If a circle has two of its diameters along the lines x+y=5 and x-y=1 and has area 9π, then the equation of the circleis

Two diameters x+y=5 and x−y=1 intersect at(3, 2) which is the centre of the circle. Let r be the radius of thecircle. Then,Area =9π⇒πr2=9π⇒r=3.So, the equation of the circle is(x−3)2+(y−2)2=32 or, x2+y2−6x−4y+4=0