AQA C1 2007 January — Question 4 14 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2007
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeLine-circle intersection points
DifficultyModerate -0.8 This is a routine multi-part question testing standard circle techniques: completing the square (algorithmic), reading off centre/radius (trivial), checking intersection with x-axis (substitute y=0), and finding line-circle intersection points (substitute linear equation into circle). All parts follow textbook procedures with no problem-solving or novel insight required. Easier than average A-level question.
Spec1.02f Solve quadratic equations: including in a function of unknown1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle
4 A circle with centre \(C\) has equation \(x ^ { 2 } + y ^ { 2 } + 2 x - 12 y + 12 = 0\).
  1. By completing the square, express this equation in the form $$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = r ^ { 2 }$$
  2. Write down:
    1. the coordinates of \(C\);
    2. the radius of the circle.
  3. Show that the circle does not intersect the \(x\)-axis.
  4. The line with equation \(x + y = 4\) intersects the circle at the points \(P\) and \(Q\).
    1. Show that the \(x\)-coordinates of \(P\) and \(Q\) satisfy the equation $$x ^ { 2 } + 3 x - 10 = 0$$
    2. Given that \(P\) has coordinates (2,2), find the coordinates of \(Q\).
    3. Hence find the coordinates of the midpoint of \(P Q\).
4(a)
\((x + 1)^2 + (y - 6)^2\)
AnswerMarks Guidance
\((1 + 36 - 12 = 25)\) RHS \(= 5^2\)B2, B1 B1 for one term correct or missing \(+\) sign; Condone 25
4(b)(i)
Centre \((-1, 6)\)
AnswerMarks Guidance
Radius \(= 5\)B1✓, B1✓ FT their \(a\) and \(b\) from part (a) or correct; FT their \(r\) from part (a) RHS must be \(> 0\)
4(b)(ii)
4(c)
AnswerMarks Guidance
Attempt to solve "their" \(x^2 + 2x + 12 = 0\)M1 Or comparing "their" \(y_c = 6\) and their \(r = 5\) may use a diagram with values shown; \(r < y_c\) does not intersect; condone \(\pm 1\) or \(\pm 6\) in centre for A1
(all working correct) so no real roots or statement that does not intersectA1 2 marks
4(d)(i)
\((4 - x)^2 = 16 - 8x + x^2\)
\(x^2 + (4 - x)^2 + 2x - 12(4 - x) + 12 = 0\)
AnswerMarks Guidance
or \((x + 1)^2 + (-2 - x)^2 = 25\)B1, M1 Or \((-2 - x)^2 = 4 + 4x + x^2\); Sub \(y = 4 - x\) in circle eqn (condone slip) or "their" circle equation; AG CSO (must have \(= 0\))
\(\Rightarrow 2x^2 + 6x - 20 = 0 \Rightarrow x^2 + 3x - 10 = 0\)A1 3 marks
4(d)(ii)
\((x + 5)(x - 2) = 0 \Rightarrow x = -5, x = 2\)
AnswerMarks Guidance
\(Q\) has coordinates \((-5, 9)\)M1, A1 Correct factors or unsimplified solution to quadratic (give credit if factorised in part (i)); SC2 if \(Q\) correct. Allow \(x = -5\) \(y = 9\)
4(d)(iii)
Mid point of "their" \((-5, 9)\) and \((2,2)\)
AnswerMarks Guidance
\((-1\frac{1}{2}, 5\frac{1}{2})\)M1, A1 Arithmetic mean of either \(x\) or \(y\) coords; Must follow from correct value in (ii)
Total: 14
## 4(a)
$(x + 1)^2 + (y - 6)^2$
$(1 + 36 - 12 = 25)$ RHS $= 5^2$ | B2, B1 | B1 for one term correct or missing $+$ sign; Condone 25

## 4(b)(i)
Centre $(-1, 6)$
Radius $= 5$ | B1✓, B1✓ | FT their $a$ and $b$ from part (a) or correct; FT their $r$ from part (a) RHS must be $> 0$

## 4(b)(ii)
| | |

## 4(c)
Attempt to solve "their" $x^2 + 2x + 12 = 0$ | M1 | Or comparing "their" $y_c = 6$ and their $r = 5$ may use a diagram with values shown; $r < y_c$ does not intersect; condone $\pm 1$ or $\pm 6$ in centre for A1

(all working correct) so no real roots or statement that does not intersect | A1 | 2 marks

## 4(d)(i)
$(4 - x)^2 = 16 - 8x + x^2$
$x^2 + (4 - x)^2 + 2x - 12(4 - x) + 12 = 0$
or $(x + 1)^2 + (-2 - x)^2 = 25$ | B1, M1 | Or $(-2 - x)^2 = 4 + 4x + x^2$; Sub $y = 4 - x$ in circle eqn (condone slip) or "their" circle equation; AG CSO (must have $= 0$)

$\Rightarrow 2x^2 + 6x - 20 = 0 \Rightarrow x^2 + 3x - 10 = 0$ | A1 | 3 marks

## 4(d)(ii)
$(x + 5)(x - 2) = 0 \Rightarrow x = -5, x = 2$
$Q$ has coordinates $(-5, 9)$ | M1, A1 | Correct factors or unsimplified solution to quadratic (give credit if factorised in part (i)); SC2 if $Q$ correct. Allow $x = -5$ $y = 9$

## 4(d)(iii)
Mid point of "their" $(-5, 9)$ and $(2,2)$
$(-1\frac{1}{2}, 5\frac{1}{2})$ | M1, A1 | Arithmetic mean of either $x$ or $y$ coords; Must follow from correct value in (ii)

**Total: 14**

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4 A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } + 2 x - 12 y + 12 = 0$.
\begin{enumerate}[label=(\alph*)]
\item By completing the square, express this equation in the form

$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = r ^ { 2 }$$
\item Write down:
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $C$;
\item the radius of the circle.
\end{enumerate}\item Show that the circle does not intersect the $x$-axis.
\item The line with equation $x + y = 4$ intersects the circle at the points $P$ and $Q$.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinates of $P$ and $Q$ satisfy the equation

$$x ^ { 2 } + 3 x - 10 = 0$$
\item Given that $P$ has coordinates (2,2), find the coordinates of $Q$.
\item Hence find the coordinates of the midpoint of $P Q$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2007 Q4 [14]}}
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