Let O be the center of the circle x ^ 2 + y ^ 2 = r ^ 2 , where r >...

Circumcenter $C$ of $\delta OPQ$ lies on $x + 2y = 4$

Let $C ≡ \left(4 - 2\alpha, \alpha\right)$

$∵ OC ⊥ PQ ⇒ \left(slopeofOC\right) \cdot \left(slopeofPQ\right) = - 1$

$⇒ \left(\dfrac{\alpha}{4 - 2\alpha}\right)\left(- \dfrac{2}{4}\right) = - 1 ⇒ \alpha = \dfrac{8}{5} ⇒ C\left(\dfrac{4}{5}, \dfrac{8}{5}\right)$

$OM = \dfrac{| 0 + 0 - 5 |}{\sqrt{{2}^{2} + {4}^{2}}} = \dfrac{\sqrt{5}}{2}$

$CM = \dfrac{\left|2\left(\dfrac{4}{5}\right) + 4\left(\dfrac{8}{5}\right) - 5\right|}{\sqrt{{2}^{2} + {4}^{2}}} = \dfrac{3}{2\sqrt{5}}$

$OC = PC = \sqrt{{\left(\dfrac{4}{5}\right)}^{2} + {\left(\dfrac{8}{5}\right)}^{2}} = \dfrac{4}{\sqrt{5}}$

Now, $P{M}^{2} = O{P}^{2} - O{M}^{2} = P{C}^{2} - C{M}^{2} ⇒ {r}^{2} - \dfrac{5}{4} = \dfrac{16}{5} - \dfrac{9}{20} ⇒ r = 2$