| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Range of parameter for intersection |
| Difficulty | Standard +0.8 This question requires completing the square to find centre and radius (standard), but part (b) demands geometric insight to determine when a circle intersects axes at exactly three points—requiring case analysis of tangency conditions at either x or y axis. The problem-solving element and need to consider multiple geometric scenarios elevates this above routine circle questions. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Centre \(=(3,4)\) | B1 (1.1b) | Accept \(a=3\), \(b=4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2+y^2-6x-8y-p=0 \Rightarrow (x-3)^2+(y-4)^2-9-16-p=0\) | M1 (1.1a) | Rearranges into standard form with \((x\pm3)^2+(y\pm4)^2\) seen; or forms expression for radius of form \(\sqrt{(\pm3)^2+(\pm4)^2\pm p}\) |
| \((x-3)^2+(y-4)^2=25+p\) or \(\sqrt{3^2+4^2+p}\) | A1 (1.1b) | |
| Radius \(=\sqrt{25+p}\) | A1 (1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Circle passes through origin: \(\sqrt{25+p}=5 \Rightarrow p=0\); Circle just touches \(x\)-axis: \(\sqrt{25+p}=4 \Rightarrow p=-9\) | M1 (3.1a) | Begins to solve by sketching a circle through origin, or touching an axis, or substituting \(x=0\) or \(y=0\) into circle equation involving \(p\) |
| Forms equation to find \(p\) by equating radius to 5 or greater value of \(a\) and \(b\), or substituting both \(x=0\) and \(y=0\) | M1 (3.1a) | |
| \(p=0\) | R1 (2.2a) | |
| \(p=-9\) | R1 (2.2a) |
## Question 7(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre $=(3,4)$ | B1 (1.1b) | Accept $a=3$, $b=4$ |
## Question 7(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2+y^2-6x-8y-p=0 \Rightarrow (x-3)^2+(y-4)^2-9-16-p=0$ | M1 (1.1a) | Rearranges into standard form with $(x\pm3)^2+(y\pm4)^2$ seen; or forms expression for radius of form $\sqrt{(\pm3)^2+(\pm4)^2\pm p}$ |
| $(x-3)^2+(y-4)^2=25+p$ or $\sqrt{3^2+4^2+p}$ | A1 (1.1b) | |
| Radius $=\sqrt{25+p}$ | A1 (1.1b) | |
## Question 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Circle passes through origin: $\sqrt{25+p}=5 \Rightarrow p=0$; Circle just touches $x$-axis: $\sqrt{25+p}=4 \Rightarrow p=-9$ | M1 (3.1a) | Begins to solve by sketching a circle through origin, or touching an axis, or substituting $x=0$ or $y=0$ into circle equation involving $p$ |
| Forms equation to find $p$ by equating radius to 5 or greater value of $a$ and $b$, or substituting both $x=0$ and $y=0$ | M1 (3.1a) | |
| $p=0$ | R1 (2.2a) | |
| $p=-9$ | R1 (2.2a) | |7 A circle has equation
$$x ^ { 2 } + y ^ { 2 } - 6 x - 8 y = p$$
7
\begin{enumerate}[label=(\alph*)]
\item (i) State the coordinates of the centre of the circle.\\
7 (a) (ii) Find the radius of the circle in terms of $p$.\\
7
\item The circle intersects the coordinate axes at exactly three points. Find the two possible values of $p$.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2021 Q7 [8]}}