The circle C has centre (3, 1) and passes through the point P(8, 3) - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 2

The tangent to the circle at point P(8, 3) can be found using the gradient. First, we calculate the gradient of the radius at that point:

1. Gradient of the radius:
   The radius connects the center (3, 1) to the point P(8, 3). The gradient (m) is calculated as follows:
   $m_{radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{8 - 3} = \frac{2}{5}$

2. Gradient of the tangent:
   Using the fact that the tangent is perpendicular to the radius, the gradient of the tangent ($m_{tangent}$) is the negative reciprocal of the gradient of the radius:
   $m_{tangent} = -\frac{1}{m_{radius}} = -\frac{5}{2}$

3. Equation of the tangent line:
   Using the point-slope form of the equation of a line:
   $y - y_1 = m (x - x_1)$
   Substituting in the values we have:
   $y - 3 = -\frac{5}{2}(x - 8)$
   This can be rearranged to find the standard form:
   $2(y - 3) = -5(x - 8)$
   $2y - 6 = -5x + 40$
   $5x + 2y - 46 = 0$
   This gives us the required equation in the form $ax + by + c = 0$. Here, a = 5, b = 2, and c = -46.