If the lines 3x-4y-7=0 and  2x-3y-5=0 are diameters of a circle of area  49π square units, the equation of the circle, is

The centre of the circle is the point of intersection of its diameters 3x−4y−7=0 and  2x−3y−5=0  So, its coordinates are (1 -1). Let r be the radius of the circle. Then πr2=49π⇒r=7Hence, the equation of the circle is (x−1)2+(y+1)2=72or ,x2+y2−2x−2y−47=0