Equation of a circle which passes through (3, 6) and touches the axes, is
x2+y2+6x+6y+3=0
x2+y2−6x−6y−9=0
x2+y2−6x−6y+9=0
none of these
The equation of a circle touching the co-ordinate
axes is
(x−a)2+(y−a)2=a2 or, x2+y2−2ax−2ay+a2=0
This passes through (3, 6).
∴ 9+36−6a−12a+a2=0⇒a=3,15.
Hence, the required circle are
x2+y2−6x−6y+9=0 or, x2+y2−30x−30y+225=0.