| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Chord length calculation |
| Difficulty | Standard +0.3 This is a standard C1 circle question testing completing the square, reading centre/radius, substituting a point, and applying the perpendicular-from-centre-to-chord theorem. Part (d) requires recognizing that the perpendicular distance, half-chord, and radius form a right triangle, then using Pythagoras - a routine application once the setup is identified. Slightly above average due to the multi-step nature and part (d) requiring geometric insight, but all techniques are standard C1 material. |
| 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 | Mark | Guidance |
| \((x+1)^2 - 1 + (y-3)^2 - 9 - 40 = 0\) | M1 | Attempt to complete the square on both \(x\) and \(y\) terms |
| \((x+1)^2 + (y-3)^2 = 50\) | A1 A1 | A1 for each completed square bracket correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(C = (-1, 3)\) | B1 | Both coordinates required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r = \sqrt{50}\) | M1 | Correct method to find radius from \(d=50\) |
| \(r = 5\sqrt{2}\) | A1 | Must be in form \(n\sqrt{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((4+1)^2 + (k-3)^2 = 50\) | M1 | Substituting \(x=4\) into circle equation |
| \((k-3)^2 = 25\) | A1 | Correct simplification |
| \(k = 8\) or \(k = -2\) | A1 | Both values required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using Pythagoras: \(d^2 = r^2 - 1^2 = 50 - 1\) | M1 | \(r^2 - \left(\frac{QR}{2}\right)^2\), i.e. \(50-1\) |
| \(d = \sqrt{49} = 7\) | A1 | cao |
## Question 4:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(x+1)^2 - 1 + (y-3)^2 - 9 - 40 = 0$ | M1 | Attempt to complete the square on both $x$ and $y$ terms |
| $(x+1)^2 + (y-3)^2 = 50$ | A1 A1 | A1 for each completed square bracket correct |
**Part (b)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $C = (-1, 3)$ | B1 | Both coordinates required |
**Part (b)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $r = \sqrt{50}$ | M1 | Correct method to find radius from $d=50$ |
| $r = 5\sqrt{2}$ | A1 | Must be in form $n\sqrt{2}$ |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(4+1)^2 + (k-3)^2 = 50$ | M1 | Substituting $x=4$ into circle equation |
| $(k-3)^2 = 25$ | A1 | Correct simplification |
| $k = 8$ or $k = -2$ | A1 | Both values required |
**Part (d):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Using Pythagoras: $d^2 = r^2 - 1^2 = 50 - 1$ | M1 | $r^2 - \left(\frac{QR}{2}\right)^2$, i.e. $50-1$ |
| $d = \sqrt{49} = 7$ | A1 | cao |
---\begin{enumerate}[label=(\alph*)]
\item Express this equation in the form
$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = d$$
\item \begin{enumerate}[label=(\roman*)]
\item State the coordinates of $C$.
\item Find the radius of the circle, giving your answer in the form $n \sqrt { 2 }$.
\end{enumerate}\item The point $P$ with coordinates $( 4 , k )$ lies on the circle. Find the possible values of $k$.
\item The points $Q$ and $R$ also lie on the circle, and the length of the chord $Q R$ is 2 . Calculate the shortest distance from $C$ to the chord $Q R$.\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2015 Q4 [11]}}