By property of circle: The rectangle circumscribed by a  circle is a square and diagonals of square bisect each other.
By solving x-3y=S and 3x+y=P  simultaneously gives
the point[ (S+3P)/10 , (P-3S)/10] and other two lines x-3y=Q and 3x+y=R give the point [ (Q+3R)/10 ,(R-3Q)/10] which are diagonally opposite points of the rectangle as the first and the third lines as well as the second and the forth lines are parallel.
Diameter of the circle is the diagonal of the square and the centre of square is the centre of the circle.

Hence by mid point formula we get
[ (S+3P+Q+3R)/20 , (P-3S+R-3Q)/20] AS THE CENTRE OF THE CIRCLE