| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard circle techniques (writing equations, finding intersections, perpendicular gradients) with straightforward algebraic manipulation. Part (d)(i) is a 'show that' requiring substitution and simplification, while (d)(ii) uses the quadratic formula—all routine C1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)^2 + (y+8)^2 = 100\) | B1 | accept \((y - (-8))^2\) or \(k = 10^2\) |
| B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 0 \Rightarrow \text{'their'}(x - a)^2 + b^2 = k\) | M1 | |
| \((x-3)^2 = 36\) or \(x^2 - 6x - 27(= 0)\) (PI) | A1 | |
| \(\Rightarrow x = -3, 9\) | A1 | |
| Alternative: \(d^2 = 10^2 - 8^2\) | M1 | \((d^2) = 10^2 - 8^2\) |
| \(d^2 = 36\) | A1 | or \(d = 6\) |
| \(\Rightarrow x = -3, 9\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Line CA has gradient \(-\frac{2}{5}\) | M1 | |
| CA has equation \((y+8) = -\frac{2}{5}(x-3)\) | A1 | any form of correct equation e.g. \(y = -\frac{2}{5}x + c, c = \frac{34}{5}\) |
| \(2x + 5y + 34 = 0\) | A1 also | integer coefficients - all terms on 1 side |
| Answer | Marks | Guidance |
|---|---|---|
| their \((x-3)^2 + (2x+1+8)^2\) or \(x^2 + (2x+1)^2 - 6x + 16(2x+1)(+73)\) | M1 | substituting \(y = 2x+1\) correctly into LHS of 'their' circle equation and attempt to expand in terms of \(x\) only |
| \(x^2 - 6x + 9 + 4x^2 + 36x + 81 = 100\) or \(x^2 + 4x^2 + 4x + 1 - 6x + 16 + 32x + 16 + 73 = 100\) | A1 | any correct equation (with brackets expanded) |
| \(\Rightarrow 5x^2 + 30x - 10 = 0\) | ||
| \(\Rightarrow x^2 + 6x - 2 = 0\) | A1 also | AG; all algebra must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| \((x+3)^2 = 11\) | M1 | or correct use of formula |
| \(x = -3 \pm \sqrt{11}\) | A1 also | exactly this |
**8(a)**
$(x-3)^2 + (y+8)^2 = 100$ | B1 | accept $(y - (-8))^2$ or $k = 10^2$
| B1 |
**8(b)**
$y = 0 \Rightarrow \text{'their'}(x - a)^2 + b^2 = k$ | M1 |
$(x-3)^2 = 36$ or $x^2 - 6x - 27(= 0)$ (PI) | A1 |
$\Rightarrow x = -3, 9$ | A1 |
**Alternative:** $d^2 = 10^2 - 8^2$ | M1 | $(d^2) = 10^2 - 8^2$
$d^2 = 36$ | A1 | or $d = 6$
$\Rightarrow x = -3, 9$ | A1 |
**8(c)**
Line CA has gradient $-\frac{2}{5}$ | M1 |
CA has equation $(y+8) = -\frac{2}{5}(x-3)$ | A1 | any form of correct equation e.g. $y = -\frac{2}{5}x + c, c = \frac{34}{5}$
$2x + 5y + 34 = 0$ | A1 also | integer coefficients - all terms on 1 side
**8(d)(i)**
their $(x-3)^2 + (2x+1+8)^2$ or $x^2 + (2x+1)^2 - 6x + 16(2x+1)(+73)$ | M1 | substituting $y = 2x+1$ correctly into LHS of 'their' circle equation and attempt to expand in terms of $x$ only
$x^2 - 6x + 9 + 4x^2 + 36x + 81 = 100$ or $x^2 + 4x^2 + 4x + 1 - 6x + 16 + 32x + 16 + 73 = 100$ | A1 | any correct equation (with brackets expanded)
$\Rightarrow 5x^2 + 30x - 10 = 0$ | |
$\Rightarrow x^2 + 6x - 2 = 0$ | A1 also | AG; all algebra must be correct
**8(d)(ii)**
$(x+3)^2 = 11$ | M1 | or correct use of formula
$x = -3 \pm \sqrt{11}$ | A1 also | exactly this
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**TOTAL: 75 marks**8 A circle has centre $C ( 3 , - 8 )$ and radius 10 .
\begin{enumerate}[label=(\alph*)]
\item Express the equation of the circle in the form
$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$
\item Find the $x$-coordinates of the points where the circle crosses the $x$-axis.
\item The tangent to the circle at the point $A$ has gradient $\frac { 5 } { 2 }$. Find an equation of the line $C A$, giving your answer in the form $r x + s y + t = 0$, where $r , s$ and $t$ are integers.
\item The line with equation $y = 2 x + 1$ intersects the circle.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinates of the points of intersection satisfy the equation
$$x ^ { 2 } + 6 x - 2 = 0$$
\item Hence show that the $x$-coordinates of the points of intersection are of the form $m \pm \sqrt { n }$, where $m$ and $n$ are integers.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2011 Q8 [13]}}