Radius of Circle Passing Through Given Line Chord

To find the value of $r$, we can use the family of circles passing through the intersection of the given circle and the line chord.

The equation of the given circle is: $S \equiv x^2 + y^2 - r^2 = 0$

The equation of the line passing through $P$ and $Q$ is: $L \equiv 2x + 4y - 5 = 0$

Any circle passing through the intersection points $P$ and $Q$ of $S$ and $L$ can be represented as: $S + \lambda L = 0$ $x^2 + y^2 - r^2 + \lambda(2x + 4y - 5) = 0$

Since this circle is the circumcircle of triangle $OPQ$, it must pass through the origin $O(0,0)$. Substituting $(0,0)$ into the equation gives: $-r^2 - 5\lambda = 0 \implies \lambda = -\frac{r^2}{5}$

Substituting $\lambda$ back into the circumcircle equation, we get: $x^2 + y^2 - r^2 - \frac{r^2}{5}(2x + 4y - 5) = 0$ $x^2 + y^2 - \frac{2r^2}{5}x - \frac{4r^2}{5}y = 0$

The centre $(h, k)$ of this circumcircle is given by: $h = \frac{r^2}{5}, \quad k = \frac{2r^2}{5}$

We are given that the centre of the circumcircle lies on the line $x + 2y = 4$. Substituting $(h, k)$ into this line equation gives: $\frac{r^2}{5} + 2\left(\frac{2r^2}{5}\right) = 4$ $\frac{5r^2}{5} = 4 \implies r^2 = 4$

Since $r > \frac{\sqrt{5}}{2}$, taking the positive square root yields: $r = 2$