Considering the given information:
The Cartesian equation is given as,
\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 2(1)}}\)
Putting\({\rm{rcos\theta }}\)for x and\({\rm{rsin\theta }}\) for y in equation 1 .
\(\begin{aligned}{c}{{\rm{(rcos\theta )}}^{\rm{2}}}{\rm{ + (rsin\theta }}{{\rm{)}}^{\rm{2}}}{\rm{ = 2}}\\{{\rm{r}}^{\rm{2}}}\left( {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta + si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right){\rm{ = 2}}\\{{\rm{r}}^{\rm{2}}}{\rm{(1) = 2}}\\{{\rm{r}}^{\rm{2}}}{\rm{ = 2}}\end{aligned}\)
Taking the square root on both sides of the equation,
\(\begin{aligned}{c}\sqrt {{{\rm{r}}^{\rm{2}}}} {\rm{ = }}\sqrt {\rm{2}} \\{\rm{r = }}\sqrt {\rm{2}} \end{aligned}\)
Therefore, the required polar equation for the curve \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 2}}\) is \({\rm{r = }}\sqrt {\rm{2}} \).