The diameter of the circle, whose Centre lies on the line x + y = 2...

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

Any point on the line can be assumed as \left(\alpha, 2 - \alpha\right)$.</p><p>So let Centre of circle is \left(\alpha, 2 - \alpha\right)$.

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 $x = 3 ⇒ x - 3 = 0$

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

For the line $y = 2 ⇒ y - 2 = 0$

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

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

\left|\alpha\right| = \left|\alpha - 3\right|$</p><p>Squaring both side</p><p>$⇒ {\alpha}^{2} = {\left(\alpha - 3\right)}^{2} ⇒ {\alpha}^{2} = {\alpha}^{2} - 6\alpha + 9$</p><p>$⇒ 6\alpha = 9 ⇒ \alpha = \dfrac{3}{2}$</p><p>We know$r = \left|\alpha\right| ⇒ r = \dfrac{3}{2}$.</p><p>Diameter$= 2r = 2\left(\dfrac{3}{2}\right) = 3.