The point P(-2, 1) lies on the circle $x^2 + y^2 - 8x - 6y - 15 = 0$ - Scottish Highers Maths - Question 2 - 2017

To find the gradient of the radius at point $P(-2, 1)$, we calculate the slope of the line connecting the centre $(4, 3)$ and the point $P$:

Gradient = ( \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 3}{-2 - 4} = \frac{-2}{-6} = \frac{1}{3} )

The slope of the tangent line is the negative reciprocal of the radius gradient:

Perpendicular gradient = $-\frac{1}{ \frac{1}{3}} = -3$.

With the slope of the tangent known, we can use the point-slope form of the equation of a line:

$y - y_1 = m (x - x_1)$

Substituting in the values: $m = -3$, $x_1 = -2$, and $y_1 = 1$:

$y - 1 = -3(x + 2)$

This simplifies to:

$y - 1 = -3x - 6$

Thus, the equation of the tangent line is:

$y = -3x - 5$