Consider a circle C which touches the y-axis at (0,5) and cuts off an intercept $3\sqrt{2}$ on the x-axis. Then the centre and the radius of the circle C are:
 

Since the circle touches $y-$ axis at $\left(0,5\right)$, equation can be written as 

$x^2+y^2-2ax-10y+25=0$

Given that $x-$ Intercept is $3\sqrt{2}$

it implies that 

$2\sqrt{g^2-c}=3\sqrt{2}$

Squaring on both sides and then substitute the appropriate values 

$\begin{array}{rcl}4\left(g^2-c\right)& =& 18\\ g^2-c& =& \frac{9}{2}\\ g^2& =& \frac{9}{2}+25\\ & =& \frac{59}{2}\end{array}$

Hence, the circle is $x^2+y^2\pm \sqrt{118}x-10y=25=0$

Hence, the center is $\mathrm{C}\boxed{\left(\mp \frac{\sqrt{118}}{2},5\right)},$ and the radius is $\mathrm{r}=\sqrt{\frac{118}{4}+25-25}=\boxed{\sqrt{\frac{59}{2}}}$