find the equation of circles which touch both the axes and pass through the point (2,1)?

Let r be the radius of the circle.
Since the circle touches both the axes, then the center of the circle = (r, r)

Equation of the circle : (x - r)² + (y - r)² = r²

The circle passes through the point (2, 1) :
(2 - r)² + (1 - r)² = r²
4 - 4r + r² + 1 - 2r + r² = r²
r² - 6r + 5 = 0
(r - 1)(r - 5) = 0
r = 1 or r = 5

Equation of the circle :
(x - 1)² + (y - 1)² = 1² or (x - 5)² + (y - 5)² = 5²
x² - 2x + 1 + y² - 2y + 1 = 1 or x² - 10x + 25 + y² - 10y + 25 = 25
x² + y² - 2x - 2y + 1 = 0 or x² + y²- 10x - 10y + 25 = 0

(x-1)^2 + (y-1)^2 = 1

There are two such circles, one that is mostly to the left of (2,1), and the other that is mostly above (2,1).

Each of these satisfies the equations
(2-h)^2 + (1-k)^2 = r^2,
h^2 = k^2 = r^2. Therefore
(2-h)^2 + (1-h)^2 = h^2 =>
4 - 4h + h^2 + 1 - 2h + h^2 = h^2 =>
h^2 - 6h + 5 = 0.
The two possible values of h are 1 and 5.

One circle is
(x-1)^2 + (y-1)^2 = 1; the other is
(x-5)^2 + (y-5)^2 = 25.