The circum centre of the triangle whose sides are 3x−y−5=0,x+2y−4=0,5x+3y+1=0  is α,β then  α2+β2=

Given lines are  3x−y−5=0       …… (1) x+2y−4=0       …… (2)   5x+3y+1=0     …… (3)The point of intersections of above lines are vertices of triangle Hence the vertices are   A2,1,B−2,3,C1,−2Suppose that the point Sα,β is the circum center of the triangle, so that  SA=SB=SCSolving the equation SA=SB=SC  . We get  α,β=−67,27Therefore,  α2+β2=3649+449=4049=0.816