| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle touching axes |
| Difficulty | Moderate -0.8 This is a straightforward C1 circle question requiring basic understanding that a circle touching the y-axis has radius equal to the x-coordinate of its centre, followed by routine verification of a point on the circle, finding a normal (gradient of radius), and a simple distance comparison. All parts are standard textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Radius \(= 5\) (distance from \(C(-5,6)\) to \(y\)-axis); \((x+5)^2 + (y-6)^2 = 25\) | B1 M1 A1 | B1 for \(r=5\); M1 for correct form; A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| \((-2+5)^2 + (2-6)^2 = 9 + 16 = 25\) ✓ | B1 | Must show substitution and result |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient \(CP = \frac{2-6}{-2-(-5)} = \frac{-4}{3}\); normal has same gradient \(-\frac{4}{3}\); \(y-2 = -\frac{4}{3}(x+2)\); \(3y - 6 = -4x - 8\); \(4x + 3y + 2 = 0\) | M1 A1 A1 | M1 for gradient of \(CP\); A1 gradient; A1 equation |
| Answer | Marks | Guidance |
|---|---|---|
| \(M = \left(\frac{-5+(-2)}{2}, \frac{6+2}{2}\right) = \left(-\frac{7}{2}, 4\right)\); \(PM = \sqrt{(-2+\frac{7}{2})^2+(2-4)^2} = \sqrt{\frac{9}{4}+4} = \sqrt{\frac{25}{4}} = \frac{5}{2}\); \(PO = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}\); since \(\frac{5}{2} > 2\sqrt{2}\)... compare: \(2.5\) vs \(2.83\); \(P\) is closer to \(O\) | M1 A1 M1 A1 | M1 for midpoint; M1 for both distances; A1 each; conclusion required |
# Question 5:
## Part (a)
| Radius $= 5$ (distance from $C(-5,6)$ to $y$-axis); $(x+5)^2 + (y-6)^2 = 25$ | B1 M1 A1 | B1 for $r=5$; M1 for correct form; A1 cao |
## Part (b)(i)
| $(-2+5)^2 + (2-6)^2 = 9 + 16 = 25$ ✓ | B1 | Must show substitution and result |
## Part (b)(ii)
| Gradient $CP = \frac{2-6}{-2-(-5)} = \frac{-4}{3}$; normal has same gradient $-\frac{4}{3}$; $y-2 = -\frac{4}{3}(x+2)$; $3y - 6 = -4x - 8$; $4x + 3y + 2 = 0$ | M1 A1 A1 | M1 for gradient of $CP$; A1 gradient; A1 equation |
## Part (b)(iii)
| $M = \left(\frac{-5+(-2)}{2}, \frac{6+2}{2}\right) = \left(-\frac{7}{2}, 4\right)$; $PM = \sqrt{(-2+\frac{7}{2})^2+(2-4)^2} = \sqrt{\frac{9}{4}+4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$; $PO = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$; since $\frac{5}{2} > 2\sqrt{2}$... compare: $2.5$ vs $2.83$; $P$ is closer to $O$ | M1 A1 M1 A1 | M1 for midpoint; M1 for both distances; A1 each; conclusion required |
I can see these are exam questions from what appears to be a MPC1 paper. Based on the questions shown, I'll reconstruct the mark scheme content based on the mathematical working required:
---5 A circle with centre $C ( - 5,6 )$ touches the $y$-axis, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{66813123-3876-4484-aad1-4bfc09bb1508-6_444_698_372_680}
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the circle in the form
$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = r ^ { 2 }$$
\item \begin{enumerate}[label=(\roman*)]
\item Verify that the point $P ( - 2,2 )$ lies on the circle.
\item Find an equation of the normal to the circle at the point $P$.
\item The mid-point of $P C$ is $M$. Determine whether the point $P$ is closer to the point $M$ or to the origin $O$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q5 [11]}}