Let C(3,3) is the centre of the circle passing through the points P(6,2),Q(0,4) and R(4,6).
CP=CQ=CR [radii of same circle]
By distance formula, CP = √[(x2−x1)2+(y2−y1)2]
CP =√[(6−3)2+(2−3)2]
CP = √[(3)2+(−1)2]
CP = √[9+1]
CP = √10
By distance formula, CQ = √[(x2−x1)2+(y2−y1)2]
CQ = √[(0−3)2+(4−3)2]
CQ = √[(3)2+(1)2]
CQ = √[9+1]
CQ = √10
By distance formula, CR =√[(x2−x1)2+(y2−y1)2]
CR = √[(4−3)2+(6−3)2]
CR = √[(1)2+(3)2]
CR = √[1+9]
CR = √10
Since CP=CQ=CR,
C(3,3) is the centre of the circle passing through the points P(6,2),Q(0,4) and R(4,6).
Hence proved.