| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic circle equations, verification by substitution, gradient calculations, and perpendicular gradients. All parts are routine applications of standard formulas with no problem-solving required. Slightly easier than average due to the step-by-step scaffolding and elementary C1 content. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((x-2)^2 + (y+1)^2\) | M1 | \((x \pm a)^2 + (y \pm b)^2\) |
| \(= 5^2\) or \(25\) | B1 | |
| A1 | Correct equation for circle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Sub \(x=6\), \(y=2\) into their circle equation | M1 | Or distance \(PC^2 = 4^2 + 3^2\) |
| \((6-2)^2 + (2+1)^2 = 16 + 9 = 25\) | A1 | Shown to equal radius\(^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Gradient \(CP = (2 - -1)/(6-2)\) | M1 | Must be \(y\) on top and subtraction \((6-2)\) |
| \(= \frac{3}{4} = 0.75\) | A1 | Any correct equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Grad of perp \(= -1/\text{their gradient } CP\) | M1 | Or \(m_1 m_2 = -1\) used or stated |
| \(= -4/3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y - 2 = \text{their } (d)(i)\text{ gradient}(x - 6)\) | B1\(\checkmark\) | OE such as \(3y + 4x = 30\) |
## Question 3:
**Part (a)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $(x-2)^2 + (y+1)^2$ | M1 | $(x \pm a)^2 + (y \pm b)^2$ |
| $= 5^2$ or $25$ | B1 | |
| | A1 | Correct equation for circle |
**Part (b)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Sub $x=6$, $y=2$ into their circle equation | M1 | Or distance $PC^2 = 4^2 + 3^2$ |
| $(6-2)^2 + (2+1)^2 = 16 + 9 = 25$ | A1 | Shown to equal radius$^2$ |
**Part (c)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Gradient $CP = (2 - -1)/(6-2)$ | M1 | Must be $y$ on top and subtraction $(6-2)$ |
| $= \frac{3}{4} = 0.75$ | A1 | Any correct equivalent |
**Part (d)(i)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Grad of perp $= -1/\text{their gradient } CP$ | M1 | Or $m_1 m_2 = -1$ used or stated |
| $= -4/3$ | A1 | |
**Part (d)(ii)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $y - 2 = \text{their } (d)(i)\text{ gradient}(x - 6)$ | B1$\checkmark$ | **OE** such as $3y + 4x = 30$ |
**Total: 10 marks**
---3 A circle has centre $C ( 2 , - 1 )$ and radius 5 . The point $P$ has coordinates $( 6,2 )$.
\begin{enumerate}[label=(\alph*)]
\item Write down an equation of the circle.
\item Verify that the point $P$ lies on the circle.
\item Find the gradient of the line $C P$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of a line which is perpendicular to $C P$.
\item Hence find an equation for the tangent to the circle at the point $P$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2005 Q3 [10]}}