October 2021 Paper 1 Q7
7. The circle \(C\) has equation
\[x^2 + y^2 - 10x + 4y + 11 = 0\]The line \(l\) has equation \(y = 3x + k\) where \(k\) is a constant.
Given that \(l\) is a tangent to \(C\),
| Scheme | Marks | AO |
|---|---|---|
| (i) \((x - 5)^2 + (y + 2)^2 = \ldots\) | M1 | 1.1b |
| \((5,\ -2)\) | A1 | 1.1b |
| (ii) \(r = \sqrt{\text{``}5\text{''}^2 + \text{``}{-}2\text{''}^2 - 11}\) | M1 | 1.1b |
| \(r = 3\sqrt{2}\) | A1 | 1.1b |
| (4) |
Notes
(a)(i) M1: Attempts to complete the square on by halving both \(x\) and \(y\) terms.
Award for sight of \((x \pm 5)^2,\ (y \pm 2)^2 = \ldots\) This mark can be implied by a centre of \((\pm 5,\ \pm 2)\).
A1: Correct coordinates. (Allow \(x = 5,\ y = -2\))
(a)(ii) M1: Correct strategy for the radius or radius\(^2\). For example award for \(r = \sqrt{\text{``}\pm 5\text{''}^2 + \text{``}\pm 2\text{''}^2 - 11}\)
or an attempt such as \((x - a)^2 - a^2 + (y - b)^2 - b^2 + 11 = 0 \Rightarrow (x - a)^2 + (y - b)^2 = k \Rightarrow r^2 = k\)
A1: \(r = 3\sqrt{2}\). Do not accept for the A1 either \(r = \pm 3\sqrt{2}\) or \(\sqrt{18}\)
The A1 can be awarded following sign slips on \((5,\ -2)\) so following \(r^2 = \text{``}\pm 5\text{''}^2 + \text{``}\pm 2\text{''}^2 - 11\)
| Scheme | Marks | AO |
|---|---|---|
| \(y = 3x + k \Rightarrow x^2 + (3x + k)^2 - 10x + 4(3x + k) + 11 = 0\) \(\Rightarrow x^2 + 9x^2 + 6kx + k^2 - 10x + 12x + 4k + 11 = 0\) | M1 | 2.1 |
| \(\Rightarrow 10x^2 + (6k + 2)x + k^2 + 4k + 11 = 0\) | A1 | 1.1b |
| \(b^2 - 4ac = 0 \Rightarrow (6k + 2)^2 - 4 \times 10 \times \left(k^2 + 4k + 11\right) = 0\) | M1 | 3.1a |
| \(\Rightarrow 4k^2 + 136k + 436 = 0 \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(k = -17 \pm 6\sqrt{5}\) | A1 | 2.2a |
| (5) | ||
| (9 marks) |
Notes
(b) Main method seen
M1: Substitutes \(y = 3x + k\) into the given equation (or their factorised version) and makes progress by attempting to expand the brackets. Condone lack of = 0
A1: Correct 3 term quadratic equation.
The terms must be collected but this can be implied by correct \(a,\ b\) and \(c\)
M1: Recognises the requirement to use \(b^2 - 4ac = 0\) (or equivalent) where both \(b\) and \(c\) are expressions in \(k\). It is dependent upon having attempted to substitute \(y = 3x + k\) into the given equation
M1: Solves 3TQ in \(k\). See General Principles.
The 3TQ in \(k\) must have been found as a result of attempt at \(b^2 - 4ac \ldots 0\)
A1: Correct simplified values
Look carefully at the method used. It is possible to attempt this using gradients
(b) Alt 1
| Scheme | Marks | AO |
|---|---|---|
| \(x^2 + y^2 - 10x + 4y + 11 = 0 \Rightarrow 2x + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 10 + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 A1 | 2.1 1.1b |
| Sets \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3 \Rightarrow x + 3y + 1 = 0\) and combines with equation for \(C\) \(\Rightarrow 5x^2 - 50x + 44 = 0 \quad\) or \(\quad 5y^2 + 20y + 11 = 0\) \(\Rightarrow x = \ldots\) or \(y = \ldots\) | M1 | 3.1a |
| \(x = \dfrac{25 \pm 9\sqrt{5}}{5},\ y = \dfrac{-10 \pm 3\sqrt{5}}{5},\ k = y - 3x \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(k = -17 \pm 6\sqrt{5}\) | A1 | 2.2a |
M1: Differentiates implicitly condoning slips but must have two \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)’s coming from correct terms
A1: Correct differentiation.
M1: Sets \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3\), makes \(y\) or \(x\) the subject, substitutes back into \(C\) and attempts to solve the resulting quadratic in \(x\) or \(y\).
M1: Uses at least one pair of coordinates and \(l\) to find at least one value for \(k\). It is dependent upon having attempted both M’s
A1: Correct simplified values
(b) Alt 2
| Scheme | Marks | AO |
|---|---|---|
| \(x^2 + y^2 - 10x + 4y + 11 = 0 \Rightarrow 2x + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - 10 + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 A1 | 2.1 1.1b |
| Sets \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3 \Rightarrow x + 3y + 1 = 0\) and combines with equation for \(l\) \(y = 3x + k,\ x + 3y = 1\) \(\Rightarrow x = \ldots\) and \(y = \ldots\) in terms of \(k\) | M1 | 3.1a |
| \(x = \dfrac{-3k - 1}{10},\ y = \dfrac{k - 3}{10},\ x^2 + y^2 - 10x + 4y + 11 = 0 \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(k = -17 \pm 6\sqrt{5}\) | A1 | 2.2a |
Very similar except it uses equation for \(l\) instead of \(C\) in mark 3
M1 A1: Correct differentiation (See alt 1)
M1: Sets \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3\), makes \(y\) or \(x\) the subject, substitutes back into \(l\) to obtain \(x\) and \(y\) in terms of \(k\)
M1: Substitutes for \(x\) and \(y\) into \(C\) and solves resulting 3TQ in \(k\)
A1: Correct simplified values
(b) Alt 3
| Scheme | Marks |
|---|---|
| \(y = 3x + k \Rightarrow m = 3 \Rightarrow m_r = -\dfrac{1}{3}\) | M1 |
| \(y + 2 = -\dfrac{1}{3}(x - 5)\) | A1 |
| \((x - 5)^2 + (y + 2)^2 = 18,\ y + 2 = -\dfrac{1}{3}(x - 5)\) \(\Rightarrow \dfrac{10}{9}(x - 5)^2 = 18 \Rightarrow x = \ldots\) or \(\Rightarrow 10(y + 2)^2 = 18 \Rightarrow y = \ldots\) | M1 |
| \(x = \dfrac{25 \pm 9\sqrt{5}}{5},\ y = \dfrac{-10 \pm 3\sqrt{5}}{5},\ k = y - 3x \Rightarrow k = \ldots\) | M1 |
| \(k = -17 \pm 6\sqrt{5}\) | A1 |
M1: Applies negative reciprocal rule to obtain gradient of radius
A1: Correct equation of radial line passing through the centre of \(C\)
M1: Solves simultaneously to find \(x\) or \(y\)
Alternatively solves "\(y = -\dfrac{1}{3}x - \dfrac{1}{3}\)" and \(y = 3x + k\) to get \(x\) in terms of \(k\) which they substitute in \(x^2 + (3x + k)^2 - 10x + 4(3x + k) + 11 = 0\) to form an equation in \(k\).
M1: Applies \(k = y - 3x\) with at least one pair of values to find \(k\)
A1: Correct simplified values