Let O be the center of the circle x 2 + y 2 = r 2 , where r > 5 2 . Suppose P Q is a chord of this…

Circumcenter $C$ of $\Delta OPQ$ lies on $\mathrm{x}+2\mathrm{y}=4$

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

$\because OC\perp PQ \Rightarrow \left(slope of OC\right)·\left(slope of PQ\right)=-1$

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

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

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

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

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