October 2021 Paper 1 Q7
7 In this question you must show detailed reasoning.
The points A \((-1, 4)\) and B \((7, -2)\) are at opposite ends of a diameter of a circle.
(a) Find the equation of the circle. [4]
(b) Find the coordinates of the points of intersection of the circle and the line \(y = 2x + 5\). [3]
(c) Q is the point of intersection with the larger \(y\)-coordinate.
Calculate the area of the triangle ABQ. [3]
Calculate the area of the triangle ABQ. [3]
| Scheme | Marks | AO |
|---|---|---|
| DR Midpoint of AB is \((3, 1)\) Centre C of the circle is \((3, 1)\) | B1 | 3.1a |
| and radius \(\sqrt{(7-3)^2 + (-2-1)^2} = 5\) | M1 | 3.1a |
| So circle is \((x-3)^2 + (y-1)^2 = 25\) | M1 A1 | 1.1b 1.1b |
| [4] |
Notes
B1: soi
M1: Attempt to find length of AB, AC or BC
M1: Uses their midpoint and radius (do not allow for diameter used)
A1: Need not be simplified
| Scheme | Marks | AO |
|---|---|---|
| DR Crosses \(y = 2x + 5\) where \((x-3)^2 + (2x+5-1)^2 = 25\) | M1 | 1.1b |
| \(5x^2 + 10x = 0\) giving \(x = -2, 0\) | A1 | 1.1b |
| So points are \((-2, 1)\) and \((0, 5)\) | A1 | 1.1b |
| [3] |
Notes
M1: Substituting \(y = 2x + 5\) and attempting to collect terms oe
Allow for a quadratic solved BC providing it is seen in form \(ax^2 + bx = 0\) or \(ay^2 + by + c = 0\)
A1: Both values correct
A1: Correct \(y\) coordinates FT their \(x\)-coordinates
| Scheme | Marks | AO |
|---|---|---|
| DR AQ \(= \sqrt{2}\) and BQ \(= \sqrt{7^2 + 7^2} = 7\sqrt{2}\) | M1 | 3.1a |
| Triangle ABQ has a right angle at Q (angle in a semicircle) So area of triangle is \(\frac{1}{2} \times \text{AQ} \times \text{BQ}\) | M1 | 2.1 |
| Area = 7 | A1 | 1.1b |
| [3] |
Notes
Note QAB \(= 81.9^\circ\) and QBA \(= 8.1^\circ\)
M1: Attempt to find two lengths to be used in their area calculation (excluding AB)
M1: Correct method for finding the area
A1: FT their Q