A variable circle passes through the fixed point $A(p,q)$ and touches $x$ -axis. The locus of the other end of the diameter through $A$ is

Let the other end of the diameter be $(h,k)$.
The ends of the diameter are $A(p,q)$ and $(h,k).$
The equation of the circle is:
$(x-h)(x-p)+(y-k)(y-q)=0$.
$x^2-(h+p)x+hp+y^2-(k+q)y+kq=0.$
Since the circle touches the $x$ -axis, set $y=0$. The resulting quadratic in $x$ must have equal roots (Discriminant $D=0$):
$x^2-(h+p)x+(hp+kq)=0$.
$D=(h+p{)}^2-4(1)(hp+kq)=0$.
$h^2+p^2+2hp-4hp-4kq=0$
$h^2+p^2-2hp=4kq$
$(h-p{)}^2=4kq$.
Replacing $(h,k)$ with $(x,y)$ gives the locus:
$(x-p{)}^2=4qy$.