The point A has co-ordinates (0, 1) - Leaving Cert Mathematics - Question 3 - 2012

The diameter [AB] has endpoints A (0, 1) and B (10, 1). The centre of the circle can be found by calculating the midpoint of AB:

$\text{Centre} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{0 + 10}{2}, \frac{1 + 1}{2} \right) = (5, 1)$

Next, we find the radius of the circle, which is half the distance of AB. The distance between points A and B is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(10 - 0)^2 + (1 - 1)^2} = \sqrt{10^2} = 10$

So, the radius is:

$r = \frac{d}{2} = \frac{10}{2} = 5$

The equation of the circle is given by:

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

Substituting the centre (5, 1) and the radius 5:

$(x - 5)^2 + (y - 1)^2 = 5^2$

Thus, the equation of the circle is:

$(x - 5)^2 + (y - 1)^2 = 25$

The slope of the line DB can be calculated by using the coordinates of points D and B. Since point B is (10, 1) and point D lies on the line l:

To find the coordinates of D, we can substitute the x-coordinate of D into the equation of the line l:

$y = \frac{1}{2}x + 1$

At the point D, let's consider x-coordinate as (x_D), thus:

$y_D = \frac{1}{2}x_D + 1$

The slope of the line DB is determined by:

$\text{slope of DB} = \frac{y_D - y_B}{x_D - x_B} = \frac{\left(\frac{1}{2}x_D + 1\right) - 1}{x_D - 10} = \frac{\frac{1}{2}x_D}{x_D - 10}$

The slope of DB is useful in determining the angle and relationship between lines in the coordinate plane.