A circle C with centre at the point (2, -1) passes through the point A at (4, -5) - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 2

To find the equation of the tangent line at point A (4, -5), we follow these steps:

1. First, calculate the gradient of the radius from the center (2, -1) to the point A (4, -5):

   $m_{radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - (-1)}{4 - 2} = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$

2. The gradient of the tangent line is the negative reciprocal of the gradient of the radius:

   $m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{-2} = \frac{1}{2}$

3. Using the point-slope form of the equation of a line, the tangent line at point A (4, -5) can be represented as:

   $y - y_1 = m_{tangent}(x - x_1)$ $y + 5 = \frac{1}{2}(x - 4)$

4. Simplifying this equation:

   $2(y + 5) = x - 4$ $2y + 10 = x - 4$ $x - 2y - 14 = 0$

Thus, the equation of the tangent to the circle C at point A is:

$x - 2y - 14 = 0$