A circle of radius $ 1$ is surrounded by $ 4$ circles of radius $ r$ as shown. What is $ r$?
[asy]defaultpen(linewidth(.9pt));

real r = 1 + sqrt(2);
pair A = dir(45)*(r + 1);
pair B = dir(135)*(r + 1);
pair C = dir(-135)*(r + 1);
pair D = dir(-45)*(r + 1);

draw(Circle(origin,1));
draw(Circle(A,r));draw(Circle(B,r));draw(Circle(C,r));draw(Circle(D,r));

draw(A--(dir(45)*r + A));
draw(B--(dir(45)*r + B));
draw(C--(dir(45)*r + C));
draw(D--(dir(45)*r + D));
draw(origin--(dir(25)));

label("$r$",midpoint(A--(dir(45)*r + A)), SE);
label("$r$",midpoint(B--(dir(45)*r + B)), SE);
label("$r$",midpoint(C--(dir(45)*r + C)), SE);
label("$r$",midpoint(D--(dir(45)*r + D)), SE);
label("$1$",origin,W);[/asy]$ \textbf{(A)}\ \sqrt {2}\qquad \textbf{(B)}\ 1 \plus{} \sqrt {2}\qquad \textbf{(C)}\ \sqrt {6}\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 2 \plus{} \sqrt {2}$
1. Consider the central circle with radius \(1\) and the four surrounding circles each with radius \(r\). The centers of the surrounding circles form a square around the central circle.
2. The distance from the center of the central circle to the center of any surrounding circle is \(1 + r\).
3. The distance between the centers of two adjacent surrounding circles is \(2r\).
4. The diagonal of the square formed by the centers of the surrounding circles is equal to the distance between the centers of two opposite surrounding circles, which is \(2(1 + r)\).
5. In a square, the diagonal can also be expressed as \( \sqrt{2} \) times the side length. Here, the side length is \(2r\), so the diagonal is \(2r\sqrt{2}\).
6. Equate the two expressions for the diagonal:
   \[
   2(1 + r) = 2r\sqrt{2}
   \]
7. Simplify the equation:
   \[
   1 + r = r\sqrt{2}
   \]
8. Isolate \(r\):
   \[
   1 = r\sqrt{2} - r
   \]
   \[
   1 = r(\sqrt{2} - 1)
   \]
   \[
   r = \frac{1}{\sqrt{2} - 1}
   \]
9. Rationalize the denominator:
   \[
   r = \frac{1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1
   \]

The final answer is \(\boxed{1 + \sqrt{2}}\)