C2 January 2014 (IAL) Q8
8.
Figure 2 shows a circle \(C\) with centre \(O\) and radius 5
(a) Write down the cartesian equation of \(C\). (1)
The points \(P(-3, -4)\) and \(Q(3, -4)\) lie on \(C\).
(b) Show that the tangent to \(C\) at the point \(Q\) has equation\[3x - 4y = 25\] (4)
(c) Show that, to 3 decimal places, angle \(POQ\) is 1.287 radians. (2)
The tangent to \(C\) at \(P\) and the tangent to \(C\) at \(Q\) intersect on the \(y\)-axis at the point \(R\).
(d) Find the area of the shaded region \(PQR\) shown in Figure 2. (4)
| Scheme | Marks |
|---|---|
| \(x^2 + y^2 = 25\ \left(\text{or } 5^2\right)\) | B1 |
| (1) |
Notes
B1: Allow \((x - 0)^2 + (y - 0)^2 = 25\)
| Scheme | Marks |
|---|---|
| \(Gradient\ OQ = -\dfrac{4}{3}\) | B1 |
| \(Tangent\ Gradient = \dfrac{3}{4}\) | M1 |
| \(y + 4 = \dfrac{3}{4}(x - 3)\) | M1 |
| \(3x - 4y = 25*\) | A1 |
| (4) |
Notes
B1: Correct gradient
M1: Correct perpendicular gradient rule
M1: Correct straight line method using \((3, -4)\) and their numerical gradient.
A1: Correct completion with no errors
| Scheme | Marks |
|---|---|
| \(6^2 = 5^2 + 5^2 - 2 \times 5 \times 5\cos\theta\) or \(\tan\dfrac{1}{2}\theta = \dfrac{3}{4}\) | M1 |
| \(\theta = \cos^{-1}\left(\dfrac{5^2 + 5^2 - 6^2}{2 \times 5 \times 5}\right)\) or \(\theta = 2\tan^{-1}\left(\dfrac{3}{4}\right)\) | |
| \(\theta = 1.287*\) | A1 |
| (2) |
Notes
M1: Correct statement for angle \(POQ\)
A1: cso
| Scheme | Marks |
|---|---|
| At \(R\) \(y = -\dfrac{25}{4}\) or \(OR = \dfrac{25}{4}\) or \(QR = \dfrac{15}{4}\) | B1 |
| Area \(POQR = \dfrac{25}{4} \times 3\ (= 18.75)\) \(or\ OPQ + PQR = \dfrac{4 \times 6}{2} + \dfrac{6}{2}\left(\dfrac{25}{4} - 4\right)(= 18.75)\) \(or\ 2 \times OQR = 2 \times \dfrac{1}{2} \times 5 \times \dfrac{15}{4}(= 18.75)\) | M1 |
| \(Area\ Sector = \dfrac{1}{2} \times 5^2 \times 1.287\ \ (16.0875)\) | M1 |
| \(18.75 - \dfrac{1}{2} \times 5^2 \times 1.287 = 2.6625\) | A1 |
| (4) | |
| Total 11 |
Notes
B1: May be implied
M1: Valid attempt at kite area
M1: Attempt sector area
A1: Awrt 2.66
8(d) Alternative: \(\triangle PQR\) – segment
| Scheme | Marks |
|---|---|
| At \(R\) \(y = -\dfrac{25}{4}\) or \(OR = \dfrac{25}{4}\) | B1 |
| \(\triangle PQR = \dfrac{1}{2} \times 6 \times \left(\dfrac{25}{4} - 4\right)\left(= \dfrac{27}{4}\right)\) | M1 |
| Segment \(= \dfrac{1}{2} \times 5^2 \times 1.287 - \dfrac{1}{2} \times 6 \times 4\ (= 4.0875)\) | M1 |
| \(\dfrac{27}{4} - 4.0875 = 2.6625\) | A1 |
B1: May be implied M1: Valid attempt at triangle area M1: Valid attempt at segment area A1: Awrt 2.66