A2 June 2023 Paper 1 Q13
13


| Scheme | Marks | AO |
|---|---|---|
(i) ![]() | M1 A1 A1 | 1.1 1.1 1.1 |
| [3] | ||
(ii) ![]() | M1 A1 A1 | 1.1 1.1 1.1 |
| [3] |
Notes
(a)(i)
M1: circle, centre O
A1: radius \(\sqrt{5}\)
A1: shaded inside (oe). Candidates may shade the region that is not required, but should clearly indicate that what they have shaded is not required.
(a)(ii)
M1: \((-2, 4)\) and \((2, 6)\) identified
A1: perpendicular bisector of \((-2, 4)\) and \((2, 6)\)
A1: shaded on RHS of line (oe). Candidates may shade the region that is not required, but should clearly indicate that what they have shaded is not required.
| Scheme | Marks | AO |
|---|---|---|
| gradient \(= -2\) | M1 | 1.1 |
| passing through \((0, 5)\) | B1 | 3.1a |
| equation \(y = -2x + 5\) | A1 | 1.1 |
| circle is \(x^2 + y^2 = 5\) | B1 | 3.1a |
| \(x^2 + (5 - 2x)^2 = 5\) | M1 | 2.1 |
| \(\Rightarrow 5x^2 - 20x + 20 = 0\) | M1 | 1.1 |
| \(\Rightarrow x = 2\) [only] | A1* | 2.2a |
| [unique solution is] \(z = 2 + \mathrm{i}\) | A1dep | 3.2a |
| [8] |
Notes
A1: (1st) oe. Allow inequality. Could be obtained from diagram in (a).
B1: (2nd) allow inequality
M1: (2nd) or \(\left(\frac{5}{2} - \frac{y}{2}\right)^2 + y^2 = 5\). Must be an equation.
M1: (3rd) simplifying to a three-term quadratic equation
A1*: or \(y = 1\) [only]
Alternative method
| Scheme | Marks |
|---|---|
| \((x + 2)^2 + (y - 4)^2 = (x - 2)^2 + (y - 6)^2\) | M1 M1 |
| \(y = -2x + 5\) | A1 |
| circle is \(x^2 + y^2 = 5\) | B1 |
| \(x^2 + (5 - 2x)^2 = 5\) | M1 |
| \(\Rightarrow 5x^2 - 20x + 20 = 0\) | M1 |
| \(\Rightarrow x = 2\) [only] | A1* |
| [unique solution is] \(z = 2 + \mathrm{i}\) | A1dep |
M1: (1st) squaring both sides of equations or inequality
M1: (2nd) expanding all four sets of brackets
A1: oe. Allow inequality.
B1: allow inequality
M1: (3rd) or \(\left(\frac{5}{2} - \frac{y}{2}\right)^2 + y^2 = 5\). Must be an equation.
M1: (4th) simplifying to a three term quadratic equation
A1*: or \(y = 1\) [only]

