C2 January 2011 Q9

EdexcelOld spec10 marksCo-ordinate Geometry

9. The points \(A\) and \(B\) have coordinates \((-2, 11)\) and \((8, 1)\) respectively.

Given that \(AB\) is a diameter of the circle \(C\),

(a) show that the centre of \(C\) has coordinates \((3, 6)\), (1)

(b) find an equation for \(C\). (4)

(d) Find an equation of the tangent to \(C\) at the point \((10, 7)\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are constants. (4)

Mark scheme (a)

Scheme	Marks
\(C\left(\dfrac{-2 + 8}{2}, \dfrac{11 + 1}{2}\right) = C(3, 6)\) AG Correct method (no errors) for finding the mid-point of \(AB\) giving \((3, 6)\)	B1*
	(1)

Notes

Alternative method: \(C\left(-2 + \dfrac{8 - -2}{2}, 11 + \dfrac{1 - 11}{2}\right)\) or \(C\left(8 + \dfrac{-2 - 8}{2}, 1 + \dfrac{11 - 1}{2}\right)\)

Mark scheme (b)

Scheme	Marks
\((8 - 3)^2 + (1 - 6)^2\) or \(\sqrt{(8 - 3)^2 + (1 - 6)^2}\) or Applies distance formula in order to find the radius.	M1
\((-2 - 3)^2 + (11 - 6)^2\) or \(\sqrt{(-2 - 3)^2 + (11 - 6)^2}\) Correct application of formula.	A1
\((x - 3)^2 + (y - 6)^2 = 50\ \left(\text{or } \left(\sqrt{50}\right)^2 \text{ or } \left(5\sqrt{2}\right)^2\right)\) \((x \pm 3)^2 + (y \pm 6)^2 = k\), \(k\) is a positive value.	M1
\((x - 3)^2 + (y - 6)^2 = 50\) (Not \(7.07^2\))	A1
	(4)

Notes

You need to be convinced that the candidate is attempting to work out the radius and not the diameter of the circle to award the first M1. Therefore allow 1^st M1 generously for \(\dfrac{(-2 - 8)^2 + (11 - 1)^2}{2}\)

Award 1^st M1A1 for \(\dfrac{(-2 - 8)^2 + (11 - 1)^2}{4}\) or \(\dfrac{\sqrt{(-2 - 8)^2 + (11 - 1)^2}}{2}\).

Correct answer in (b) with no working scores full marks.

Mark scheme (c)

Scheme	Marks
{For \((10, 7)\),} \(\underline{(10 - 3)^2 + (7 - 6)^2 = 50}\), {so the point lies on \(C\).}	B1
	(1)

Notes

B1 awarded for correct verification of \(\underline{(10 - 3)^2 + (7 - 6)^2 = 50}\) with no errors.

Also to gain this mark candidates need to have the correct equation of the circle either from part (b) or re-attempted in part (c). They cannot verify \((10, 7)\) lies on \(C\) without a correct \(C\).

Also a candidate could either substitute \(x = 10\) in \(C\) to find \(y = 7\) or substitute \(y = 7\) in \(C\) to find \(x = 10\).

Mark scheme (d)

Scheme	Marks
\(\{\text{Gradient of radius}\} = \dfrac{7 - 6}{10 - 3}\) or \(\dfrac{1}{7}\) This must be seen in part (d).	B1
Gradient of tangent \(= \dfrac{-7}{1}\) Using a perpendicular gradient method.	M1
\(y - 7 = -7(x - 10)\) \(y - 7 = (\text{their gradient})(x - 10)\)	M1
\(y = -7x + 77\) \(y = -7x + 77\) or \(y = 77 - 7x\)	A1 cao
	(4)
	[10]

Notes

2^nd M1 mark also for the complete method of applying \(7 = (\text{their gradient})(10) + c\), finding \(c\).
Note: Award 2^nd M0 in (d) if their numerical gradient is either 0 or \(\infty\).

Alternative: For first two marks (differentiation):
\(2(x - 3) + 2(y - 6)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) (or equivalent) scores B1.
1^st M1 for substituting both \(x = 10\) and \(y = 7\) to find a value for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), which must contain both \(x\) and \(y\). (This M mark can be awarded generously, even if the attempted “differentiation” is not “implicit”.)

Alternative: \((10 - 3)(x - 3) + (7 - 6)(y - 6) = 50\) scores B1M1M1 which leads to \(y = -7x + 77\).