June 2024 Paper 1 Q15
15 The circle \(x^2 + y^2 + 2x - 14y + 25 = 0\) has its centre at the point C. The line \(7y = x + 25\) intersects the circle at points A and B.
Prove that triangle ABC is a right-angled triangle. [9]
| Scheme | Marks | AO |
|---|---|---|
| Circle is \((x+1)^2 + (y-7)^2 = 25\) | M1 | 3.1a |
| So C is the point \((-1, 7)\) | A1 | 2.1 |
| The intersection of the line and circle at \((7y - 25)^2 + y^2 + 14y - 50 - 14y + 25 = 0\) | M1 | 3.1a |
| Giving \(50y^2 - 350y + 600 = 0\) | M1 | 2.1 |
| So \(y = 3, 4\) | A1 | 1.1 |
| A and B are \((-4, 3)\quad (3, 4)\) | A1 | 1.1 |
| Gradients of AC and BC are \(\dfrac{4}{3}\) and \(\dfrac{3}{-4}\) | M1 A1 | 3.1a 2.1 |
| The product of the gradients is -1 so the lines are perpendicular | ||
| So the triangle is right-angled | A1 | 2.2a |
| [9] |
Notes
M1: Attempts to complete the square for either \(x\) or \(y\) terms Soi
A1: Correct coordinates of the centre seen or used. Condone incorrect or missing radius
M1: Attempt to solve the equations simultaneously
\(\left[x^2 + \left(\dfrac{x+25}{7}\right)^2 + 2x - 14\left(\dfrac{x+25}{7}\right) + 25 = 0\right]\) or oe
M1: Simplifies equation leading to two roots.
Allow arithmetic errors \([50x^2 + 50x - 600 = 0]\)
A1: Could be solved by calculator \([x = -4, 3]\)
A1: FT their \(y\) values
M1: Attempt to find gradient(s) of at least one of these lines
A1: Both correct gradients (not OA or OB)
A1: Clear argument based on perpendicular lines www
Alternative for last three marks
| Scheme | Marks | AO |
|---|---|---|
| Distance AB\(^2 = (-4-3)^2 + (3-4)^2 = 50\) | M1 | |
| AC\(^2\) = BC\(^2\) = radius\(^2\) = 25 | A1 | |
| So AC\(^2\) + BC\(^2\) = 50 = AB\(^2\) | ||
| So by Pythagoras the triangle is right-angled | A1 |
M1: Attempt to find the length of one of the sides of the triangle
FT their coordinates
A1: All three correct lengths found
A1: Clear argument based on Pythagoras’ theorem or the cosine rule leading to a value for angle ACB www