Given, centre of the circle lies on x-axis. ∴ Centre of the circle =(−g,0). Then, equation of the circle is ‌(x+g)2+y2=(√g2−c)2 ‌⇒‌‌x2+g2+2‌xg+y2=g2−c ‌⇒‌‌x2+2xg+y2+c=0 . . . (i) This circle passes through the point (0,2). ‌∴‌0+2(0)g+(2)2+c‌=0 ⇒c‌=−4 On putting the value of c in Eq. (i), we get x2+2xg+y2−4=0 . . . (ii) This circle also passes through the point (3,3). ‌∴‌(3)2+2(3)g+(3)2−4‌=0 ⇒6g‌=−14
⇒‌‌g=−‌
14
6
=‌
−7
3
On putting the value of g in Eq. (ii), we get x2+2(−‌
7
3
)x+y2−4‌=0 ⇒‌‌3x2+3y2−14x−17‌=0 which is required equation of circle.