Let O be the centre of the circle $x^{2} + y^{2} = r^{2},$ where $r > \sqrt{5}/2.$ Suppose PQ is a chord of this circle and the equation of the line passing through P and Q is $2x + 4y = 5.$ If the centre of the circumcircle of the triangle OPQ lies on the line $x + 2y = 4,$ then the value of r is ____.

Let O be the centre of the circle x² + y² = r², where r > √5/2. Suppose PQ is a chord of this circle and the equation of the line passing through P and Q is 2x + 4y = 5. If the centre of the circumcircle of the triangle OPQ lies on the line x + 2y = 4, then the value of r is ______ $\text{Circle: } x^2 + y^2 = r^2, \quad O = (0,0)$ $r > \frac{\sqrt{5}}{2}$ $PQ: \ 2x + 4y = 5$ $\text{Circumcircle of } \triangle OPQ, \text{ centre } C$ $C \in x + 2y = 4$ $\text{Find: } r$ $2x + 4y = 5 \ \Leftrightarrow \ x + 2y = \frac{5}{2}$ $d(O, PQ) = \frac{|0 + 0 - 5/2|}{\sqrt{1^2+2^2}} = \frac{\sqrt{5}}{2}$ $r > \frac{\sqrt{5}}{2} \Rightarrow \text{real chord } PQ$ $M = \text{foot of } \perp \text{ from } O = \left(\frac{1}{2}, 1\right)$ $M = \text{midpoint of } PQ$ $OP = OQ = r \Rightarrow C \text{ on perpendicular bisector of } PQ$ $C = t(1, 2) = (t, 2t)$ $CO = CP$ $P \in PQ \Rightarrow x_P + 2y_P = \frac{5}{2}$ $(t-x_P)^2 + (2t-y_P)^2 = t^2 + (2t)^2$ $x_P^2 + y_P^2 - 2t(x_P + 2y_P) = 0$ $x_P^2 + y_P^2 = r^2, \quad x_P + 2y_P = \frac{5}{2}$ $r^2 - 2t \cdot \frac{5}{2} = 0 \ \Rightarrow \ r^2 = 5t$ $C = (t, 2t) \in x + 2y = 4$ $\Rightarrow t + 2(2t) = 4 \ \Rightarrow \ 5t = 4$ $t = \frac{4}{5}, \quad r^2 = 5t = 4$ $r = 2 \quad (r > 0)$ $r = 2 > \frac{\sqrt{5}}{2} \ \checkmark$ $\boxed{r = 2}$