Existential threat

If PQ is perpendicular to the given line, then the line segment joining the centers of the two circles must be parallel to the the given line.

Since the first circle is tangent to $x + y - 2 = 0$ at $(1,1)$ , then its center lies on the line y = x, thus its coordinates are $(t, t) , t \ne 1$ . The center of the second circle is $(-2, -\frac{5}{2} )$ . The line joining them has a slope of $m = \dfrac{ t + \frac{5}{2} }{ t + 2 }$ . Setting $m = -1$ and solving for $t$ , we obtain, $- (t+2) = t + \frac{5}{2}$ , resulting in $t = -\frac{9}{4}$ . So our first circle must be chosen as $(x + \frac{9}{4} )^2 + (y + \frac{9}{4} )^2 = \dfrac{25}{8}$ , which when expanded, results in $x^2 + \frac{9}{2} x + \frac{81}{16} + y^2 + \frac{9}{2} y + \frac{81}{16} = \frac{25}{8}$

Subtracting the equation of this circle from the equation of the second circle yields, $-\frac{1}{2} x + \frac{1}{2} y = 13$ . This line is the line on which P and Q lie, it is easy to check that this line is perpendicular to the given line $x + y - 2 = 0$ . Hence, the answer is $\boxed{Yes}$ .