Given, sides of triangles x+y=6 . . . (i) 2x+y=4 . . . (ii) x+2y=5... (iii) From Eqs. (i) and (ii), x=−2,y=8 From Eqs. (i) and (iii), x=7,y=−1 From Eqs. (ii) and (iii), x=1,y=2 Let vertices are A(−2,8),B(7,−1),C(1,2). Let coordinate of circumcentre be P(h,k). Then, PA=PB=PC 7)2+(k+1)2 ⇒(h+2)2+(k−8)2=(h−7)2+(k+1)2 ⇒h−k+1=0 . . . (iv) PB=PC PB2=PC2 ⇒(h−7)2+(k+1)2=(h−1)2+(k−2)2 ⇒−12h+6k+45=0 . . . (v) From Eqs. (iv) and (v), we get h=