Since, the lines 2x−3y=5 and 3x−4y=7 are the diameters of a circle. Therefore, the point of intersection is the centre of the circle. On solving the given equations, we get x=1 and y=−1ie, the centre of the circle. ∴ Required equation of circle is (x−1)2+(y+1)2=72 ⇒x2+y2−2x+2y+2=49 ⇒x2+y2−2x+2y−47=0