圆心为 $C$ 的圆的方程为 $x^2 + y^2 -10x -4y +25 = 0$。

1. Group the $x$ and $y$ terms to prepare for completing the square

$x^2 - 10x + y^2 - 4y = -25$

2. Complete the square for both $x$ and $y$

$(x - 5)^2 - 25 + (y - 2)^2 - 4 = -25$
$(x - 5)^2 + (y - 2)^2 = 4$

3. Identify the coordinates of the centre $C$

Comparing with $(x - a)^2 + (y - b)^2 = r^2$,
Centre $C = (5, 2)$

4. Identify the radius of the circle

$r^2 = 4$
$r = 2$

1. State the perpendicular distance from the centre to the x-axis using the results from part (a)

From part (a), the centre of the circle is $C(5, 2)$ and the radius is $r = 2$.
The perpendicular distance from the centre $(5, 2)$ to the x-axis ($y=0$) is the y-coordinate, which is $2$ units.

2. Compare the distance with the radius to conclude

Since the perpendicular distance from the centre to the x-axis ($2$ units) is exactly equal to the radius of the circle ($2$ units), the x-axis touches the circle at exactly one point. Therefore, the x-axis is a tangent to the circle.