The diameter of the circle, whose centre lies on the line $$x + y = 2$$ in the first quadrant and which touches both the lines $$x = 3$$ and $$y = 2$$ is __________

Let the centre of the required circle be the point $$(h,\,k)$$. Because the centre lies on the straight line $$x + y = 2$$ we can write

$$h + k = 2 \quad\quad (1)$$

We are told that the circle touches the vertical line $$x = 3$$. For any point $$(h,\,k)$$ the perpendicular distance to the line $$x = 3$$ equals the absolute difference of their $$x$$-coordinates, so

$$\text{Distance from }(h,k)\text{ to }x = 3 = |h - 3|.$$

This distance is exactly the radius $$r$$ of the circle, hence

$$r = |h - 3|.$$

Similarly, the circle also touches the horizontal line $$y = 2$$. The perpendicular distance from $$(h,\,k)$$ to this line is the absolute difference of their $$y$$-coordinates, giving

$$\text{Distance from }(h,k)\text{ to }y = 2 = |k - 2|.$$

This distance must be the same radius $$r$$, so we also have

$$r = |k - 2|.$$

Because the centre lies in the first quadrant and we already know from (1) that $$h + k = 2,$$ each of $$h$$ and $$k$$ is positive and less than $$2.$$ That means $$h \lt 2$$ and $$k \lt 2,$$ placing the centre to the left of $$x = 3$$ and below $$y = 2.$$ Therefore the absolute value signs can be removed by inserting negative signs inside, yielding

$$r = 3 - h \quad\text{and}\quad r = 2 - k.$$

Equating these two expressions for the same radius gives

$$3 - h = 2 - k.$$

Simplifying, we obtain

$$-h + k = -1$$

or, more neatly,

$$k - h = -1 \quad\quad (2)$$

Now we have the pair of linear equations

$$\begin{aligned} h + k &= 2 \quad\quad&(1)\\ k - h &= -1 \quad\quad&(2) \end{aligned}$$

Adding equations (1) and (2) eliminates $$h$$:

$$\bigl(h + k\bigr) + \bigl(k - h\bigr) = 2 + (-1) \implies 2k = 1 \implies k = \dfrac12.$$

Substituting $$k = \dfrac12$$ back into equation (1) gives

$$h + \dfrac12 = 2 \implies h = 2 - \dfrac12 = \dfrac32.$$

Thus the centre is $$(h,\,k) = \left(\dfrac32,\,\dfrac12\right).$$

Using either expression for $$r,$$ say $$r = 3 - h,$$ we find

$$r = 3 - \dfrac32 = \dfrac32.$$

The diameter $$D$$ of the circle is twice the radius:

$$D = 2r = 2\left(\dfrac32\right) = 3.$$

Hence, the correct answer is Option 3.