Eight points are equally spaced around a circle of radius $r$. If we draw a circle of radius $1$ centered at each of the eight points, then each of these circles will be tangent to two of the other eight circles that are next to it. IF $r^2=a+b\sqrt{2}$, where $a$ and $b$ are integers, then what is $a+b$?

$\text{(A) }3\qquad\text{(B) }4\qquad\text{(C) }5\qquad\text{(D) }6\qquad\text{(E) }7$
1. We start by noting that the eight points are equally spaced around a circle of radius \( r \). This means they form the vertices of a regular octagon inscribed in the circle.
2. Each of the smaller circles has a radius of 1 and is centered at one of these eight points. Since each smaller circle is tangent to its two neighboring circles, the distance between the centers of any two adjacent smaller circles is \( 2 \) (the sum of their radii).
3. The distance between the centers of two adjacent smaller circles is also the side length of the regular octagon formed by the eight points. Let's denote this side length by \( s \). Therefore, \( s = 2 \).
4. To find the relationship between the side length \( s \) and the circumradius \( r \) of a regular octagon, we use the formula for the side length of a regular octagon inscribed in a circle:
   \[
   s = r \sqrt{2 - 2 \cos \left( \frac{360^\circ}{8} \right)}
   \]
5. Simplifying the angle, we get:
   \[
   \cos \left( \frac{360^\circ}{8} \right) = \cos 45^\circ = \frac{\sqrt{2}}{2}
   \]
6. Substituting this into the formula for \( s \), we have:
   \[
   s = r \sqrt{2 - 2 \cdot \frac{\sqrt{2}}{2}} = r \sqrt{2 - \sqrt{2}}
   \]
7. Given that \( s = 2 \), we set up the equation:
   \[
   2 = r \sqrt{2 - \sqrt{2}}
   \]
8. Solving for \( r \), we square both sides:
   \[
   4 = r^2 (2 - \sqrt{2})
   \]
9. Isolating \( r^2 \), we get:
   \[
   r^2 = \frac{4}{2 - \sqrt{2}}
   \]
10. To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator:
    \[
    r^2 = \frac{4 (2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} = \frac{4 (2 + \sqrt{2})}{4 - 2} = \frac{4 (2 + \sqrt{2})}{2} = 2 (2 + \sqrt{2}) = 4 + 2\sqrt{2}
    \]
11. Therefore, we have:
    \[
    r^2 = 4 + 2\sqrt{2}
    \]
12. Comparing this with the given form \( r^2 = a + b\sqrt{2} \), we identify \( a = 4 \) and \( b = 2 \).
13. Thus, \( a + b = 4 + 2 = 6 \).

The final answer is \( \boxed{6} \)