Let A $$\equiv$$ (t$$_1^2$$, 2t1) and B $$\equiv$$ (t$$_2^2$$, 2t2)
The centre of the circle = $$\left( {{{t_1^2 + t_2^2} \over 2},{t_1} + {t_2}} \right)$$
As the circle touches the x-axis thus $${t_1} + {t_2} = \pm \,r$$
Slope of $$AB = {2 \over {{t_1} + {t_2}}} = \pm \,{2 \over r}$$
If $$a, b$$ and $$c$$ denote the lengths of the sides of the triangle opposite to the angles $$A, B$$ and $$C$$, respectively, then
