A circle of radius 3 unit passes through the point (7,3)  and its centre lies on the straight line x−y−1=0 . Then its equation is:

Let centre be (h,k) . Then (h−7)2+(k−3)2=9  and h−k−1=0 Solving, (h,k)=(7,6) and (4,3) ,∴     Required circles are     x2+y2−8x−6y+16=0  and x2+y2−14x−12y+76=0 .