Given that we need to find the equation of the circle which touches both axes and passes through the point (2, 1).

Let us assume the circle touches the x-axis at the point (a, 0) and y-axis at the point (0,a).

Then we get to the centre of the circle as (a, a).

We have a circle with centre (a, a) and passing through the point (2, 1) and having radius |a|.

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

We know that the distance between the two points (x₁,y₁) and (x₂,y₂) is .

Let us assume r is the radius of the circle.

⇒

⇒ a² = a² - 4a + 4 + a² - 2a + 1

⇒ a² - 6a + 5 = 0

⇒ a² - 5a - a + 5 = 0

⇒ a(a - 5) - 1(a - 5) = 0

⇒ (a - 1)(a - 5) = 0

⇒ a - 1 = 0 or a - 5 = 0

⇒ a = 1 or a = 5 ..... (1)

Case (i)

We have got the centre at (5, 5) and having radius 5 units.

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:

⇒ (x - p)² + (y - q)² = r²

Now we substitute the corresponding values in the equation:

⇒ (x - 5)² + (y - 5)² = 5²

⇒ x² - 10x + 25 + y² - 10y + 25 = 25

⇒ x² + y² - 10x - 10y + 25 = 0.

∴The equation of the circle is x² + y² - 10x - 10y + 25 = 0.

Case (ii)

We have got the centre at (1, 1) and having a radius 1 unit.

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:

⇒ (x - p)² + (y - q)² = r²

Now we substitute the corresponding values in the equation:

⇒ (x - 1)² + (y - 1)² = 1²

⇒ x² - 2x + 1 + y² - 2y + 1 = 1

⇒ x² + y² - 2x - 2y + 1 = 0

∴ The equation of the circle is x² + y² - 2x - 2y + 1 = 0.