The lines 2x−3y−5=0 and 3x−4y=7 arediameters of a circle of area 154(=49π) sq. units, then the equation ofthe circle is

The centre of the required circle lies at theintersection of 2x-3y-5=0 and 3x-4y-7=0. Thus, thecoordinates of the centre are (1, -1).Let r be the radius of the circle. Then, by hypothesis, we haveπr2=154⇒227r2=154⇒r=7Hence, the equation of the required circle is(x−1)2+(y+1)2=72⇒x2+y2−2x+2y−47=0