AC is diameter of circle. AB is a tangent. BC meets the circle again at D. AC = 1, AB= a, CD = b, then

Assume point B such that $\mathrm{OD}\perp \mathrm{AC}$
$\mathrm{△}\mathrm{ABC}\text{ and }\mathrm{△}\mathrm{ODC}$are similar triangles
$\Rightarrow \frac{\mathrm{AC}}{\mathrm{OC}}-\frac{\mathrm{AB}}{\mathrm{OD}}$
Hence AB = 1 $\Rightarrow \mathrm{BC}=\sqrt{2}$
Assume BD $=\mathrm{x}\Rightarrow \mathrm{CD}=\mathrm{y}$
$\therefore$using similar${\mathrm{\Delta }}^{\text{'}}s$
$\therefore$ D is mid point$\Rightarrow \mathrm{DC}=\frac{1}{\sqrt{2}}$
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