The lines 2x−3y=5  and 3x−4y=7  are the diameters of a circle of area 154 square units. An equation of this circle is (π=22/7)

The centre of the circle is the point of intersection of the given diameters 2x−3y=5  and 3x−4y=7,  which is (1, −1)  and if r is the radius of the circle, then πr2=154 ⇒r2=154×722 ⇒  r=7 Hence an equation of the required circle is(x−1)2+(y+1)2=72 ⇒x2+y2−2x+2y=47.