Each of the points P and Q lie on a circle with its center at O and having a radius of √50.
Point R is inside the circle such that:

∠PQR = 90^{o}, PQ = 6, QR = 2

Find OR

Let S be the opposite endpoint of the diameter containing P. The angle PQS is then a right angle, which means that R is on QS.

By Pythagoras QS = 2*sqrt(41) Then RS = 2*sqrt(41)-2

Angle PSQ is the same as angle OSR, and cos(angle PSQ) = sqrt(41/50)

Then by the law of cosines OR^2 = 50 + (2*sqrt(41)-2)^2 - 2*sqrt(50)*(2*sqrt(41)-2)*sqrt(41/50).

Then OR = sqrt(54-4*sqrt(41)) = 5.328