Let the equation of circle be x2+y2+2gx+2fy+c=0, whose centre (−g,−f) since, the circle passing through (0,0) and (4,0)
∴02+02+2g(0)+2f(0)+c=0⇒c=0
and 42+02+2g(4)+2f(0)+0=0 ⇒16+8g=0⇒g=−2 Also, the centre (−g,−f) lies on line y=x ∴−f=−g⇒f=g ∴f=−2 Now, the equation of circle having centre (2,2) and radius √8 is (x−2)2+(y−2)2=(√8)2 ⇒(x−2)2+(y−2)2=8