C2 January 2010 Q8

EdexcelOld spec12 marksCo-ordinate Geometry Trigonometry

Figure 3: circle C with centre N, a horizontal chord AB of length 12 below the x-axis, and tangents at A and B meeting at P — Figure 3

Figure 3 shows a sketch of the circle \(C\) with centre \(N\) and equation\[(x - 2)^2 + (y + 1)^2 = \frac{169}{4}\]

(a) Write down the coordinates of \(N\). (2)

(b) Find the radius of \(C\). (1)

The chord \(AB\) of \(C\) is parallel to the \(x\)-axis, lies below the \(x\)-axis and is of length 12 units as shown in Figure 3.

(d) Show that angle \(ANB = 134.8^\circ\), to the nearest 0.1 of a degree. (2)

The tangents to \(C\) at the points \(A\) and \(B\) meet at the point \(P\).

(e) Find the length \(AP\), giving your answer to 3 significant figures. (2)

Scheme	Marks
\(N\,(2,\ \text{-}1)\)	B1, B1
	(2)

Scheme	Marks
\(r = \sqrt{\dfrac{169}{4}} = \dfrac{13}{2} = 6.5\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
Complete Method to find \(x\) coordinates, \(x_2 - x_1 = 12\) and \(\dfrac{x_1 + x_2}{2} = 2\) then solve	M1
To obtain \(\ x_1 = -4,\quad x_2 = 8\)	A1ft A1ft
Complete Method to find \(y\) coordinates, using equation of circle or Pythagoras i.e. let \(d\) be the distance below \(N\) of \(A\) then \(d^2 = 6.5^2 - 6^2\ \Rightarrow\ d = 2.5 \Rightarrow y = ..\)	M1
So \(\ y_2 = y_1 = -3.5\)	A1
	(5)

Notes

1^st M1 for finding \(x\) coordinates – may be awarded if either \(x\) co-ord is correct
A1ft,A1ft are for \(\alpha - 6\) and \(\alpha + 6\) if \(x\) coordinate of \(N\) is \(\alpha\)
2^nd M1 for a method to find \(y\) coordinates – may be given if \(y\) co-ordinate is correct
A marks is for –3.5 only.

Mark scheme (d)

Scheme	Marks
Let \(A\hat{N}B = 2\theta\ \Rightarrow\ \sin\theta = \dfrac{6}{\text{"}6.5\text{"}}\ \Rightarrow\ \theta = (67.38)\ldots\)	M1
So angle \(ANB\) is 134.8 \(*\)	A1
	(2)

Notes

M1 for a full method to find \(\theta\) or angle \(ANB\) (eg sine rule or cosine rule directly or finding another angle and using angles of triangle.) ft their 6.5 from radius or wrong \(\boldsymbol{y}\).
(cos \(AN\)B \(= \dfrac{\text{"}6.5\text{"}^2 + \text{"}6.5\text{"}^2 - 12^2}{2 \times \text{"}6.5\text{"} \times \text{"}6.5\text{"}} = \)-0.704)
A1 is a printed answer and must be 134.8 – do not accept 134.76.

Mark scheme (e)

Scheme	Marks
\(AP\) is perpendicular to \(AN\) so using triangle \(ANP\quad \tan\theta = \dfrac{AP}{\text{"}6.5\text{"}}\)	M1
Therefore \(\quad AP = 15.6\)	A1cao
	(2)
	[12]

Notes

M1 for a full method to find \(AP\)

Alternative Methods
N.B. May use triangle \(AXP\) where \(X\) is the mid point of \(AB\). Or may use triangle ABP. From circle theorems may use angle \(BAP\) = 67.38 or some variation.
Eg \(\dfrac{AP}{\sin 67.4} = \dfrac{12}{\sin 45.2}\), \(AP = \dfrac{6}{\sin 22.6}\) or \(AP = \dfrac{6}{\cos 67.4}\) are each worth M1