June 2023 Paper 3 Q4
4 A circle \(C\) has equation \(x^2 + y^2 - 6x + 10y + k = 0\).
Determine the coordinates of the two points on \(C\) at which the gradient of the tangent is \(\frac{1}{2}\). [5]
| Scheme | Marks | AO |
|---|---|---|
| \((x - 3)^2 + (y + 5)^2 = -k + 9 + 25\) | M1 | 3.1a |
| \(-k + 34 \gt 0 \Rightarrow k \lt 34\) | A1 | 2.3 |
| [2] |
Notes
M1: Complete the square (for \(x\) and \(y\)) to obtain \((x \pm 3)^2 + (y \pm 5)^2 + \ldots\)
or for \(\ldots \pm k \pm 3^2 \pm 5^2\)
A1: cao – www
allow equivalent in either set notation e.g. \(\{k : k \lt 34\}\) or interval notation e.g. \((-\infty, 34)\) or \((-\infty, 34]\) but not \([-\infty, 34)\) or \([-\infty, 34]\) unless \(k \lt 34\) already seen
If implying a lower limit then A0
Allow \(k \leqslant 34\)
| Scheme | Marks | AO |
|---|---|---|
| \(x^2 + y^2 - 6x + 10y - 46 = 0\) (for reference) | ||
| \(2x + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 6 + 10\dfrac{\mathrm{d}y}{\mathrm{d}x}\ (= 0)\) | M1* | 3.1a |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{6 - 2x}{2y + 10} \Rightarrow \dfrac{3 - x}{y + 5} = \dfrac{1}{2}\) | M1dep* | 1.1 |
| \(2x + y = 1\) \(\Rightarrow x^2 + (1 - 2x)^2 - 6x + 10(1 - 2x) - 46\ (= 0)\) | M1 | 2.1 |
| \(5x^2 - 30x - 35\ (= 0)\) | M1 | 1.1 |
| \((7, -13), (-1, 3)\) | A1 | 2.2a |
| [5] |
Notes
M1*: Attempt to differentiate the equation for \(C\) implicitly – must be four terms including a \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) term and two other terms correct (condone either \(-46\) or \(k\) appearing in their deriv. as a 5th term) but if the derivative of \(x^2 + y^2 - 6x + 10y - 46 = 0\) is put equal to \(\frac{1}{2}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (and used subsequently) then M0
or applying the chain rule to an expression of the form \(\pm 5 \pm \sqrt{\lambda \pm (x \pm 3)^2}\) for some non-zero \(\lambda\), so must be of the form \(\frac{1}{2}(\mathrm{f}(x))^{-\frac{1}{2}}\mathrm{g}(x)\) where \(\mathrm{f}(x)\) is quadratic and \(\mathrm{g}(x)\) is linear
M1dep*: Sets derivative equal to \(\frac{1}{2}\)
or substitutes \(\frac{1}{2}\) for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Substitutes their linear expression into the given equation of \(C\) with \(k = -46\) (must be five terms with two quadratic terms and two linear terms in \(x\) oe if using completing the square form from (a)) to obtain an expression/equation in \(x\) (or \(y\)) only
Dependent on first two M marks – if chain rule used this mark is implied by setting \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) equal to \(\frac{1}{2}\)
M1: Simplify to a 3TQ in \(x\) (or \(y\)) (allow sign errors only when simplifying from their five term equation/expression or completing the square equation/expression)
Dependent on all M marks
For \(y\) if correct: \(y^2 + 10y - 39\ (= 0)\)
A1: BC - do not need to be stated as coordinates
Two values of \(x\) and \(y\) only
Alternative method for first two marks
| Scheme | Marks |
|---|---|
| \(m_r = \dfrac{y + 5}{x - 3}\) or \(y + 5 = m_r(x - 3)\) | M1* |
| \(\dfrac{y + 5}{x - 3} = -2\) or \(y + 5 = -2(x - 3)\) | M1dep* |
M1*: Finding an expression for the gradient of the line segment that passes through the centre and any point \((x, y)\) on the circumference of \(C\)
Follow through their centre from (a) – gradient expression must be correct for their centre
M1dep*: Equating gradient with \(-2\) (oe)