| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle touching axes |
| Difficulty | Standard +0.3 This is a straightforward multi-part circle question requiring standard techniques: finding radius from geometry (touching y-axis means r=5), verifying a point lies on the circle by substitution, finding tangent using perpendicular gradient, and using the perpendicular from centre to chord property. All parts follow routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents1.07m Tangents and normals: gradient and equations1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((x-5)^2 + (y-8)^2\) | B1 | |
| \(= 25\) | B1 (2 marks) | Condone \(5^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((2-5)^2 + (12-8)^2 = 9 + 16 = 25 \Rightarrow A\) lies on circle | B1 (1 mark) | or \(AC^2 = 3^2 + 4^2\); hence \(AC = 5\) (also radius \(= 5\)); CSO; \((\Rightarrow \text{radius} = AC) \Rightarrow A\) lies on circle; must have concluding statement & RHS of circle equation correct or \(r = 5\) stated if Pythagoras is used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{grad } AC = -\frac{4}{3}\) | B1 | |
| Gradient of tangent is \(\frac{3}{4}\) | B1\(\checkmark\) | FT their \(-1/\text{grad } AC\) |
| \(y - 12 = \text{'their tangent grad'}(x-2)\) | M1 | or \(y = \text{'their tangent grad'} \cdot x + c\) & attempt to find \(c\) using \(x=2,\ y=12\) |
| \(y - 12 = \frac{3}{4}(x-2)\) or \(y = \frac{3}{4}x + \frac{21}{2}\) etc | A1 | Correct equation in any form |
| \(3x - 4y + 42 = 0\) | A1 (5 marks) | CSO; must have integer coefficients with all terms on one side of equation; accept \(0 = 8y - 6x - 84\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((CM^2 =)\ (7-5)^2 + (12-8)^2\) | M1 | or \((CM^2 =)\ 20\) |
| \((\Rightarrow CM = \sqrt{20})\ \Rightarrow CM = 2\sqrt{5}\) | A1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(PM^2 = PC^2 - CM^2 = 25 - 20\) | M1 | Pythagoras used correctly; e.g. \(d^2 + \left(2\sqrt{5}\right)^2 = 5^2\) |
| \(\Rightarrow PM = \sqrt{5}\) | A1 | |
| Area \(\triangle PCQ = \sqrt{5} \times 2\sqrt{5}\) | ||
| \(= 10\) | A1 (3 marks) | CSO |
## Question 6:
**Part 6(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(x-5)^2 + (y-8)^2$ | B1 | |
| $= 25$ | B1 (2 marks) | Condone $5^2$ |
**Part 6(b)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(2-5)^2 + (12-8)^2 = 9 + 16 = 25 \Rightarrow A$ lies on circle | B1 (1 mark) | or $AC^2 = 3^2 + 4^2$; hence $AC = 5$ (also radius $= 5$); CSO; $(\Rightarrow \text{radius} = AC) \Rightarrow A$ lies on circle; must have concluding statement & RHS of circle equation correct or $r = 5$ stated if Pythagoras is used |
**Part 6(b)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{grad } AC = -\frac{4}{3}$ | B1 | |
| Gradient of tangent is $\frac{3}{4}$ | B1$\checkmark$ | FT their $-1/\text{grad } AC$ |
| $y - 12 = \text{'their tangent grad'}(x-2)$ | M1 | **or** $y = \text{'their tangent grad'} \cdot x + c$ & attempt to find $c$ using $x=2,\ y=12$ |
| $y - 12 = \frac{3}{4}(x-2)$ **or** $y = \frac{3}{4}x + \frac{21}{2}$ etc | A1 | Correct equation in any form |
| $3x - 4y + 42 = 0$ | A1 (5 marks) | CSO; must have integer coefficients with all terms on one side of equation; accept $0 = 8y - 6x - 84$ etc |
**Part 6(c)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(CM^2 =)\ (7-5)^2 + (12-8)^2$ | M1 | or $(CM^2 =)\ 20$ |
| $(\Rightarrow CM = \sqrt{20})\ \Rightarrow CM = 2\sqrt{5}$ | A1 (2 marks) | |
**Part 6(c)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $PM^2 = PC^2 - CM^2 = 25 - 20$ | M1 | Pythagoras used correctly; e.g. $d^2 + \left(2\sqrt{5}\right)^2 = 5^2$ |
| $\Rightarrow PM = \sqrt{5}$ | A1 | |
| Area $\triangle PCQ = \sqrt{5} \times 2\sqrt{5}$ | | |
| $= 10$ | A1 (3 marks) | CSO |
---6 The circle with centre $C ( 5,8 )$ touches the $y$-axis, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{dbc25177-4a28-480f-93d5-41acb2a2d28c-5_485_631_370_715}
\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 Verify that the point $A ( 2,12 )$ lies on the circle.
\item Find an equation of the tangent to the circle at the point $A$, giving your answer in the form $s x + t y + u = 0$, where $s , t$ and $u$ are integers.
\end{enumerate}\item The points $P$ and $Q$ lie on the circle, and the mid-point of $P Q$ is $M ( 7,12 )$.
\begin{enumerate}[label=(\roman*)]
\item Show that the length of $C M$ is $n \sqrt { 5 }$, where $n$ is an integer.
\item Hence find the area of triangle $P C Q$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2012 Q6 [13]}}