June 2025 Paper 2 Q5
5
In this question you must show detailed reasoning.
A circle has diameter \(PQ\) where \(P\) is \((-5, 1)\) and \(Q\) is \((5, 1)\). The line \(x + 2y = 12\) meets the circle at \(A\) and \(B\).
Find the exact length \(AB\). [8]
| Scheme | Marks | AO |
|---|---|---|
| DR | ||
| Centre \((0,1)\) and radius \(= 5\) | B1 | 3.1a |
| \(x^2 + (y - 1)^2 = 25\) | M1 | 1.1 |
| \((12 - 2y)^2 + (y - 1)^2 = 25\) or \(x^2 + \left(6 - \frac{x}{2} - 1\right)^2 = 25\) or \(x^2 + 25 - 5x + \frac{1}{4}x^2 = 25\) oe | M1 | 1.1 |
| \(5y^2 - 50y + 120 = 0\) or \(y^2 - 10y + 24 = 0\) or \(\frac{5}{4}x^2 - 5x = 0\) oe | M1 | 1.1 |
| \(y = 4\) or \(y = 6\) | A1 | 1.1 |
| \(x = 4\) or \(x = 0\) | A1 | 1.1 |
| \(AB^2 = (4 - 0)^2 + (6 - 4)^2\,[= 20]\) | M1 | 3.1a |
| \(AB = 2\sqrt{5}\) or \(\sqrt{20}\) | A1 | 1.1 |
| [8] |
Notes
B1: Both soi (i.e. may be seen embedded in equation)
M1: FT their centre and radius. OR allow 5 instead of 25
M1: Substitute from line into their circle equation in any form to reach an equation in \(x\) or \(y\) only (substitution must be correct for this mark).
M1: Rearrange their quadratic to a 2- or 3-term quadratic \(= 0\) (this mark may be implied by correct values for \(x\) or \(y\) i.e. either of the next two A marks). The \(= 0\) may be implied by correct solutions to their quadratic. Condone one error in the terms.
A1: Both seen
A1: Both seen
M1: FT their coordinates (must see this step oe)
A1: From fully correct working (dep on all previous marks)
Alternative method
| Scheme | Marks | AO |
|---|---|---|
| Centre \(C(0,1)\) and radius \(= 5\) | B1 | |
| Perp from \((0,1)\) to mid-pt of chord is | M1 | |
| \(y = 2x + 1\) | A1 | |
| Solve \(y = 2x + 1\) and \(x + 2y = 12\) | M1 | |
| Meet at \(M(2,5)\) | A1 | |
| \(CM = \sqrt{20}\) | M1 | |
| \(AM = \sqrt{5^2 - 20}\,\left[= \sqrt{5}\right]\) | M1 | |
| \(AB = 2\sqrt{5}\) or \(\sqrt{20}\) | A1 |
B1: Both soi
M1: Attempt find equation of perpendicular
A1: Correct equation
A1: From fully correct working (dep on all previous marks)