Equal circles with centres O and O' touch each other at X. OO' produced to meet a circle with centre O' , at A. AC is a tangent to the circle where centre O. O'D is perpendicular to $AC$. Then $\large\frac{DO'}{CO}$

Question

$\begin{array}{1 1} \frac{1}{3} \\ \frac{1}{6} \\\frac{1}{9} \\ \frac{1}{5} \end{array}$