Let the equation of circle be
$(x-3)^{2}+(y-0)^{2}+\lambda y=0$
As it passes through $(1,-2)$
$\therefore (1-3)^{2}+(-2)^{2}+\lambda(-2)=0 $
$\Rightarrow 4+4-2 \lambda=0 \Rightarrow \lambda=4$
$\therefore$ Equation of circle is
$(x-3)^{2}+y^{2}+4 y=0$
By hit and trial method, we see that point $(5,-2)$ satisfies equation of circle.
