C2 June 2008 Q5
5. The circle \(C\) has centre \((3, 1)\) and passes through the point \(P(8, 3)\).
| Scheme | Marks |
|---|---|
| \((8 - 3)^2 + (3 - 1)^2\) or \(\sqrt{(8 - 3)^2 + (3 - 1)^2}\) | M1 A1 |
| \((x \pm 3)^2 + (y \pm 1)^2 = k\) or \((x \pm 1)^2 + (y \pm 3)^2 = k\) (\(k\) a positive value) | M1 |
| \((x - 3)^2 + (y - 1)^2 = 29\) (Not \(\left(\sqrt{29}\right)^2\) or \(5.39^2\)) | A1 |
| (4) |
Notes
For the M mark, condone one slip inside a bracket, e.g. \((8 - 3)^2 + (3 + 1)^2\), \((8 - 1)^2 + (1 - 3)^2\)
The first two marks may be gained implicitly from the circle equation.
| Scheme | Marks |
|---|---|
| Gradient of radius \(= \dfrac{2}{5}\) (or exact equiv.) Must be seen or used in (b) | B1 |
| Gradient of tangent \(= \dfrac{-5}{2}\) (Using perpendicular gradient method) | M1 |
| \(y - 3 = \dfrac{-5}{2}(x - 8)\) (ft gradient of radius, dependent upon both M marks) | M1 A1ft |
| \(5x + 2y - 46 = 0\) (Or equiv., equated to zero, e.g. \(92 - 4y - 10x = 0\)) (Must have integer coefficients) | A1 |
| (5) | |
| 9 |
Notes
2nd M: Eqn. of line through \((8, 3)\), in any form, with any grad.(except 0 or \(\infty\)).
If the 8 and 3 are the ‘wrong way round’, this M mark is only given if a correct general formula, e.g. \(y - y_1 = m(x - x_1)\), is quoted.
Alternative: 2nd M: Using \((8, 3)\) and an \(m\) value in \(y = mx + c\) to find a value of \(c\).
A1ft: as in main scheme.
(Correct substitution of 8 and 3, then a wrong \(c\) value will still score the A1ft)
Alternatives for the first 2 marks: (but in these 2 cases the 1st A mark is not ft)
(i) Finding gradient of tangent by implicit differentiation
\(2(x - 3) + 2(y - 1)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) (or equivalent) B1
Subs. \(x = 8\) and \(y = 3\) into a ‘derived’ expression to find a value for \(\mathrm{d}y/\mathrm{d}x\) M1
(ii) Finding gradient of tangent by differentiation of \(y = 1 + \sqrt{20 + 6x - x^2}\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\left(20 + 6x - x^2\right)^{-\frac{1}{2}}(6 - 2x)\) (or equivalent) B1
Subs. \(x = 8\) into a ‘derived’ expression to find a value for \(\mathrm{d}y/\mathrm{d}x\) M1
Another alternative:
Using \(xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0\)
\(x^2 + y^2 - 6x - 2y - 19 = 0\) B1
\(8x + 3y,\quad -3(x + 8) - (y + 3) - 19 = 0\) M1, M1 A1ft (ft from circle eqn.)
\(5x + 2y - 46 = 0\) A1