The area of the triangle formed by joining the origin to the point...

Length of perpendicular from origin to the line $x\sqrt{5} + 2y = 3\sqrt{5}$ is

$OL = \dfrac{3\sqrt{5}}{\left(\sqrt{5}\right)^{2} + {2}^{2}} = \dfrac{3\sqrt{5}}{9} = \sqrt{5}$

Radius of the given circle $= \sqrt{10} = OQ = OP$

$PQ = 2QL$

$= 2\sqrt{OQ^{2} - OL^{2}}$

$= 2\sqrt{10 - 5} = 2\sqrt{5}$

Thus, area of $\Delta OPQ = \dfrac{1}{2} \times PQ \times OL$

$= \dfrac{1}{2} \times 2\sqrt{5} \times \sqrt{5} = 5$& sq units

Hence, option (c) is correct.