| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - intersection or geometric properties |
| Difficulty | Moderate -0.8 This is a straightforward C2 circle question with standard techniques: writing a circle equation (trivial), finding a tangent equation using perpendicular gradients (routine), calculating an angle using cosine rule or dot product (standard), and finding an area by subtracting a sector from a triangle (bookwork method). All parts are guided 'show that' questions requiring no problem-solving or insight, making it easier than the average A-level question. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 + y^2 = 25\) (or \(5^2\)) | B1 | Allow \((x-0)^2 + (y-0)^2 = 25\) |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient \(OQ = -\frac{4}{3}\) | B1 | Correct gradient |
| Tangent Gradient \(= \frac{3}{4}\) | M1 | Correct perpendicular gradient rule |
| \(y + 4 = \frac{3}{4}(x-3)\) | M1 | Correct straight line method using \((3,-4)\) and their numerical gradient |
| \(3x - 4y = 25\) | A1 | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| \(6^2 = 5^2 + 5^2 - 2\times5\times5\cos\theta\) or \(\tan\frac{1}{2}\theta = \frac{3}{4}\) | M1 | Correct statement for angle \(POQ\) |
| \(\theta = \cos^{-1}\left(\frac{5^2+5^2-6^2}{2\times5\times5}\right)\) or \(\theta = 2\tan^{-1}\left(\frac{3}{4}\right)\) | — | — |
| \(\theta = 1.287\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| At \(R\): \(y = -\frac{25}{4}\) or \(OR = \frac{25}{4}\) or \(QR = \frac{15}{4}\) | B1 | May be implied |
| Area \(POQR = \frac{25}{4}\times3\ (=18.75)\) or \(OPQ + PQR = \frac{4\times6}{2} + \frac{6}{2}\left(\frac{25}{4}-4\right)(=18.75)\) or \(2\times OQR = 2\times\frac{1}{2}\times5\times\frac{15}{4}(=18.75)\) | M1 | Valid attempt at kite area |
| Area Sector \(= \frac{1}{2}\times5^2\times1.287\ (=16.0875)\) | M1 | Attempt at sector area |
| \(18.75 - \frac{1}{2}\times5^2\times1.287 = 2.6625\) | A1 | Awrt 2.66 |
| Answer | Marks | Guidance |
|---|---|---|
| At \(R\): \(y = -\frac{25}{4}\) or \(OR = \frac{25}{4}\) | B1 | May be implied |
| \(\triangle PQR = \frac{1}{2}\times6\times\left(\frac{25}{4}-4\right) = \frac{27}{4}\) | M1 | Valid attempt at triangle area |
| Segment \(= \frac{1}{2}\times5^2\times1.287 - \frac{1}{2}\times6\times4\ (=4.0875)\) | M1 | Valid attempt at segment area |
| \(\frac{27}{4} - 4.0875 = 2.6625\) | A1 | Awrt 2.66 |
## Question 8(a):
| $x^2 + y^2 = 25$ (or $5^2$) | B1 | Allow $(x-0)^2 + (y-0)^2 = 25$ |
|---|---|---|
## Question 8(b):
| Gradient $OQ = -\frac{4}{3}$ | B1 | Correct gradient |
|---|---|---|
| Tangent Gradient $= \frac{3}{4}$ | M1 | Correct perpendicular gradient rule |
| $y + 4 = \frac{3}{4}(x-3)$ | M1 | Correct straight line method using $(3,-4)$ and their numerical gradient |
| $3x - 4y = 25$ | A1 | Correct completion with no errors |
## Question 8(c):
| $6^2 = 5^2 + 5^2 - 2\times5\times5\cos\theta$ or $\tan\frac{1}{2}\theta = \frac{3}{4}$ | M1 | Correct statement for angle $POQ$ |
|---|---|---|
| $\theta = \cos^{-1}\left(\frac{5^2+5^2-6^2}{2\times5\times5}\right)$ or $\theta = 2\tan^{-1}\left(\frac{3}{4}\right)$ | — | — |
| $\theta = 1.287$ | A1 | cso |
## Question 8(d):
| At $R$: $y = -\frac{25}{4}$ or $OR = \frac{25}{4}$ or $QR = \frac{15}{4}$ | B1 | May be implied |
|---|---|---|
| Area $POQR = \frac{25}{4}\times3\ (=18.75)$ or $OPQ + PQR = \frac{4\times6}{2} + \frac{6}{2}\left(\frac{25}{4}-4\right)(=18.75)$ or $2\times OQR = 2\times\frac{1}{2}\times5\times\frac{15}{4}(=18.75)$ | M1 | Valid attempt at kite area |
| Area Sector $= \frac{1}{2}\times5^2\times1.287\ (=16.0875)$ | M1 | Attempt at sector area |
| $18.75 - \frac{1}{2}\times5^2\times1.287 = 2.6625$ | A1 | Awrt 2.66 |
## Question 8(d) △PQR–segment method (Appendix):
| At $R$: $y = -\frac{25}{4}$ or $OR = \frac{25}{4}$ | B1 | May be implied |
|---|---|---|
| $\triangle PQR = \frac{1}{2}\times6\times\left(\frac{25}{4}-4\right) = \frac{27}{4}$ | M1 | Valid attempt at triangle area |
| Segment $= \frac{1}{2}\times5^2\times1.287 - \frac{1}{2}\times6\times4\ (=4.0875)$ | M1 | Valid attempt at segment area |
| $\frac{27}{4} - 4.0875 = 2.6625$ | A1 | Awrt 2.66 |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e7043e7a-2c8f-425a-8471-f647828cc297-22_1015_1542_267_185}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a circle $C$ with centre $O$ and radius 5
\begin{enumerate}[label=(\alph*)]
\item Write down the cartesian equation of $C$.
The points $P ( - 3 , - 4 )$ and $Q ( 3 , - 4 )$ lie on $C$.
\item Show that the tangent to $C$ at the point $Q$ has equation
$$3 x - 4 y = 25$$
\item Show that, to 3 decimal places, angle $P O Q$ is 1.287 radians.
The tangent to $C$ at $P$ and the tangent to $C$ at $Q$ intersect on the $y$-axis at the point $R$.
\item Find the area of the shaded region $P Q R$ shown in Figure 2.\\
\includegraphics[max width=\textwidth, alt={}, center]{e7043e7a-2c8f-425a-8471-f647828cc297-25_177_154_2576_1804}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q8 [11]}}