Answer
$$\kappa \left( x \right) = \frac{1}{a}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {{a^2} - {x^2}} \cr
& {\text{Calculate }}f'\left( x \right){\text{ and }}f''\left( x \right) \cr
& f'\left( x \right) = \frac{{ - 2x}}{{2\sqrt {{a^2} - {x^2}} }} = - \frac{x}{{\sqrt {{a^2} - {x^2}} }} \cr
& f''\left( x \right) = - \frac{{{a^2}}}{{{{\left( {{a^2} - {x^2}} \right)}^{3/2}}}} \cr
& {\text{Use the formula }}\kappa \left( x \right) = \frac{{\left| {f''\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \left| { - \frac{{{a^2}}}{{{{\left( {{a^2} - {x^2}} \right)}^{3/2}}}}} \right|\frac{1}{{{{\left( {1 + \frac{{{x^2}}}{{{a^2} - {x^2}}}} \right)}^{3/2}}}} \cr
& {\text{Simplifying}} \cr
& \kappa \left( x \right) = \frac{{{a^2}}}{{{{\left( {{a^2} - {x^2}} \right)}^{3/2}}}}\frac{1}{{{{\left( {\frac{{{a^2}}}{{{a^2} - {x^2}}}} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \frac{{{a^2}}}{{{{\left( {{a^2} - {x^2}} \right)}^{3/2}}}}\frac{{{{\left( {{a^2} - {x^2}} \right)}^{3/2}}}}{{{a^3}}} \cr
& \kappa \left( x \right) = \frac{1}{a} \cr} $$
