If a circle touches x axis at (3,0) and passing through the point (2,1) then its equation is

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