find the eqn of the circle which passes through the points (3,-2) , (-2,0) and has the centre on the line 2x-y =3

Lemme give you some hints.
See, for a circle, the perpendicular drawn to any of its chord passes through its center. Also, it bisects the chord. So if we find a perpendicular line which is passing through the mid-pt of (3,-2) and (-2,0) i.e., (1/2,-1). That line would be 5x-2y=9/2. On solving this equation with 2x-y=3, we’ll get the co-ordinates of the center. Center will come out as (-3/2,-6). Thus, radius is sqrt(145/4).
Hence, the final eq of the req circle is:- 4x^2 + 4y^2 + 12x + 48y + 8 = 0