October 2020 Paper 2 Q14
14. A circle \(C\) with radius \(r\)
- lies only in the 1st quadrant
- touches the \(x\)-axis and touches the \(y\)-axis
The line \(l\) has equation \(2x + y = 12\)
Given also that \(l\) is a tangent to \(C\),
| Scheme | Marks | AO |
|---|---|---|
| \(C\) is \((x - r)^2 + (y - r)^2 = r^2\) or \(x^2 + y^2 - 2rx - 2ry + r^2 = 0\) | B1 | 2.2a |
| \(y = 12 - 2x,\ x^2 + y^2 - 2rx - 2ry + r^2 = 0\) \(\Rightarrow x^2 + (12 - 2x)^2 - 2rx - 2r(12 - 2x) + r^2 = 0\) or \(y = 12 - 2x,\ (x - r)^2 + (y - r)^2 = r^2\) \(\Rightarrow (x - r)^2 + (12 - 2x - r)^2 = r^2\) | M1 | 1.1b |
| \(x^2 + 144 - 48x + 4x^2 - 2rx - 24r + 4rx + r^2 = 0\) \(\Rightarrow 5x^2 + (2r - 48)x + (r^2 - 24r + 144) = 0\) * | A1* | 2.1 |
| (3) |
Notes
B1: Deduces the correct equation of the circle
M1: Attempts to form an equation with terms of the form \(x^2\), \(x\), \(r^2\), and \(xr\) only using \(y = 12 \pm 2x\) and their circle equation which must be of an appropriate form. I.e. includes or implies an \(x^2\), \(y^2\), \(r^2\) such as \(x^2 + y^2 = r^2\)
If their circle equation starts off as e.g. \((x \pm a)^2 + (y \pm b)^2 = r^2\) then the B mark and the M mark can be awarded when the “\(a\)” and “\(b\)” are replaced by \(r\) or \(-r\) as appropriate for their circle equation.
A1*: Uses correct and accurate algebra leading to the given solution.
| Scheme | Marks | AO |
|---|---|---|
| \(b^2 - 4ac = 0 \Rightarrow (2r - 48)^2 - 4 \times 5 \times (r^2 - 24r + 144) = 0\) | M1 | 3.1a |
| \(r^2 - 18r + 36 = 0\) or any multiple of this equation | A1 | 1.1b |
| \(\Rightarrow (r - 9)^2 - 81 + 36 = 0 \Rightarrow r = \ldots\) | dM1 | 1.1b |
| \(r = 9 \pm 3\sqrt{5}\) | A1 | 1.1b |
| (4) | ||
| (7 marks) |
Notes
M1: Attempts to use \(b^2 - 4ac \ldots 0\) o.e. with \(a = 5, b = 2r - 48, c = r^2 - 24r + 144\) and where ... is "=" or any inequality
Allow minor slips when copying the \(a\), \(b\) and \(c\) provided it does not make the work easier and allow their \(a\), \(b\) and \(c\) if they are similar expressions.
FYI \((2r - 48)^2 - 4 \times 5 \times (r^2 - 24r + 144) = 4r^2 - 192r + 2304 - 20r^2 + 480r - 2880 = -16r^2 + 288r - 576\)
A1: Correct quadratic equation in \(r\) (or inequality). Terms need not be all one side but must be collected.
E.g. allow \(r^2 - 18r = -36\) and allow any multiple of this equation (or inequality).
dM1: Correct attempt to solve their 3TQ in \(r\). Dependent upon previous M
A1: Careful and accurate work leading to both answers in the required form (must be simplified surds)