The work done on a particle of mass $m$ by a force,

            $K \left[\frac{x}{{\left(x^2+y^2\right)}^{3/2}}\hat{i}+\frac{y}{{\left(x^2+y^2\right)}^{3/2}}\hat{j}\right]$

($K$ being a constant of appropriate dimensions), when the particle is taken from the point $(a, 0)$ to the point $(0, a)$ along a circular path of radius a about the origin in the $x – y$ plane is

Let us consider a point on the circle 

The equation of circle is $x^2 + y^2 = a^2$

The force is

$\overrightarrow{F}= K \left[\frac{x\hat{i}}{{\left(x^2+y^2\right)}^{3/2}}+\frac{y\hat{j}}{{\left(x^2+y^2\right)}^{3/2}}\right]$

$\overrightarrow{F}= K \left[\frac{x\hat{i}}{{\left(a^2\right)}^{3/2}}+\frac{y\hat{j}}{{\left(a^2\right)}^{3/2}}\right]$

$\overrightarrow{F}=\frac{K}{a^3}\left[x\hat{i}+y\hat{j}\right]$

The force acts radially outwards as shown in the figure and the displacement is tangential to the circular path. Therefore the angle between the force and
displacement is $90°$ and $W = 0$