In the figure above, we have $$\mathrm{OA}=\mathrm{a}, \mathrm{OB}=\mathrm{b}, \mathrm{BA}=\sqrt{\mathrm{a}^2+\mathrm{b}^2}, \mathrm{BC}=\mathrm{k}, \mathrm{CD}=\mathrm{x}$$ \& $$\mathrm{BD}=\sqrt{\mathrm{x}^2+\mathrm{y}^2}$$ (Note: - $$\mathrm{BA} \& \mathrm{BD}$$ are found using Pythagoras theorem in triangle ABC \& triangle $$B C D$$ respectively)

Clearly $$\triangle \mathrm{BOA} \sim \triangle \mathrm{BCD}$$ (as $$C D$$ is parallel with $$O A$$ ), so using the similar triangle property (i.e. in similar triangles, the ratio of the corresponding sides is same), we get the $$\frac{O B}{B C}=\frac{O A}{C D}=\frac{B A}{B D}$$

i.e. $$\frac{b}{k}=\frac{a}{x}=\frac{\sqrt{a^2+b^2}}{\sqrt{x^2+y^2}}$$.

Now, $$\frac{b}{k}=\frac{a}{x} \Rightarrow x=\frac{a k}{b}$$ \& from the figure $$C O=y=B O-B C=b-k$$
Hence we get $$x=\frac{a k}{b} \& y=b-k$$, so the answer is $$(A)$$.