$$
\((x-a)^2+(y-b)^2=r^2\)
$$


Starting with the first equation:

$$
\(\begin{aligned}
x^2+y^2 & =2 x+14 y+1 \\
x^2-2 x+y^2-14 y & =1 \\
x^2-2 x+1-1+y^2-14 y+49-49 & =1 \\
(x-1)^2-1+(y-7)^2-49 & =1 \\
(x-1)^2+(y-7)^2 & =51
\end{aligned}\)
$$


You have your first center point: (\(1, 7)\). Now for the second equation:

$$
\(\begin{aligned}
\frac{8 y-6 x}{x^2+y^2} & =1 \\
8 y-6 x & =x^2+y^2 \\
x^2+6 x+y^2-8 y & =0 \\
x^2+6 x+9-9+y^2-8 y+16-16 & =0 \\
(x+3)^2-9+(y-4)^2-16 & =0 \\
(x+3)^2+(y-4)^2 & =25
\end{aligned}\)
$$



You have your second center point: $(-3,4)$. Now apply the Distance between Points formula:

$$
\(\text { distance }=\sqrt{(1-(-3))^2+(7-4)^2}=\sqrt{(16+9)}=\sqrt{25}=5\)
$$


You can discard the negative square root because distance is always positive. The correct answer choice is $\(\mathbf{B}\)$.