C2 January 2008 Q8
8. A circle \(C\) has centre \(M(6, 4)\) and radius 3.
Figure 3 shows the circle \(C\). The point \(T\) lies on the circle and the tangent at \(T\) passes through the point \(P(12, 6)\). The line \(MP\) cuts the circle at \(Q\).
The shaded region \(TPQ\) is bounded by the straight lines \(TP\), \(QP\) and the arc \(TQ\), as shown in Figure 3.
| Scheme | Marks |
|---|---|
| \((x - 6)^2 + (y - 4)^2 =\ ;\ 3^2\) | B1; B1 |
| (2) |
Notes
Allow 9 for \(3^2\).
| Scheme | Marks |
|---|---|
| Complete method for \(MP\): \(= \sqrt{(12 - 6)^2 + (6 - 4)^2}\) | M1 |
| \(= \sqrt{40}\) (= 6.325) [These first two marks can be scored if seen as part of solution for (c)] | A1 |
| Complete method for \(\cos\theta\), \(\sin\theta\) or \(\tan\theta\) e.g. \(\cos\theta = \dfrac{\mathrm{MT}}{\mathrm{MP}} = \dfrac{3}{\text{candidate’s }\sqrt{40}}\) (= 0.4743) (\(\theta = 61.6835^\circ\)) [If TP = 6 is used, then M0] | M1 |
| \(\theta = 1.0766\) rad AG | A1 |
| (4) |
Notes
First M1 can be implied by \(\sqrt{40}\)
For second M1:
May find TP \(= \sqrt{(\sqrt{40})^2 - 3^2} = \sqrt{31}\), then either
\(\sin\theta = \dfrac{TP}{MP} = \dfrac{\sqrt{31}}{\sqrt{40}}\) (= 0.8803...) or \(\tan\theta = \dfrac{\sqrt{31}}{3}\) (1.8859..) or cos rule
NB. Answer is given, but allow final A1 if all previous work is correct.
| Scheme | Marks |
|---|---|
| Complete method for area \(TMP\); e.g. \(= \dfrac{1}{2}\times 3\times\sqrt{40}\sin\theta\) | M1 |
| \(= \dfrac{3}{2}\sqrt{31}\) (= 8.3516..) allow awrt 8.35 | A1 |
| Area (sector) \(MTQ = 0.5\times 3^2\times 1.0766\) (= 4.8446…) | M1 |
| Area \(TPQ\) = candidate’s (8.3516.. − 4.8446..) | M1 |
| \(= 3.507\) awrt [Note: 3.51 is A0] | A1 |
| (5) | |
| [11] |
Notes
First M1: (alternative) \(\dfrac{1}{2}\times 3\times\sqrt{40 - 9}\)