P(6,-6) ,Q(3,-7) and R(3,3)
Let O(x,y) be the centre of the circle
passing through the points P,Q and R .
OP = OQ …(Radii of the same circle )
`sqrt((x-6)^(2) + [y-(-6)]^(2))= sqrt((x-3)^(2) + [y -(-7)]^(2))" "` ... (Distance formula)
Squaring both the sides ,we get
`(x-6)^(2) + (y + 6)^(2) = (x-3)^(2) + (y+7)^(2)`
` x^(2) - 12 x + 36 + y^(2) + 12y + 36 + 36 = x^(2) - 6x + 9 + y^(2) + 14y + 49`
` therefore - 12x + 12y + 72 = - 6x + 14y + 58`
` therefore - 12x + 6x + 12y - 14y = 58 -72`
` therefore -6x -2y =-14`
` therefore -2 (3x +y) = -14 `
` therefore 3x +y = (-14)/(-2)`
` therefore 3x +y = 7` ...(1)
OQ = OR ...(Radii of the same circle )
` therefore sqrt((x-3)^(2) + [y -(-7)]^(2)) = sqrt((x-3)^(2) +(y-3)^(2))`
Squaring both the sides
`(x-3)^(2) + (y+7)^(2) = (x-3)^(3) + 9y-3)^(3)`
` therefore (y+7)^(2) = (y-3)^(3)`
`y^(2) + 14y + 49 = y^(2) -6y + 9`
` therefore 14y + 6y = 9 - 49`
` therefore 20y = - 40`
`therefore 20y = -40`
`therefore y = (-40)/(20)`
` therefore y = (-40)/(20) therefore y = -2`
Substituting `y = - 2 ` in (1) we get
` 3x + (-2) = 7`
`therefore 3x - 2 = 7`
`therefore 3x = 7+ 2`
` therefore 3x = 9 `
` therefore x = (9)/(3)`
` therefore x = 3`
Coordinates of centre of the circle are (3,-2) .
