| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Find equations of tangent lines with given gradient or from external point using discriminant |
| Difficulty | Standard +0.3 This is a standard AS-level circle-tangent question requiring substitution of a line into a circle equation, applying the discriminant condition for tangency (b²-4ac=0), and finding points of contact. All steps are routine applications of well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Circle equation: \((x-0)^2 + (y-10)^2 = (\sqrt{20})^2\) | B1 (1.1b) | Any correct form |
| Substitute \(y = mx\): \(x^2 + m^2x^2 - 20mx + 80 = 0\) | M1 (1.1a) | Substitutes \(mx\) for \(y\) |
| \((1 + m^2)x^2 - 20mx + 80 = 0\) | R1 (2.1) | Simplifies to given quadratic AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using \(b^2 = 4ac\): \(400m^2 = 4 \times (1+m^2) \times 80\) | M1 (3.1a) | Uses discriminant of given equation |
| \(5m^2 = 4(1+m^2)\), so \(m^2 = 4\) | A1 (1.1b) | Correct equation in \(m\) |
| \(m = \pm 2\) | A1 (1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use \(m = 2\): \(5x^2 - 40x + 80 = 0\), \(x = 4\) | M1 (3.1a) | Uses one \(m\) value from (a)(ii) |
| \(x = 4\) (or \(x = -4\)) | A1 (1.1b) | One correct \(x\) value |
| Use line equation to find \(y\): \(y = 8\) | M1 (1.1a) | |
| \((4, 8)\) and \((-4, 8)\) | A1 (1.1b) | Both correct coordinate sets |
## Question 11(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Circle equation: $(x-0)^2 + (y-10)^2 = (\sqrt{20})^2$ | B1 (1.1b) | Any correct form |
| Substitute $y = mx$: $x^2 + m^2x^2 - 20mx + 80 = 0$ | M1 (1.1a) | Substitutes $mx$ for $y$ |
| $(1 + m^2)x^2 - 20mx + 80 = 0$ | R1 (2.1) | Simplifies to given quadratic AG |
## Question 11(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $b^2 = 4ac$: $400m^2 = 4 \times (1+m^2) \times 80$ | M1 (3.1a) | Uses discriminant of given equation |
| $5m^2 = 4(1+m^2)$, so $m^2 = 4$ | A1 (1.1b) | Correct equation in $m$ |
| $m = \pm 2$ | A1 (1.1b) | |
## Question 11(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $m = 2$: $5x^2 - 40x + 80 = 0$, $x = 4$ | M1 (3.1a) | Uses one $m$ value from (a)(ii) |
| $x = 4$ (or $x = -4$) | A1 (1.1b) | One correct $x$ value |
| Use line equation to find $y$: $y = 8$ | M1 (1.1a) | |
| $(4, 8)$ and $(-4, 8)$ | A1 (1.1b) | Both correct coordinate sets |11 A circle $C$ has centre $( 0,10 )$ and radius $\sqrt { 20 }$
A line $L$ has equation $y = m x$\\
11
\begin{enumerate}[label=(\alph*)]
\item (i) Show that the $x$-coordinate of any point of intersection of $L$ and $C$ satisfies the equation
$$\left( 1 + m ^ { 2 } \right) x ^ { 2 } - 20 m x + 80 = 0$$
11 (a) (ii) Find the values of $m$ for which the equation in part (a)(i) has equal roots.\\
11
\item Two lines are drawn from the origin which are tangents to $C$.
Find the coordinates of the points of contact between the tangents and $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2021 Q11 [10]}}