The points A(1, 8) and B(9, 0) are the end-points of a diameter of the circle w, as shown in the diagram - Leaving Cert Mathematics - Question 4 - 2018

The length of the radius can be found using the distance formula between the center and one of the endpoints of the diameter. Using point A(1, 8):

$$|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Substituting the coordinates:

$$|AO| = \sqrt{(5 - 1)^2 + (4 - 8)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}.$$

Thus, the radius is $4\sqrt{2}$.

For a circle with center (h, k) and radius r, the equation is given by:

$$(x - h)^2 + (y - k)^2 = r^2.$$

Substituting h = 5, k = 4, and r = 4\sqrt{2}:

$$(x - 5)^2 + (y - 4)^2 = (4\sqrt{2})^2 = 32.$$

Hence, the equation of the circle w is:

$$(x - 5)^2 + (y - 4)^2 = 32.$$

To find the tangent line at point A(1, 8), we first calculate the slope of the radius OA. The slope can be calculated as:

$$\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 8}{5 - 1} = \frac{-4}{4} = -1.$$

The slope of the tangent line will be the negative reciprocal of the slope of the radius:

$$\text{slope of tangent} = 1.$$

Using point-slope form of a line equation:

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

Substituting m = 1 and point A(1, 8):

$$(y - 8) = 1(x - 1) \\ y - 8 = x - 1 \\ y = x + 7.$$

Rearranging to standard form:

$$-x + y - 7 = 0.$$

Thus, the equation of the tangent line is $-x + y - 7 = 0$.