Prove that the distance between the circumcenter and the incenter of triangle ABC is`sqrt(R^2-2R r)`

Explanation

Let O be the circumcenter and OF be the perpendicular to AB.

Let I be the incenter and IE be the perpendicular to AC. Then,
`angleOAF = 90^(@) -C`
`rArr angleOAI = angleIAF - angleOAF`
`=(A)/(2)-(90^(@) -C)`
`=(A)/(2) + C -(A+ B + C)/(2) = (C -B)/(2)`
Also,
`AI = (IE)/(sin.(A)/(2)) = (r)/(sin.(A)/(2)) = 4R sin.(B)/(2) sin.(C)/(2)`
Hence, in `DeltaOAI, OI^(2) = OA^(2) - 2OA " " AI cos angle OAI`
`R^(2) + 16R^(2) sin^(2).(B)/(2) sin^(2).(C)/(2) -8R^(2) sin.(B)/(2) sin.(C)/(2) cos.(C -B)/(2)`
`rArr (OI^(2))/(R^(2)) = 1+ 16 sin^(2).(B)/(2) sin^(2).(B)/(2) sin^(2).(C)/(2) - 8 sin.(B)/(2) sin.(C)/(2) (cos.(B)/(2) cos.(C)/(2) + sin.(B)/(2) sin.(C)/(2))`
`= 1 - 8sin.(B)/(2) sin.(C)/(2) (cos.(B)/(2) cos.(C)/(2) -sin.(B)/(2) sin.(C)/(2))`
`= 1-8 sin.(B)/(2) sin.(C)/(2) cos.(B + C)/(2)`
`=1-8 sin.(B)/(2) sin.(C)/(2) sin.(A)/(2)`
`:. OI = R sqrt(1- 8 sin.(A)/(2) sin.(B)/(2) sin.(C)/(2)) = sqrt(R^(2) -2Rr)`