A circle touching the $x-$ axis at $(3, 0)$ and making an intercept of length $8$ on the $y-$ axis passes through the point:

Since circle touches the x-axis at $\left(3,\mathrm{ }0\right),$ hence centre can be taken as $\left(3,\mathrm{\alpha }\right)$ and radius $=\left|\mathrm{\alpha }\right|$
Since y-intercept is 8, by applying Pythagorean theorem
$3^2+4^2={\alpha }^2\Rightarrow \alpha =5,-5$
Hence two circles are possible
${\left(x-3\right)}^2+{\left(y-5\right)}^2=25$ and ${\left(x-3\right)}^2+{\left(y+5\right)}^2=25.$
Clearly $(3,\mathrm{ }10)$ satisfies the given equation of circle.