Solution: Instead of A,B,C,D, number the circles 1,2,3,4, respectively. Descarte's Theorem states that if four circles are all mutually tangent, then: $$(k_1 + k_2 + k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2)$$ Where $$k_i$$ is the curvature (reciprocal of the radius) of circle $$i$$. This is usually stated in a form solved for $$k_4$$: $$k_4 = k_1 + k_2 + k_3 \pm 2 \sqrt{k_1 k_2 + k_2 k_3 + k_3 k_1}$$ Relative to circle 4, the three circles have curvatures $$\frac11, \frac12, -\frac13.$$ Then, $$k_4 = \frac11 + \frac12 - \frac13 \pm 2 \sqrt{\frac11(\frac12) - \frac12(\frac13) - \frac13(\frac11)}$$ $$k_4 = \frac76 \pm 2 \sqrt{0}$$ $$k_4 = \frac76$$ Therefore the radius of the fourth circle is $$\frac67$$.