The circle passing through $(1,-2)$ and touching the axis of $x$ at $(3,0)$ also passes through the point:

Let the equation of the circle touching the $x$ -axis at $(3,0)$ be:
$$(x-3{)}^2+(y-r{)}^2=r^2$$
$$(x-3{)}^2+y^2-2ry+r^2=r^2\Rightarrow (x-3{)}^2+y^2-2ry=0$$
Since it passes through $(1,-2)$:
$$(1-3{)}^2+(-2{)}^2-2r(-2)=0$$
$$4+4+4r=0\Rightarrow 4r=-8\Rightarrow r=-2$$
Equation of circle: $(x-3{)}^2+y^2+4y=0$.Now check the options:For $(5,-2)$: $(5-3{)}^2+(-2{)}^2+4(-2)=4+4-8=0$.The point $(5,-2)$ satisfies the equation.