1. Identify the curves

   The two curves are:

   • Parabola: $y^2=8x$
   • Circle: $x^2+y^2+12y+35=0$

2. Rewrite the circle in standard form

   Complete the square in $y$: $x^2+y^2+12y+35=0$ $x^2+(y+6)^2-36+35=0$ $x^2+(y+6)^2=1$

   So this is a circle with:

   • Centre $C=(0,-6)$
   • Radius $r=1$

3. Reduce the problem

   The shortest distance between the parabola and the circle equals:

   $\text{(shortest distance from centre of circle to parabola)} - r$

   because the nearest point on the circle to an external point lies along the radius.

   So we first find the minimum distance from $(0,-6)$ to the parabola $y^2=8x$.

4. Take a general point on the parabola

   Since $y^2=8x=4ax$ with $a=2$, a standard parametric point on the parabola is $P=(2t^2,4t)$

5. Distance from $(0,-6)$ to $P$

   Let the centre be $C=(0,-6)$. Then $CP^2=(2t^2-0)^2+(4t+6)^2$ $CP^2=4t^4+16t^2+48t+36$

   We minimize $f(t)=4t^4+16t^2+48t+36$.

6. Differentiate and find critical points

   $f'(t)=16t^3+32t+48$ $f'(t)=16(t^3+2t+3)$

   Set this equal to zero: $t^3+2t+3=0$

   Try $t=-1$: $(-1)^3+2(-1)+3=-1-2+3=0$

   Hence, $t^3+2t+3=(t+1)(t^2-t+3)$

   Since $t^2-t+3>0$ for all real $t$, the only real critical point is $t=-1$

7. Check minimum distance

   Since the quartic has positive leading coefficient and only one real critical point, this gives the minimum.

   At $t=-1$, $P=(2, -4)$

   Then $CP=\sqrt{(2-0)^2+(-4+6)^2}=\sqrt{4+4}=2\sqrt{2}$

8. Distance from circle to parabola

   The circle has radius $1$, so the shortest distance between the circle and parabola is $2\sqrt{2}-1$

9. Match with options

   $2\sqrt{2}-1$ corresponds to Option B.

10. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

Hence they agree.