Step 1: Set up the centre.
Let the centre be $(h,k)$. It lies on $x-4y=1$, so $h-4k=1$, giving $h=4k+1$.
Step 2: Use equal radii.
The centre is equally far from $(3,7)$ and $(5,5)$, so the squared distances are equal: $(h-3)^2+(k-7)^2=(h-5)^2+(k-5)^2$.
Step 3: Expand and cancel.
Expanding gives $-6h+9-14k+49=-10h+25-10k+25$, that is $-6h-14k+58=-10h-10k+50$.
Step 4: Simplify to a line.
Bring terms together: $4h-4k+8=0$, so $h-k+2=0$, giving $h=k-2$.
Step 5: Solve the two conditions.
From $h=4k+1$ and $h=k-2$: $4k+1=k-2$, so $3k=-3$, $k=-1$, and then $h=-3$. Centre is $(-3,-1)$.
Step 6: Find the radius and write the circle.
$r^2=(-3-5)^2+(-1-5)^2=64+36=100$. So $(x+3)^2+(y+1)^2=100$, which expands to $x^2+y^2+6x+2y-90=0$, option (4). \[ \boxed{x^2+y^2+6x+2y-90=0} \]