The points A(1, 8) and B(0, 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 (0.5, 4) and one of the endpoints, say A(1, 8):

$$\text{Radius} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

Substituting the values:

$$\text{Radius} = \sqrt{(1 - 0.5)^2 + (8 - 4)^2} = \sqrt{(0.5)^2 + (4)^2}$$

This simplifies to:

$$\sqrt{0.25 + 16} = \sqrt{16.25} = \sqrt{65/4} = \frac{\sqrt{65}}{2},$$

Thus $p=1$, $q=65$. The radius is $\frac{1\sqrt{65}}{2}$.

The equation of a circle is given by:

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

where (h, k) is the center and r is the radius. From part (a), we have:

• Center (h, k) = (0.5, 4)
• Radius $r = \frac{\sqrt{65}}{2}$. We calculate the radius squared:

$$\left(\frac{\sqrt{65}}{2}\right)^2 = \frac{65}{4}.$$

Thus, substituting these values into the equation, we have:

$$(x - 0.5)^2 + (y - 4)^2 = \frac{65}{4}.$$

To find the equation of the tangent line at point A(1, 8), we first find the slope of the radius at this point, which is given by:

$$m_r = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 4}{1 - 0.5} = \frac{4}{0.5} = 8.$$

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

$$m_t = -\frac{1}{m_r} = -\frac{1}{8}.$$

Using point-slope form:

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

substituting for point A(1, 8):

$$y - 8 = -\frac{1}{8}(x - 1).$$

Simplifying gives:

$$8(y - 8) = -(x - 1)\n\Rightarrow 8y - 64 = -x + 1\n\Rightarrow x + 8y - 65 = 0.$$

Thus the equation of the tangent line in the form $ax + by + c = 0$ is $x + 8y - 65 = 0$.