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

Centre of circle lies on the line $x+y=2$.

Any point on the line can be assumed as $\left(\alpha , 2-\alpha \right)$.

So let Centre of circle is $\left(\alpha , 2-\alpha \right)$.

Since both lines $x=3$ and $y=2$ touches the circle means distance from the Centre of circle to the line is equal to radius.

Let $r$ is radius of circle.

For the line $x=3 \Rightarrow x-3=0$

$r=\left|\frac{\alpha -3}{1}\right|=\left|\alpha -3\right| ...\left(1\right)$

For the line $y=2 \Rightarrow y-2=0$

$r=\left|\frac{\left(2-\alpha \right)-2}{1}\right|=\left|\alpha \right| ...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\left|\alpha \right|=\left|\alpha -3\right|$

Squaring both side

$\Rightarrow {\alpha }^2={\left(\alpha -3\right)}^2 \Rightarrow {\alpha }^2={\alpha }^2-6\alpha +9$

$\Rightarrow 6\alpha =9 \Rightarrow \alpha =\frac{3}{2}$

We know $r=\left|\alpha \right| \Rightarrow r=\frac{3}{2}$.

Diameter$=2r=2\left(\frac{3}{2}\right)=3$.