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Let $A, B, C, D$ be the respective centers of the radius $2, 3, 1, r$ circles.

Then $AB = 1, BC = 2, AD = 2 + r, BD = 3 - r$ and $CD = 1 + r$ .

Let $\angle ABD = x$ ; then $\angle CBD = \pi - x$ .

Now use the Cosine Law on triangles $ABD$ and $CBD$ to find that

(i) $(2 + r)^{2} = 1 + (3 - r)^{2} - 2*(3 - r)*\cos(x)$ and

(ii) $(1 + r)^{2} = 4 + (3 - r)^{2} + 4*(3 - r)*\cos(x)$ .

Multiply equation (i) by $2$ and add to equation (ii) and simplify to find that $r = \dfrac{6}{7}$ .

Thus $a = 6, b = 7$ and $a + b = \boxed{13}$ .

Obviously the largest circle should be tangent to all the other three circles which are given hence by Descartes'rule we can solve for the curvature of the circle and finally it's radius.Here we go $(k_1 +k_2+k_3+k_4)^2= 2({k_1}^2+{k_2}^2+{k_3}^2+{k_4}^2)$ where $k_1=-\dfrac{1}{3} and \quad k_2=\dfrac{1}{2}\quad k_3=1\quad k_4=x$ where k is the curvature i.e. reciporcal of the radius of the circles pluging in these values and after solving we get $r=\boxed{\dfrac{6}{7}}$