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Let theta = angle CAB. Then the area of triangle ABC can be written in two ways and equated. That is, $\frac{1}{2}base . height = semiperimeter.InCircleRadius$ $\frac{26 \cos \theta . 26 \sin \theta}{2} = \frac{26\cos\theta + 26\sin\theta+26}{2}r$ On line OT'T $2r = 13 - 13\sin\theta$ equating the two expressions for r $5 \sin \theta = 1 + \cos \theta$ Giving $\tan \theta = \frac{5}{12}$ Hence OT' = 5 and T'T = 2r = 8