Given that we need to find the equation of the circle which touches both the axes at a distance of 6 units from the origin.

A circle touches the axes at the points (±6, 0) and (0,±6).

From the figure, we can see that the centre of the circle is (±6,±6).

We have a circle with centre (±6,±6) and passing through the point (0,6).

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.

⇒

⇒

⇒ r = √36

⇒ r = 6 ..... (1)

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±6)² + (y±6)² = ()²

⇒ x²±12x + 36 + y²±12y + 36 = 36

⇒ x² + y²±12x±12y + 36 = 0.

∴The equation of the circle is x² + y²±12x±12y + 36 = 0. We get four circles which satisfy this condition.