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

Let the equation of circle be

$(x-3{)}^2+(y-0{)}^2+\lambda y=0$

As passes through $(1,-2)$

$\therefore (1-3{)}^2+(-2{)}^2+\lambda (-2)=0$

$\Rightarrow \lambda =4$

$\therefore$ Equation of circle is $(x-3{)}^2+y^2-8=0$

So, $(5,-2)$ satisfies equation of circle