Explanation

(a)
In a circle $AB$ is as a diameter where the coordinates of $A$ are $(p,\ q)$ and let the coordinates of $B$ are $\left( x_{1},\ y_{1} \right)$
Equation of circle in diameter form is
$(x - p)\left( x - x_{1} \right) + (y - q)\left( y - y_{1} \right) = 0$
$\Rightarrow x^{2} - \left( p + x_{1} \right)x + y^{2} - \left( y_{1} - q \right)y + px_{1} + qy_{1} = 0$
Since, the circle touches $x$-axis
$\therefore\ \ \ y = 0$
$\Rightarrow \ \ \ x^{2} - \left( p + x_{1} \right)x + px_{1} + qy_{1} = 0$
Also, the discriminant of above equation will be equal to zero because circle touches $x$-axis
$\therefore\ \ \ \left( p + x_{1} \right)^{2} = 4\left( px_{1} + qy_{1} \right)$
$\Rightarrow \ \ \left( x_{1} - p \right)^{2} = 4qy_{1}$
Therefore, the locus of point $B$ is $(x - p)^{2} = 4qy$