Let $A (2,3)$ & $B (4,5)$ be the points through which circle is passing $\&$ radius lies on the line $y -4 x +3=0$. Le centre be $(n, k)$, this point must satisfy the line $y-4 x-+3=0$. Hence
$k-4 n+3=0$
$k=4 n-3 .$
So, Centre coordinates be $0( n , 4 n -3)$
Now, $O A=O B$ (both one radius)
$A (2,3) B (4,5)$
$\Rightarrow OA = OB$
$\Rightarrow OA ^{2}= OB ^{2}$
$(n-2)^{2}+(4 n-3-3)^{2}=(n-4)^{2}+(4 n-3-5)^{2}$
$n^{2}+4-4 n+16 n^{2}+36-48 n=n^{2}+16-8 n+16 n^{2}+64-64 n$
$40-52 n=80-72 n$
$72 n-52 n=80-40$
$20 n =40$
$n=2$
Hence $k=4 n-3$
$k=8-3$
$k=5$
Centre coordinates $O(7,5)$ & $A (7,3)$
radius $= OA =(2-2)^{2}+(5-3)^{2}=2$
Hence = radius $=2$