June 2017 Paper 2 Q22
22 The equation of a circle is \(\quad (x - 2)^2 + (y - 1)^2 = 16\)
The equation of a line is \(\quad y = 2x + 1\)
The circle and the line intersect at two points.
Work out the coordinates of the two points.
You must show your working.
Do not use trial and improvement. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((x - 2)^2 + (2x + 1 - 1)^2 = 16\) | M1 | oe Eliminates \(y\) |
| \(x^2 - 2x - 2x + 4 + 4x^2 = 16\) or \(5x^2 - 4x - 12\ (= 0)\) | M1dep | oe Expands both brackets correctly |
| \((5x + 6)(x - 2) \quad (= 0)\) or \(\dfrac{--4 \pm \sqrt{(-4)^2 - 4 \times 5 \times -12}}{2 \times 5}\) | M1 | oe eg \(\dfrac{2}{5} \pm \sqrt{\dfrac{64}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(4^2\) for \((-4)^2\) Implied by correct solutions to their 3-term quadratic seen |
| (\(x\) =) \(-1.2\) and (\(x\) =) 2 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5x^2 - 4x - 12\ (= 0)\) seen | A1 | oe eg (\(x\) =) \(-\dfrac{6}{5}\) and (\(x\) =) 2 with \(5x^2 - 4x - 12\ (= 0)\) seen |
| (\(-1.2\), \(-1.4\)) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen | A1 | oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen |
| Alternative method 2 | ||
| \(x^2 - 2x - 2x + 4 + y^2 - y - y + 1 = 16\) | M1 | oe Expands both brackets correctly |
| \(x^2 - 2x - 2x + 4 + (2x + 1)^2\) \(- (2x + 1) - (2x + 1) + 1 = 16\) or \(5x^2 - 4x - 12\ (= 0)\) | M1dep | oe Eliminates \(y\) |
| \((5x + 6)(x - 2) \quad (= 0)\) or \(\dfrac{--4 \pm \sqrt{(-4)^2 - 4 \times 5 \times -12}}{2 \times 5}\) | M1 | oe eg \(\dfrac{2}{5} \pm \sqrt{\dfrac{64}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(4^2\) for \((-4)^2\) Implied by correct solutions to their 3-term quadratic seen |
| (\(x\) =) \(-1.2\) and (\(x\) =) 2 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5x^2 - 4x - 12\ (= 0)\) seen | A1 | oe eg (\(x\) =) \(-\dfrac{6}{5}\) and (\(x\) =) 2 with \(5x^2 - 4x - 12\ (= 0)\) seen |
| (\(-1.2\), \(-1.4\)) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen | A1 | oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen |
| Alternative method 3 | ||
| \(\left(\left(\dfrac{y - 1}{2}\right) - 2\right)^2 + (y - 1)^2 = 16\) | M1 | oe Eliminates \(x\) |
| \(\left(\dfrac{y - 1}{2}\right)^2 - 2\left(\dfrac{y - 1}{2}\right) - 2\left(\dfrac{y - 1}{2}\right) + 4\) \(+ y^2 - y - y + 1 = 16\) or \(5y^2 - 18y - 35\ (= 0)\) | M1dep | oe Expands \(\left(\left(\dfrac{y - 1}{2}\right) - 2\right)^2\) and \((y - 1)^2\) correctly |
| \((5y + 7)(y - 5) \quad (= 0)\) or \(\dfrac{--18 \pm \sqrt{(-18)^2 - 4 \times 5 \times -35}}{2 \times 5}\) | M1 | oe eg \(\dfrac{9}{5} \pm \sqrt{\dfrac{256}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(18^2\) for \((-18)^2\) Implied by correct solutions to their 3-term quadratic seen |
| (\(y\) =) \(-1.4\) and (\(y\) =) 5 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen | A1 | oe eg (\(y\) =) \(-\dfrac{7}{5}\) and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen |
| (\(-1.2\), \(-1.4\)) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen | A1 | oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen |
| Alternative method 4 | ||
| \(x^2 - 2x - 2x + 4 + y^2 - y - y + 1 = 16\) | M1 | oe Expands both brackets correctly |
| \(\left(\dfrac{y - 1}{2}\right)^2 - 2\left(\dfrac{y - 1}{2}\right) - 2\left(\dfrac{y - 1}{2}\right) +\) \(4 + y^2 - y - y + 1 = 16\) or \(5y^2 - 18y - 35\ (= 0)\) | M1dep | oe Eliminates \(x\) |
| \((5y + 7)(y - 5) \quad (= 0)\) or \(\dfrac{--18 \pm \sqrt{(-18)^2 - 4 \times 5 \times -35}}{2 \times 5}\) | M1 | oe eg \(\dfrac{9}{5} \pm \sqrt{\dfrac{256}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(18^2\) for \((-18)^2\) Implied by correct solutions to their 3-term quadratic seen |
| (\(y\) =) \(-1.4\) and (\(y\) =) 5 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen | A1 | oe eg (\(y\) =) \(-\dfrac{7}{5}\) and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen |
| (\(-1.2\), \(-1.4\)) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen | A1 | oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen |
Additional guidance
| Answers only (no valid working) | Zero |
| Both solutions from scale drawing | 5 marks |
| (2, 5) is often seen without seeing any correct method | Zero |
| Allow one miscopy for up to M3A0A0 |