Equation of diameters of the circle, x+2y−5=0 ...(i) 3x−y−1=0 ...(ii) Since, intersection of lines (i) and (ii) is the centre of the circle. On solving Eqs. (i) and (ii), we get x=1 and y=2 ∴ Centre =(1,2) and radius =5 ∴ Equation of required circle (x−1)2+(y−2)2=(5)2 ⇒ x2+y2−2x−4y−20=0