1. Write the given circles in standard form

   Let the required circle $S$ have centre $(h,k)$ and radius $r$. Then $S:\ (x-h)^2+(y-k)^2=r^2.$

   The given circles are:

   • $C_1:(x-1)^2+y^2=16$ so its centre is $(1,0)$ and radius is $4$.

   • $C_2:x^2+y^2=1$ so its centre is $(0,0)$ and radius is $1$.

2. Use orthogonality condition

   Two circles with centres distance $d$ and radii $r_1,r_2$ are orthogonal if $d^2=r_1^2+r_2^2.$

   So for circle $S(h,k,r)$:

   With $C_2$

   Distance squared between centres: $h^2+k^2=r^2+1. \qquad (1)$

   With $C_1$

   Distance squared between centres: $(h-1)^2+k^2=r^2+16. \qquad (2)$

3. Subtract (1) from (2)

   $[(h-1)^2+k^2] - [h^2+k^2] = (r^2+16)-(r^2+1)$ $h^2-2h+1-h^2=15$ $-2h+1=15$ $-2h=14$ $h=-7.$

4. Use the fact that the circle passes through $(0,1)$

   Since $(0,1)$ lies on $S$, $(0-h)^2+(1-k)^2=r^2.$ Substituting $h=-7$, $49+(1-k)^2=r^2. \qquad (3)$

5. Use equation (1)

   From (1), with $h=-7$: $(-7)^2+k^2=r^2+1$ $49+k^2=r^2+1$ $r^2=48+k^2. \qquad (4)$

6. Equate (3) and (4)

   $49+(1-k)^2=48+k^2$ $49+1-2k+k^2=48+k^2$ $50-2k=48$ $-2k=-2$ $k=1.$

   Hence the centre is $(-7,1).$

7. Find the radius

   Using (4): $r^2=48+1=49$ $r=7.$

8. Check the options

   • A: radius of $S$ is 8 → False
   • B: radius of $S$ is 7 → True
   • C: centre of $S$ is $(-7,1)$ → True
   • D: centre of $S$ is $(-8,1)$ → False

Therefore the correct options are: $\boxed{B, C}$

9. Compare with stored correct answer

   Stored correct answer: C, B

   This is the same set of correct options as $\{B,C\}$, so the derived answer agrees.