Answer : $6x^2+6y^2-13x=0$
The given circle passes through the point $(2,3)$ is
$x^2+(y-k)^2=k^2$
(ie) $2^2+(3-k)^2=k^2$
On expanding we get,
$4+9-6k+k^2=k^2$
$6k=13$
$k=\large\frac{13}{6}$
Substituting the value of $k$ we get,
$x^2+(y-\large\frac{13}{6})^2=(\large\frac{13}{6})^2$
On simplifying we get,
$6x^2+6y^2-13x=0$