Let's find the equation of the circle which satisfies the given conditions:

The circle touches the $$x$$-axis.
The circle passes through the point $$(1,1)$$.
The centre of the circle lies on the line $$x + y = 3$$.

Let the centre of the circle be $$(h, k)$$. Since the circle touches the $$x$$-axis, the distance from the centre to the $$x$$-axis (which is $$k$$) is equal to the radius $$r$$ of the circle. Thus, we have:

$$r = k$$

Next, since the line $$x + y = 3$$ is known to contain the centre $$(h, k)$$, we can write:

$$h + k = 3$$

Now, the circle also passes through the point $$(1, 1)$$. Using the distance formula, the distance from the centre $$(h, k)$$ to this point should also equal the radius $$r$$, so we can write:

$$\sqrt{(h - 1)^2 + (k - 1)^2} = r$$

Since $$r = k$$, we substitute this in:

$$\sqrt{(h - 1)^2 + (k - 1)^2} = k$$

Squaring both sides, we get:

$$(h - 1)^2 + (k - 1)^2 = k^2$$

Expanding and simplifying:

$$h^2 - 2h + 1 + k^2 - 2k + 1 = k^2$$

This simplifies further:

$$h^2 - 2h + 2 = 2k$$

Remember, from $$h + k = 3$$, we can express $$k$$ as:

$$k = 3 - h$$

Substitute $$k = 3 - h$$ back into the equation:

$$h^2 - 2h + 2 = 2(3 - h)$$

Simplify:

$$h^2 - 2h + 2 = 6 - 2h$$

$$h^2 + 2 = 6$$

$$h^2 = 4$$

So, we have:

$$h = 2$$ or $$h = -2$$

Since the centre lies in the first quadrant, $$h$$ must be positive. Thus:

$$h = 2$$

And, from $$h + k = 3$$:

$$2 + k = 3$$

Therefore:

$$k = 1$$

Thus, the centre of the circle is $$(2, 1)$$ and its radius is $$1$$. The standard equation of the circle is:

$$ (x - h)^2 + (y - k)^2 = r^2 $$

Substituting for $$(h, k)$$ and $$r$$:

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

Expanding this equation, we have:

$$x^2 - 4x + 4 + y^2 - 2y + 1 = 1$$

Combine like terms:

$$x^2 + y^2 - 4x - 2y + 4 = 0$$

Thus, the equation of the circle is:

$$x^2 + y^2 - 4x - 2y + 4 = 0$$

The correct option is:

Option B: $$x^2 + y^2 - 4x - 2y + 4 = 0$$

Circle

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