Solution:
Centre of circle which touches x-axis at (3,0) is at (3,k) and
radius of circle is k,
∴(x−3)2+(y−k)2=k2
x2+y2−6x+9−2ky=0
This circle passes through (2,1),
5−12+9−2×5×k×1=0
2k=2
k=1
Hence equation of circle is
⇒ x2+y2−6x−2y+9=0
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