| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Moderate -0.3 This is a structured multi-part question covering standard C1 circle techniques: completing the square (routine), reading center/radius, substitution to find intersection, using discriminant for tangency, and distance comparison. All parts are guided with no novel insight required, making it slightly easier than average but still requiring multiple techniques. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02q Use intersection points: of graphs to solve equations1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 + (y-5)^2\) RHS \(= 5\) | B1, B1 | \(b = 5\); \(k = 5\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Centre \((0, 5)\) | B1\(\uparrow\) | FT their \(b\) from part (a) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Radius \(= \sqrt{5}\) | B1\(\uparrow\) | FT their \(k\) from part (a); RHS must be \(> 0\) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 + 4x^2 - 20x + 20 = 0\) \(\Rightarrow x^2 - 4x + 4 = 0\) | M1, A1 | May substitute into original or "their (a)"; CSO; AG |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-2)^2 = 0\) or \(x = 2\) Repeated root implies tangent Point of contact is \(P(2, 4)\) | M1, E1, A1 | Or \(b^2 - 4ac\) shown \(= 0\) plus statement |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \((CQ^2) = 1^2 + 1^2\) \(\sqrt{2} < \sqrt{5} \Rightarrow Q\) lies inside circle | M1, A1 CSO | FT their \(C\); \(CQ\) or \(CQ^2\) OE must appear for A1 |
| 2 | ||
| 11 |
## Part (a)
$x^2 + (y-5)^2$ RHS $= 5$ | B1, B1 | $b = 5$; $k = 5$
| | 2 |
## Part (b)(i)
Centre $(0, 5)$ | B1$\uparrow$ | FT their $b$ from part (a)
| | 1 |
## Part (b)(ii)
Radius $= \sqrt{5}$ | B1$\uparrow$ | FT their $k$ from part (a); RHS must be $> 0$
| | 1 |
## Part (c)(i)
$x^2 + 4x^2 - 20x + 20 = 0$ $\Rightarrow x^2 - 4x + 4 = 0$ | M1, A1 | May substitute into original or "their (a)"; CSO; AG
| | 2 |
## Part (c)(ii)
$(x-2)^2 = 0$ or $x = 2$ Repeated root implies tangent Point of contact is $P(2, 4)$ | M1, E1, A1 | Or $b^2 - 4ac$ shown $= 0$ plus statement
| | 3 |
## Part (d)
$(CQ^2) = 1^2 + 1^2$ $\sqrt{2} < \sqrt{5} \Rightarrow Q$ lies inside circle | M1, A1 CSO | FT their $C$; $CQ$ or $CQ^2$ OE must appear for A1
| | 2 |
| | 11 |
---4 A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } - 10 y + 20 = 0$.
\begin{enumerate}[label=(\alph*)]
\item By completing the square, express this equation in the form
$$x ^ { 2 } + ( y - b ) ^ { 2 } = k$$
\item Write down:
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $C$;
\item the radius of the circle, leaving your answer in surd form.
\end{enumerate}\item A line has equation $y = 2 x$.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of any point of intersection of the line and the circle satisfies the equation $x ^ { 2 } - 4 x + 4 = 0$.
\item Hence show that the line is a tangent to the circle and find the coordinates of the point of contact, $P$.
\end{enumerate}\item Prove that the point $Q ( - 1,4 )$ lies inside the circle.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2008 Q4 [11]}}