Solving the given equation in pairs taken in order, the coordinates of the vertices of the quadrilateral ABCD are
`A((3)/(2),(1)/(2)),B(0,0),C(-(2)/(9),-(4)/(9)),` and `D((42)/(17),-(19)/(17))`
First, we will find the equation of the circle passing through A,B and C. Let the equation of thic circle be
`x^(2)+y^(2)+2gx+2fy+c=0`
The coordinates of the points A,B, and C must satisfyl (1). So, substituting in (1), we have
`3g+f+c+(5)/(2)=0` (2)
`c=0`
and `4g+8f-9c=(20)/(9)` (4)
Solving (2),(3), and (4), we get `g=-10//9,f=5//6` , and `c=0`
Substituting in (1), the equation of the circle through the vertices A,B, and C is
`9x^(2)+9y^(2)-20x+15y=0`
Since the coodinates of the vertex `D(42//17,-19//17)` also satisfy (5), a cyclic quadrilateral ABCD is formed by the given lines, and (5) is the equation of the circumcircle of the quadrilateral.
