Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$

So centre will be $\left(-g,-f\right)$ and given centre lie on $x-y+3=0$,

So $g-f=3 \text{or} g=f+3 .......\left(1\right)$

Aslo given cirlce is passing through $\left(1,2\right) \& \left(3,4\right)$,

So $2g+4f+c=-5 ......\left(2\right)$

And $6g+8f+c=-25 ........\left(3\right)$

Now solving equation $\left(1\right), \left(2\right) \& \left(3\right)$ we get,

$g=-1, f=-4 \& c=13$

So, radius is given by $\sqrt{g^2+f^2-c}=\sqrt{1+16-13}=\sqrt{4}=2$