c)(ii) is slightly less mechanical than parts (a)-(b), placing it just above average difficulty. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 0 \Rightarrow y^2 - 4y - 12 (= 0)\) | M1 | sub \(x = 0\) & correct quadratic in \(y\) or \((y - 2)^2 = 16\) or \((y - 2)^2 - 16 = 0\) |
| \((y - 6)(y + 2) (= 0)\) | A1 | correct factors or formula as far as \(\frac{4 \pm \sqrt{64}}{2}\) or \(y - 2 = \pm\sqrt{16}\) |
| \(\Rightarrow y = -2, 6\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \((x + 3)^2 - 9 + (y - 2)^2 - 4 (= 12)\) | M1 | correct sum of square terms and attempt to complete squares (or multiply out); PI by next line |
| \((r^2 =) 9 + 4 + 12\) | A1 | \((r^2 =) 25\) seen on RHS |
| \((\Rightarrow r =) 5\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \((CP^2 =) (2 - (-3))^2 + (5 - 2)^2\) | M1 | condone one sign slip within one bracket |
| \(\Rightarrow (CP =) \sqrt{34}\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(PQ^2 = CP^2 - r^2 = 34 - 25\) | M1 | Pythagoras used correctly with values; FT 'their' \(r\) and \(CP\) |
| \((\Rightarrow PQ =) 3\) | A1 | 2 marks |
**7(a)**
$x = 0 \Rightarrow y^2 - 4y - 12 (= 0)$ | M1 | sub $x = 0$ & correct quadratic in $y$ or $(y - 2)^2 = 16$ or $(y - 2)^2 - 16 = 0$
$(y - 6)(y + 2) (= 0)$ | A1 | correct factors or formula as far as $\frac{4 \pm \sqrt{64}}{2}$ or $y - 2 = \pm\sqrt{16}$
$\Rightarrow y = -2, 6$ | A1 | 3 marks | condone $(0, -2)$ & $(0, 6)$
**7(b)**
$(x + 3)^2 - 9 + (y - 2)^2 - 4 (= 12)$ | M1 | **correct** sum of square terms and attempt to complete squares (or multiply out); PI by next line
$(r^2 =) 9 + 4 + 12$ | A1 | $(r^2 =) 25$ seen on RHS
$(\Rightarrow r =) 5$ | A1 | 3 marks | $r = \sqrt{25}$ or $r = \pm 5$ scores A0
**7(c)(i)**
$(CP^2 =) (2 - (-3))^2 + (5 - 2)^2$ | M1 | condone one sign slip within one bracket
$\Rightarrow (CP =) \sqrt{34}$ | A1 | 2 marks | $n = 34$
**7(c)(ii)**
$PQ^2 = CP^2 - r^2 = 34 - 25$ | M1 | Pythagoras used correctly with values; FT 'their' $r$ and $CP$
$(\Rightarrow PQ =) 3$ | A1 | 2 marks |
---7 A circle with centre $C ( - 3,2 )$ has equation
$$x ^ { 2 } + y ^ { 2 } + 6 x - 4 y = 12$$
\begin{enumerate}[label=(\alph*)]
\item Find the $y$-coordinates of the points where the circle crosses the $y$-axis.
\item Find the radius of the circle.
\item The point $P ( 2,5 )$ lies outside the circle.
\begin{enumerate}[label=(\roman*)]
\item Find the length of $C P$, giving your answer in the form $\sqrt { n }$, where $n$ is an integer.
\item The point $Q$ lies on the circle so that $P Q$ is a tangent to the circle. Find the length of $P Q$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q7 [10]}}