KEAM 2014 Math Paper

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)
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